Newton's Laws Of Motion Archives

Newton’s Laws Of Motion

December 18, 2020December 18, 2020 by Veerendra

Newton’s Laws Of Motion

(A) Newton’s First Law of Motion

A body can not change its state of motion by itself. If the object is at rest it will remain at rest and if it is in uniform motion, it continues to be in motion unless some external force is applied on it.

Newton’s First Law of Motion states that if there is no net force acting on a body, its state of motion will be unchanged.
If the body is at rest, it will remain at rest. If the body is moving, it keeps on moving at a constant speed in a straight line.
Newton’s first law is sometimes called the Law of Inertia.
The object is not necessarily at rest. The object may be moving at constant speed in a straight line.
To remember Newton’s First Law of Motion
When F = 0 (No resultant force)

Inertia:
There is an inherent property of an object by virtue of which it cannot change its state of motion or rest by itself. This property is called ‘inertia’.
Inertia is of two types– inertia of rest and inertia of motion.

(a) Inertia of rest: If the body is at rest, it will continue to be at rest unless some external force is applied on it. Examples are following.
Examples:
1. When a train at rest starts moving suddenly, a passenger standing inside the compartment tends to fall backward.
2. When a carpet is beaten up with a stick, the dust particles are detached.
3. When a bullet is fired into a glass pane, it pierces a hole only at the pt where the bullet hits the glass without breaking the entire glass pane into pieces.

(b) Inertia of motion: When a body is in uniform motion, it will continue to remain in its uniform motion, i.e. it resists any change in its state of motion due to inertia of motion.
Examples:
1. when a person jumps out of a moving bus, he should run in the direction in which bus is moving otherwise he will fall down.
2. A train moving with a uniform speed and if a ball is thrown upwards inside the train by a passenger, then the ball comes back to his hand.

Situations involving inertia
The inertia of an object is the tendency of the object to remain at rest or, if moving, to continue its uniform motion in a straight line.

	The inertia of an object is the tendency of the object to remain at rest or, if moving, to continue its uniform motion in a straight line. In Figure (a), when the bus moves forward suddenly, the feet of the passenger are made to move forward. The inertia of his body tends to remain at rest. Hence, the passenger falls backward. In Figure (b), when the bus slows down suddenly, the feet of the passenger are brought to rest. The inertia of his body tends to continue moving forward. Hence, the passenger falls forward.
	A 50 cent coin is placed on a cardboard covering the top of a glass. After the cardboard is pulled quickly, the 50 cent coin hovers over the top of the glass for an instant before dropping into the glass. The coin hovers for an instant because of its inertia.
	When a book placed in the middle of a stack of books is pulled out horizontally with a quick jerk, the books above it tend to stay at rest due to inertia.
	When very little tomato sauce is left in the bottle, the bottle is given a quick downward jerk to force the sauce out of the bottle. When the bottle moves, the sauce in it moves together. When the bottle is stopped suddenly, the inertia of the sauce keeps it moving downward and out of the bottle.
	Figure shows a driver crashing his car without wearing a safety belt. Both the driver and the car were travelling at a very high speed. When the car was stopped suddenly, the inertia of the driver caused him to be thrown forward, thus injuring himself.

Relationship between Inertia and Mass:
Larger the mass of the body, larger is the inertia.
Example: It is more difficult to stop a cricket ball than a tennis ball.

The mass of an object is the amount of matter in it. The SI unit for mass is kilogram (kg). Mass is a scalar quantity.
One kilogram is define’d to be the mass of a standard cylinder of a platinum-iridium alloy kept at the International Bureau of Weights and Measures in Sevres, France.

Inertia and Mass

It is more difficult to move a lorry than a bicycle when both are initially at rest.
When both are moving, it is also more difficult to stop the lorry than the bicycle.
Hence, in comparison, the lorry has a greater tendency to be at rest than the bicycle. Likewise, the lorry has a greater tendency to continue to be in motion than the bicycle.
The lorry has more inertia than the bicycle. The lorry has a bigger mass than the bicycle. Hence, quantitatively, the inertia of an object is measured by its mass.

Effects of Inertia

Stick

A stick is used to jerk the branch of a guava tree. At the end of the branch, there is a big guava. The guava tends to remain in its original state of rest due to its inertia. This will cause its stalk to snap and the guava will fall to the ground.
Tissue Paper

When a sheet of tissue paper is pulled quickly from a tissue box, the box will not move. The inertia of the box causes it to resist motion and remain at rest.
Blade of a Hoe

The blade of a hoe can be fitted tightly to its wooden handle by hitting the end of the handle against a hard surface as shown in the figure. The mass of the blade of the hoe is big and its inertia causes it to continue moving although the handle has been stopped, thus fitting it tighter to the handle. Likewise, the head of a hammer can be fitted into its handle this way too.
Ice Skater

After an initial push, an ice skater glides almost effortlessly on an icy surface because of inertia.
Safety Belt
- The mechanism in a safety belt allows the belt to unwind freely when pulled gently. The mechanism consists of a ratchet wheel, a locking bar and a pendulum.
- When the vehicle is at rest or moving with a uniform velocity, the pendulum hangs straight down with the locking bar resting horizontally on it.
- When the vehicle is slowed down suddenly as in the case of an accident, the pendulum which has a big mass keeps moving forward due to its inertia. As a result, the pendulum swings on its pivot and causes the locking bar to block the rotation of the ratchet wheel and thus preventing the safety belt from unwinding.
Ship

Due to its very large mass, a ship has a very large inertia. It cannot be stopped guickly even during an emergency. A very strict navigation system is needed to guide a ship when reaching a port, sailing near rocks and icebergs as well as in busy sea routes like the Straits of Malacca to prevent accidents from happening.
Rear-end Collision

When a rear-end collision occurs, the car and the body of the driver move forward suddenly. The headrest supports the head of the driver when it is thrown backward.
Aeroplane

An aeroplane has a large mass. It cannot be stopped easily when it lands at airports due to its large inertia. Therefore, a very long runway is required for the aeroplane to stop safely.
Lorry

A steel structure is fitted in the space between the driver and the load of a timber lorry. This steel structure prevents any log from moving forward and knocking against the driver compartment when the lorry stops suddenly.
Car

When a car stops suddenly during an accident, the driver continues to move forward because of inertia. The safety belt and airbag prevent the driver from crashing into the windscreen and injuring himself.

Experiment 1

Aim: To investigate the relationship between inertia and mass.
Problem: What is the relationship between inertia and mass?
Hypothesis: The inertia of a body increases when its mass increases.
Variables:
(a) Manipulated variable: Mass
(b) Responding variable: Inertia
(c) Fixed variable: Type of hacksaw blade (The same blade is used throughout the experiment)
Operational Definition: Period of oscillation is an indicator for inertia, the responding variable. The bigger the period of oscillation, the bigger is the inertia.
Material: Plasticine
Apparatus: Hacksaw blade, G-clamp, stopwatch
Method:

A hacksaw blade is clamped with a G-clamp to the leg of a table as shown in Figure 2.46.
A lump of plasticine with a mass of 30 g is attached to the free end of the hacksaw blade.
The hacksaw blade is displaced sideways slightly and released so that it would oscillate horizontally.
The time taken for 10 complete oscillations, t₁ is determined using a stopwatch and recorded. This step is repeated for another reading, t₂.
Steps 3 and 4 are repeated with mass of plasticine, m = 40 g, 50 g, 60 g and 70 g.
The readings of t₁ and t₂ are recorded in Table.

Results:
1. Tabulation of results.
Understanding inertia 9
2. Graph of period, T against mass m.
Understanding inertia 10
Discussion:

From the experiment, we observed that a larger mass of plasticine attached to the hacksaw blade increases the period of oscillation.
The period of oscillation is an indicator of the inertia of the plasticine whereby the bigger the period of oscillation, the bigger is the inertia of the plasticine. Hence, the bigger is the mass of the plasticine, the bigger is its inertia.

Conclusion:
When the mass of a body increases, the body becomes more reluctant to change its state of rest or motion. This means that the inertia of a body increases when its mass increases.

(B) Newton’s Second Law of Motion

We know that,
Newtons second law of motion
By defining 1 newton (N) as the unit force that causes an object with a mass of 1 kg to accelerate 1 m s^-2 and substituting into the equation,
1 = k (1) (1)
Therefore, k = 1
With this,
F = ma
Newton’s Second Law of Motion describes the relationship between the acceleration of an object and the force applied to it.
Newton’s second law states “the rate of change of momentum of a body is directly proportional to force and takes place in the direction of force.”
i.e., when a net external force acts on an object, the acceleration of the object is directly proportional to the net force and has a magnitude that is inversely proportional to its mass.

\(\text{ }F=\frac{{{P}_{2}}-{{P}_{1}}}{t}\text{ or }F=\left( \frac{v-u}{t} \right)=m\overset{\to }{\mathop{a}}\)
where p₁ = initial momentum = mu
p₂ = final momentum = mv
Unit of force in SI system is newton.
1N is equivalent to that force which can produce an acceleration of 1m/s⁵ in a body of mass 1 kg.
Unit of force in CGS system is dyne.
1 dyne = 1 gm – cm/s²
1 N = 10⁵ dynes

Example 1. A worker pulls a block of ice on a smooth surface with a force, F. The ice has a mass of 80 kg.
Newtons second law of motion 1
(a) If the force F = 160 N, calculate the acceleration of the ice.
(b) If the velocity of the ice changes from 0 to 8 m s^-2 in 5 s, calculate the force, F.
Solution:
Newtons second law of motion 2

Example 2. A hawker pushes a tank of water with a horizontal force of 45 N. The total weight of the trolley and water tank is equal to 900 N.
Newtons second law of motion 3
Calculate
(a) the total frictional force if the hawker moves with uniform velocity of 5 m s^-1,
(b) the acceleration of the hawker if the total frictional force is equal to 30 N.
Solution:
Newtons second law of motion 4

(C) Newton’s Third Law of Motion

Newton’s first law of motion gives a qualitative idea of force, while the second law provides us an idea to measure the force.

Newton’s third law of motion states that ” if a body A exerts a force on the body B, the body B will also exert an equal and opposite force on A.”
The force exerted by A on B is called action while the force exerted by B on A is called the reaction.
Newton’s third law is also stated as “to every action there is an equal and opposite reaction.”
Forces always occur in pairs.
Action and reaction always act on different bodies.

When you are practising arm wrestling with your friend, you will feel a force acting on you.
Your friend will also be feeling a force acting on him. This is because while he is applying a force on you, you are also applying a force on him.
This law is also known as the ‘action-reaction’ law. Hence, Newton’s third law of motion can be quoted as:
‘For every action, there is an equal but opposite reaction’.
A swimmer pushes the water backward with his hands and legs with a force, F, while the reaction force of the water, -F, pushes the swimmer forward.
Figure shows a fighter jet taking off from an airbase. The action of the exhaust gas rushing out from the nozzle with a force, F, causes a forward reaction force, -F, that moves the jet forward.

Applications of Newtons III law

Recoil of a gun: When the bullet is fired from a gun, an equal and opposite force is applied on the gun, due to which the gun recoils in backward direction.
Application in walking: While moving in forward direction we push the ground backwards that is the action. An equal and opposite force is applied by the ground on the man, thus the reaction due to which man moves forward.
Rowing a boat in river: When we push the water backward with the help of oars (applying a force backward), an equal and opposite force acts on the boat. This is the reaction which moves the boat forward.
Launching Rocket: In rocket, gases are produced in large amount. Due to internal combustion they come out and move backwards with an equal and opposite force which in turn acts on the rocket and moves it forward.

Example 1. The diagram shows forces F₁, and F₂ exerted on a wooden block placed on a concrete surface. The friction between the block and the concrete surface is 3 N.
Newtons Third law of motion 4
Which pair of forces for F₁, and F₂ causes the wooden block to move with an acceleration?
Newtons Third law of motion 5
Answer: C
Acceleration occurs when there is an unbalanced force, where the net force is not zero.

Example 2. The diagram shows a car with a mass of 850 kg moving with an acceleration of 2 m s^-2. There is a frictional force of 700 N acting on the car.
Newtons Third law of motion 6
What is the force exerted by the engine of the car?
Answer:
Force exerted by engine = (850 x 2) + 700 = 2400 N

What is Law of Conservation of Momentum

December 18, 2020December 18, 2020 by Veerendra

Law of Conservation of Momentum

According to law of conservation of momentum “if there is no force acting on a system, the momentum of the system remains unchanged.”
Generalizing the situation ” if a group of bodies are exerting force on each other, their total momentum remains conserved before and after the interaction provided there is no external force acting on them.”
i.e. m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Definition of Principle of Conservation of Momentum
The principle of conservation of momentum can also be defined as follows:

The total momentum in a closed system is constant.
A system consists of several objects acting on each other.
A closed system is a system where the sum of its external forces (also known as resultant force) is zero.
When the momentum is constant, we say the momentum is conserved.

Figure shows two billiard balls, A and B. When ball A collides with ball B, ball A loses momentum and ball B gains the amount of momentum lost by ball A.
Figure (a) shows a boy standing on a skating board with a bowling ball in his hands.
When he throws the ball forward, he moves backward. Initially, the respective momentums of both the boy and the ball is zero because both are at rest.
When the ball is thrown, the ball gains momentum. The boy gains an equal amount of momentum but in the opposite direction.
The above observations can be explained using the principle of conservation of momentum.
According to the principle of conservation of momentum, when two or more bodies act on each other, their total momentum remains constant, provided that there is no external force acting on them.

Experiment 1

Aim: To show that the total momentum of a closed system is conserved.
Problem: Is the total momentum in a closed system constant?
Hypothesis: The total momentum of a system after a collision/explosion is the same as the total momentum before the collision/explosion.
A. Collision
Materials: Cellophane tape, ticker tape, plasticine, two pins, two pieces of cork
Apparatus: Trolleys, plane, retort stands with clamps, power supply
Variables:
(a) Manipulated variable: Mass of trolley
(b) Responding variable: Velocity of trolley
Method:
Law of conservation momentum 1

The apparatus is set up by attaching two pins in front of trolley P and two pieces of cork at the end of trolley Q as shown in Figure.
Trolley P is placed at the higher end of the plane and trolley Q is placed halfway down the plane and is at rest. A ticker tape is passed through the ticker timer and is attached to trolley P.
After the ticker timer is switched on, trolley P is given a push towards trolley Q. Both trolleys will stick together and move down the plane after the collision.
The tapes are analysed to determine the following:
(a) velocity of trolley P before the collision, u_p.
(b) velocity of trolley P and trolley Q after the collision, v.
Using identical trolleys, the mass of each trolley can be taken as 1 unit of mass. The experiment is repeated as follows:
(a) two stacked trolleys collide with one trolley at rest,
(b) three stacked trolleys collide with one trolley at rest.

Results:
Law of conservation momentum 2
Discussion:
Before collision, velocity of trolley Q, u_Q = 0 because it is at rest.
Conclusion:
It is found that the total momentum after the collision is the same as that before the collision. The hypothesis is accepted.

B. Explosion
Materials: Cellophane tape, ticker tape, mallet
Apparatus: Trolleys, power supply
Variables:
(a) Manipulated variable: Mass of trolley
(b) Responding variable: Velocity of trolley
Method:
Law of conservation momentum 3

Trolley P and trolley Q are set up facing each other on a smooth table as shown in Figure (a). The spring-loaded plunger of trolley P is compressed.
Two ticker tapes are passed through the ticker timer, one attached to trolley P and another to trolley Q.
After the ticker timer is switched on, the plunger of trolley P is released by tapping the release dowel rod with a mallet. The trolleys will ‘explode’ and move apart in opposite directions as shown in Figure (b).
From the tapes obtained, the velocities of trolley P, vp and trolley Q, v_Q are determined.
The experiment is repeated by using two and then three stacked trolleys placed facing trolley Q.

Results:
Law of conservation momentum 4
Discussion:

In the case of an explosion system, the total momentum before the explosion is always zero because all the bodies in the system are at rest initially.
Since trolley P and trolley Q move in opposite directions, if we assign v_P to be positive, then v_Q will be negative.

Conclusion:
It is found that the total momentum before and after the explosion are the same and both are equal to zero. The hypothesis is accepted.

Applications of Conservation of Momentum

(a) Figure shows the launching of a rocket. A rocket carries liquid hydrogen fuel and liquid oxygen. When a mixture of hydrogen and oxygen burns in the combustion chamber during launching, jets of hot gas are expelled at very high speed through the exhausts. This produces a very large momentum.
Fig. A rocket
(b) According to the principle of conservation of momentum, an equal and opposite momentum is produced causing the rocket to propel upwards.
(a) Figure shows a jet engine. Air in front of the engine is sucked into the combustion chamber by the rotating compressor blades. In the combustion chamber, fuel is injected and burnt with the compressed air.
Fig. A jet engine
(b) The exploding hot jet of gases is expelled from the combustion chamber. This produces a very large momentum.
(c) According to the principle of conservation of momentum, the expulsion of the hot gases produces an equal and opposite momentum, causing the jet plane to move forward.
(a) When a bullet is fired from a submachine gun, the exploding gases from the bullet cause it to be shot out at a very high speed, producing a very big momentum.
Fig. A rifle
(b) An equal and opposite momentum is produced and this causes the submachine gun to recoil.

Law of Conservation of Momentum Example Problems with Solutions

Example 1. A rifle of mass 5 kg fires a bullet of mass 40 gm. The bullet leaves the barrel of the rifle with a velocity 200 m/s. If the bullet takes 0.004 s to move through the barrel, calculate the following:
(i) recoil velocity of the rifle and
(ii) the force experienced by the rifle due to its recoil.
Solution: (i) Given mass of the rifle, m₁ = 5 kg
Mass of the bullet, m₂ = 40 gm = 0.04 kg
Initial velocities, u₁ = 0, u₂ = 0
After firing velocity of the bullet, v₂ = 200 m/s
Velocity of the rifle, v₁ = ?
Applying the law of conservation of momentum, we get
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0 + 0 = 5 × v₁ + 0.04 × 200
\({{v}_{1~}}=\frac{0.04\times 200}{5}=-1.6\text{ m/s}\)
(ii) Initial momentum of the rifle = 0
Final momentum of the rifle = 5 kg ×(–1.6)
= – 8 kg-m/s
Time interval = 0.004 s
\( \therefore \text{Force}=\frac{\text{Change}\,\text{in}\,\text{momentum}}{\text{Time}\,\text{interval}} \)
\( =\frac{-8kg-m{{s}^{-1}}}{0.004}=-2000\text{ N} \)

Example 2. A bullet of mass 20 gm moving with a velocity 200 m/s gets embedded into a wooden block of mass 980 gm suspended by a string. Calculate velocity acquired by the combined system.
Solution: Mass of the bullet, m₁ = 20 gm = 0.02 kg
Velocity of the bullet, u₁ = 200 m/s
Momentum of the bullet = m₁u₁
= 0.02 × 200 kg-m/s = 4 kg-m/s
Now, the bullet gets embedded into a wooden block of mass 980 gm. The mass of the block and bullet = 980 + 20
= 1000 gm = 1kg
Let the velocity of the combined system = v
∴ Momentum of the combined system
= 1 × v kg-m/s = v kg-m/s
Now, applying the law of conservation of momentum,
m₁u₁ = (m₁ + m₂)v
or 4 = v
∴ v = 4 m/s = 4 kg m/s

Example 3. A rifle man, who together with his rifle has a mass of 100 kg, stands on a smooth surface fires 10 shots horizontally. Each bullet has a mass 10 gm a muzzle velocity of 800 m/s. What velocity does rifle man acquire at the end of 10 shots?
Solution: Let m₁ and m₂ be the masses of bullet and the rifleman and v₁ and v₂ their respective velocities after the first shot. Initially the rifleman and bullet are at rest, therefore initial momentum of system = 0.
As external force is zero, momentum of system is constant
i.e. initial momentum = final momentum
= m₁v₁ + m₂v₂
\(\Rightarrow {{v}_{2}}=\frac{{{m}_{1}}{{v}_{1}}}{{{m}_{2}}}=-\frac{(10\,\times \,{{10}^{-3}}kg)(800\,m/s)}{100kg}\)
= – 0.08 m/s
Velocity acquired after 10 shots
= 10 v₂ = 10 x (–0.08)
= – 0.8 m/s.
i.e, the velocity of rifle man is 0.8 m/s in a direction opposite to that of bullet.

Example 4. A body of mass 1 kg strikes elastically with another body at rest and continues to move in the same direction with one fourth of the initial velocity. What will be the mass of the other body ?
Solution: Given that,
Initial velocity = u
Final velocity = \(\frac { u }{ 4 }\)
So by conservation of momentum, we have
1 × u + 0 = 1 × \(\frac { u }{ 4 }\) + m × v₂
⇒ mv₂ = \(\frac { 3u }{ 4 }\) …(1)
and by conservation of energy, we have
\( \Rightarrow \frac{1}{2}\times 1\times {{u}^{2}}+0=~~\frac{1}{2}\times 1~{{\left( \frac{u}{4} \right)}^{2}}+~m{{v}_{2}}^{2}\)
\(\Rightarrow \text{m}{{\text{v}}_{\text{2}}}^{\text{2}}=\frac{15}{16}{{u}^{2}}\text{ }……\text{ (2)}\)
from equation (1) and (2),
\(=\frac{{{(m{{v}_{2}})}^{2}}}{mv_{2}^{2}}=\frac{(9/16){{u}^{2}}}{(15/16){{u}^{2}}}\)
or m = 0.6 kg

Example 5. Figure shows two rocks moving towards each other along a straight line. After colliding with each other, the two rocks lump together.
Calculate the speed of the rocks after the collision.
Law of conservation momentum 8
Solution:
Law of conservation momentum 9

Example 6. An astronaut with mass of 80 kg throws a 40 kg tool box in order to be able to get back to the space capsule.
If the box is thrown at a velocity of 6 m s^-1, what is the velocity of the astronaut after throwing the tool box?
Law of conservation momentum 10
Solution:

Example 7. Figure shows two identical boxcars of a train being coupled at a railway station. Each boxcar has a mass of 2.5 × 10⁴ kg. The velocities of boxcar 1 and boxcar 2 before the coupling are 1.0 m s^-1 and 0.8 m s^-1 respectively.
Find the velocity, v, of the boxcars after they are coupled. State one assumption in your calculation.
Law of conservation momentum 12
Solution:

Mastering Physics Solutions Chapter 32 Nuclear Physics and Nuclear Radiation

May 22, 2018 by Raju

Mastering Physics Solutions Chapter 32 Nuclear Physics and Nuclear Radiation

Mastering Physics Solutions

Chapter 32 Nuclear Physics and Nuclear Radiation Q.1CQ
Nucleus A and nucleus B have different numbers of protons and different numbers of neutrons. Explain how it is still possible for these nuclei to have equal radii.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-1cq

Chapter 32 Nuclear Physics and Nuclear Radiation Q.1P

Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-1p1

Chapter 32 Nuclear Physics and Nuclear Radiation Q.2CQ
When α particles are emitted in a nuclear decay, they have well-defined energies. In contrast, ß particles are found to be emitted with a range of energies. Explain this difference.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-2cq

Chapter 32 Nuclear Physics and Nuclear Radiation Q.2P

Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-2p1

Chapter 32 Nuclear Physics and Nuclear Radiation Q.3CQ
Is it possible for a form of heavy hydrogen to decay by emitting an α particle? Explain.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-3cq

Chapter 32 Nuclear Physics and Nuclear Radiation Q.3P

Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-3p1

Chapter 32 Nuclear Physics and Nuclear Radiation Q.4CQ
Which is more likely to expose film kept in a cardboard box, α particles or ß particles? Explain.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-4cq

Chapter 32 Nuclear Physics and Nuclear Radiation Q.4P
A certain chlorine nucleus has a radius of approximately 4.0 × 10−15 m. How many neutrons are in this nucleus?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-4p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.5CQ
It is not possible for a stable nucleusto contain more than one proton without also having at least one neutron. Explain why neutrons are necessary in a stable, multiparticle nucleus.
Solution:
No. It is not possible for a stable nucleus to have more than one proton and no neutrons.
If the nucleus has only protons, the nucleus would be unstable and blow apart because of the electrostatic repulsion between the protons.
If neutrons are present, they separate the protons, reducing the mutual repulsion between them. Also, the presence of neutrons adds a strong attractive nuclear force, enabling the nuclei to hold it together.

Chapter 32 Nuclear Physics and Nuclear Radiation Q.5P
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-5p
Solution:

Chapter 32 Nuclear Physics and Nuclear Radiation Q.6CQ
Different isotopes of a given element have different masses, but they have the same chemical properties. Explain why chemical properties are unaffected by a change of isotope.
Solution:
Isotopes are elements with same atomic number but different mass numbers.That means the isotopes of an element have same number of protons and electrons but different number of neutrons. The electrons are responsible for chemical reactions.

Chapter 32 Nuclear Physics and Nuclear Radiation Q.6P
IP (a) What initial kinetic energy must an alpha particle bave if it is to approach a stationary gold nucleus to within a distance of 22.5 fm?
(b) If the initial speed of the alpha particle is reduced by a factor of 2,bywhat factor is the distance of closest approach changed? Explain.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-6p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.7CQ
(a) Give three examples of objects for winch carbon-14 dating would give useful results.
(b) Give three examples of objects for which carbon-14 dating would not be useful.
Solution:
(a) Carbon-14 dating is useful to date specimen’s upto about 45,000 years old.
It is useful for dating organic materials like human and animal remains like
bones, ivory tusks, burnt bones, also cloth, pottery, paper, hide etc.
(b) Carbon dating is not very useful for dating inorganic materials like rocks and organic materials like fossils which are a million years old, inorganic carbon in shell, etc.

Chapter 32 Nuclear Physics and Nuclear Radiation Q.7P
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-7p
Solution:

Chapter 32 Nuclear Physics and Nuclear Radiation Q.8CQ
Explain why the large, stable nuclei in Figure 32-1 are found to lie above the N = Z line, ratherthan below the une.
Solution:
The nucleus that does not undergo radioactive decay is defined as the stable nuclei.
The elements above the N = Z line will have more neutrons than protons. These neutrons spread out the positive charge of the protons making the nuclei stable.
The nucleii below the N = Z line have more number of protons than the number of neutrons. These protons repel each other blowing the nucleus apart.
Therefore the elements above the N = Z line are said to be stable nuclei.

Chapter 32 Nuclear Physics and Nuclear Radiation Q.8P
Suppose a marble with a radius of 1.5 cm has the density of a nucLeus, as given in Example 32–2.
(a) What is the mass of this marble?
(b) How many of these marbles would be required to have a mass equal to the mass of Earth?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-8p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.9CQ
Suppose each of the following items is about 10,000 years old: a feather, a tooth, an obsidian arrowhead, a deer hide moccasin. Which of these items cannot be dated with carbon-14? Explain.
Solution:
Carbon dating is applicable only to matter which was once living (which has biological origin) and presumed to be in equilibrium with the atmosphere.
A feather, a tooth, a deer hide moccasin are all of biological origin. But obsidian is not of biological origin and so it cannot dated with carbon-14.

Chapter 32 Nuclear Physics and Nuclear Radiation Q.9P
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Solution:

Chapter 32 Nuclear Physics and Nuclear Radiation Q.10CQ
Can carbon-14 dating give the age of fossil dinosaur skeletons? Explain.
Solution:
No. Carbon-14 dating is useful to date specimen’s upto about 45,000 years old. But dinosaurs are thousands of times too old to be dated by this technique.

Chapter 32 Nuclear Physics and Nuclear Radiation Q.10P
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Solution:

Chapter 32 Nuclear Physics and Nuclear Radiation Q.11CQ
Two different samples contain the same radioactive isotope. Is it possible for these samples to have different activities? Explain.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-11cq

Chapter 32 Nuclear Physics and Nuclear Radiation Q.11P
IP Suppose a uraninm-236 nucleus undergoes fission by splitting into two smaller nuclei of equal size.
(a) Is the radius of each of the smaller nuclei one-half, more than one-half, or less than one-half the radius of the uranium-236 nucleus? Explain.
(b) Calculate the radius of the uranium-236 nucleus.
(c) Calculate the radii of the two smaller nuclei.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-11p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.12CQ
Two samples contain different radioactive isotopes. Is it possible for these samples to have the same activity? Explain.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-12cq

Chapter 32 Nuclear Physics and Nuclear Radiation Q.12P
A hypothetical nucleus weighs 11b.
(a) How many nucleons are in this nucleus?
(b) What is the radius of this nucleus?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-12p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.13CQ
Two different types of radiation deliver the same amount of energy to a sample of tissue. Does it follow that each of these types of radiation has the same RBE? Explain.
Solution:
No,
RBE is the acronym for relative biological effective ness. RBE is related to the extent of biological effect caused by a radiation RBE is not related to the amount of energy it gives. So the RBE of two different types of radiation giving same amount of energy are not equal.

Chapter 32 Nuclear Physics and Nuclear Radiation Q.13P
CE Predict/Explain Consider a nucleus that undergoes α decay.
(a) Is the radius of the resulting daughter nucleus greater than, less than, or equal to the radius of the original nucleus?
(b) Choose the test explanation from among the following:
I. The decay adds an alpha particle to the nucleus, causing its radius to increase.
II. When the nucleus undergoes decay it ejects two neutrons and two protons. This decreases the number of nucleons in the nucleus, and therefore its radius will decrease.
III. An α decay leaves the number of nucleons unchanged. As a result, the radius of the nucleus stays the same.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-13p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.14P
CE Predict/Explain Consider a nucleus that undergoes ß decay.
(a) Is the radius of the resulting daughter nucleus greater than, less than, or the same as that of the original nucleus?
(b) Choose the best explanation from among the following:
I. Capturing a ß particle will cause the radius of a nucleus to increase. Therefore, the daughter nucleus has the greater radius.
II. The original nucleus emits a ß particle, and anytime a particle is emitted from a nucleus the result is a smaller radius. Therefore, the radius of the daughter nucleus is less than the radius of the original nucleus,
III. When a nucleus emits a β particle a neutron is convened proton, but the number of nucleons is unchanged. As a result, the radius of the daughter nucleus is the same as that of the original nucleus.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-14p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.15P
CE Which of the three decay processes (α, ß or γ)results in a new element? Explain.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-15p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.16P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.17P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.18P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.19P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.20P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.21P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.22P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.23P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.24P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.25P
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Solution:

Chapter 32 Nuclear Physics and Nuclear Radiation Q.26P
CE The half-life of carbon-14 is 5730 y.
(a) Is it possible for a particular nucleus in a sample of carbon-14 to decay after only 1 s has passed? Explain.
(b) Is it possible for a particular nucleus to decay after 10,000 y? Explain.
Solution:
Yes. The half-life of carbon-14 nucleus is 5,730 y. The half-life of the carbon-14 nucleus represents the average time required for half of a large number of nuclei to decay. A given nucleus in a sample of carbon-14 can decay after only 1 s, because it has a random nature of radioactive decay.
Yes. A given nucleus in a sample of carbon-14 can decay after 10,000 y has passed, because it has a random nature of radioactive decay.

Chapter 32 Nuclear Physics and Nuclear Radiation Q.27P
CE Suppose we were to discover that the ratio of carbon-14 to carbon-12 in the atmosphere was significantly smaller 10,000 yearsago than it is today. How would this affect the ages we have assigned to objects on the basis of carbon-14 dating? In particular, would the true age of an object be greater than or less than the age we had previously assigned to it? Explain.
Solution:
We use carbon 14 dating method to find the ages. If the ratio C – 14 to C – 12 is small 10,000 years ago, then initial amount of carbon – 14 might have been smaller. Then less time is needed for the decay state to reduce to its present value. The age of an object measured in this case will be less than the true age of the object.

Chapter 32 Nuclear Physics and Nuclear Radiation Q.28P
CE A radioactive sample is placed in a closed container. Two days later only one-quarter of the sample is still radioactive. What is the half-life of this sample?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-28p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.29P
Radon gas has a half-life of 3.82 d. What is the decay constant for radon?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-29p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.30P
A radioactive substance has a decay constant equal to 8.9 × 10−3 s−1. What is the half-life of this substance?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-30p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.31P
The number of radioactive nuclei in a particular sample decreases over a period of 18 d to one-sixteenth the original number. What is the half-life of these nuclei?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-31p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.32P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.33P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.34P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.35P
An archeologist on a dig finds a fragment of an ancient basket woven from grass. Later, it is determined that the carbon-14 content of the grass in the basket is 9.25% that of an equal carbon sample from present-day grass. What is the age of the basket?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-35p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.36P
The bones of a saber-toothed tiger are found to have an activity per gram of carbon that is 15.0% of what would be found in a similar live animal. How old are these bones?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-36p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.37P
Charcoal from an ancient fire pit is found to have a carbon-14 content that is only 17.5% that of an equivalent sample of carbon from a living tree. What is the age of the fire pit?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-37p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.38P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.39P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.40P
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Solution:

Chapter 32 Nuclear Physics and Nuclear Radiation Q.41P
The atomic mass of gold-197 is 196.96654 u. How much energy is required to completely separate the nucleons in a gold-197 nucleus?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-41p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.42P
The atomic mass of lithrum-7 is 7.016003 u. How much energy is required to completely separate the nucleons in a lithium-7 nucleus?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-42p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.43P

Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.44P

Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.45P

Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.46P
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.47P
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Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.48P
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Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.49P
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Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-49p1

Chapter 32 Nuclear Physics and Nuclear Radiation Q.50P
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Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-50p1

Chapter 32 Nuclear Physics and Nuclear Radiation Q.51P
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Solution:

Chapter 32 Nuclear Physics and Nuclear Radiation Q.52P
Assuming a release of 173 MeV per fission reaction, calculate how many reactions must occur per second to produce a power output of 150 MW.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-52p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.53P
Consider a fusion reaction in which two deuterium nuclei fuse to form a tritium nucleus and a proton. Mow much energy is released in this reaction?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-53p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.54P
Consider a fusion reaction in which a proton fuses with a neutron to form a deuterium nucleus. How much energy is released in this reaction?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-54p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.55P
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Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.56P
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Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-56p1

Chapter 32 Nuclear Physics and Nuclear Radiation Q.57P
The Evaporating Sun The Sun radiates energy at the prodigious rate of 3.90 × 1026 W.
(a) At what rate, in kilograms per second, does the Sun convert mass into energy?
(b) Assuming that the Sun has radiated at this same rate for its entire lifetime of 4.50 × 109 y, and that the current mass of the Sun is 2.00 × 1030 kg, what percentage of its original mass has been converted to energy?.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-57p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.58P
BIO Radiation Damage A sample of tissue absorbs a 55-rad dose of α particles (RBE = 20).
How many rad of protons (RBB = 10) cause the same amount of damage to the tissue?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-58p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.59P
BIO X-ray Damage How many rad of 200-keV X-rays cause the same amount of biological damage as 50 rad of heavy ions?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-59p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.60P
IP BIO
(a) Find the energy absorbed by a 78-kg person who is exposed to 52 mrem of α particles with an “RBE of 15.
(b) If the RBE of the α particles is increased, does the energy absorbed increase, decrease, or stay the same? Explain.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-60p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.61P
BIO A patient undergoing radiation therapy for cancer receives a 225-rad dose of radiation.
(a) Assuming the cancerous growth has a mass of 0.17 kg, calculate how much energy it absorbs.
(b) Assuming the growth to have the specific heat of water, determine its increase in temperature.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-61p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.62P
BIO Alpha particles with an RBE of 13 deliver a 32-mrad whole-body radiation dose to a 72-kg patient.
(a) What dosage, in rem, does the patient receive?
(b) How much energy is absorbed by the patient?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-62p

Chapter 32 Nuclear Physics and Nuclear Radiation Q.63P
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Solution:

Chapter 32 Nuclear Physics and Nuclear Radiation Q.64GP
CE An α particle (charge + 2e)and a ß particle (charge − e) deflect in opposite directions when they pass through a mag netic field. Which particle defLects by a greater amount, give: that both particles have the same speed? Explain.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-64gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.65GP
CE Radioactive samples A and B have equal half-lives. The initial activity of sample A is twice that of sample B. What is the ratio of the activity of sample A to that of sample B after two half-lives have elapsed?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-65gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.66GP
CE The initial activity of sample A is twice that of sample B After two half-lives of sample A have elapsed, the two samples have the same activity. What is the ratio of the half-life of B to the half-life of A?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-66gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.67GP
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.68GP

Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.69GP
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.70GP
Suppose it is desired to give a cancerous tumor a dose of 3800 rem. How many rads are needed if the tumor is exposed to alpha radiation?
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-70gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.71GP
A patient is exposed to 260 rad of gamma rays. What is the dose the patient receives in rem?
Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.72GP
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Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.73GP
Moon Rocks In one of the rocks brought back from the Moon, it is found that 80.5% of the initial potassium-40 in the rock has decayed to argon-40.
(a) If the half-life for this decay is 1.20 × 109 years, how old is the rock?
(b) How much longer will it take before only 10.0% of the original potassium-40 is still present in the rock?
Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.74GP
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.75GP
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.76GP
An α particle fired head-on at a stationary nickel nucleus approaches to a radius of 15 fm
before being turned around.
(a) What is the maximum Coulomb force exerted on the α particle?
(b) What is the electric potential energy of the α particle at its point of closest approach?
(c) Find the initial kinetic energy of the α particle.
Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.77GP
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.78GP
IP Initially, a sample of radioactive nuclei of type A contains four times as many nuclei as a sample of radioactive nuclei of type B. Two days later (2.00 d) the two samples contain the same number of nuclei.
(a) Which type of nucleus has the longer half-life? Explain.
(b) Determine the half-life of type B nuclei if the half-life of type A nuclei is known to be 0.500 d.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-78gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.79GP
Stable nuclei have mass numbers that range from a minimum of 1 to a maximum of 209.
(a) Find the corresponding range in nuclear radii.
(b) Assuming all nuclei to be spherical, determine the ratio of the surface area of the largest stable nucleus to the surface area of the smallest nucleus.
(c) Repeat part (b), only this time find the ratio of the volumes.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-79gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.80GP
Radius of a Neutron Star Neutron stars are so named because they are composed of neutrons and have a density the same as that of a nucleus. Referring to Example 32–2 for the nuclear density, find the radius of a neutron star whose mass is 0.50 that of the Sun.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-80gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.81GP
A specimen taken from the wrappings of a mummy contains 7.82 g of carbon and has an activity of 1.38 Bq. How old is the mummy? (Refer to pages 1132 and 1133 for relevant information regarding the isotopes of carbon.)
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-81gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.82GP
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.83GP
IP Energy is released when three α particles fuse to form carbon-12.
(a) Is the mass of carbon-12 greater than, less than, or the same as the mass of three α particles? Explain.
(b) Calculate the energy given off in this fusion reaction.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-83gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.84GP
Find the dose of y rays that must be absorbed by a block of ice at 0 °C to convert it to water at 0 °C.-Givc the dosage inrad.
Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.85GP
IP
(a) What dosage (in rad) must a 1.0-kg sample of water absorb to increase its temperature by 1.0 C°?
(b) Tf the mass of the water sample is increased, does the dosage found in part (a) increase, decrease, or stay the same? Explain.
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-85gp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.86GP
BIO Chest X-rays A typical chest X-ray uses X-rays with an RBE of 0.85. If the radiation dosage is 35 mrem, find the energy absorbed by a 72-kg patient, assuming one-quarter of the patient’s body is exposed to the X-rays.
Solution:
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.87GP
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.88GP
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.89GP
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Chapter 32 Nuclear Physics and Nuclear Radiation Q.90PP
What is the decay constant, A, for iodine-131?
A. 9.98 ×10−7 s−1
B. 1.44 × ×10−6 s−1
C. 2.39 ×10−5 s−1
D. 5.99 ×10−5 s−1
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-90pp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.91PP
If a sample of iodine-131 contains 4.5 × 1016 nuclei, what is the activity of the sample?
Express your answer in curies.
A. 0.27 Ci
B. 1.2 Ci.
C. 1.7 Ci
D. 4.5 Ci
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-91pp

Chapter 32 Nuclear Physics and Nuclear Radiation Q.92PP
If the half-life of iodine-131 were only half of its actual value, would the activity of the sample in Problem be increased or decreased?
Problem
If a sample of iodine-131 contains 4.5 × 1016 nuclei, what is the activity of the sample? Express your answer in curies.
A. 0.27 Ci
B. 1.2 Ci.
C. 1.7 Ci
D. 4.5 Ci
Solution:
mastering-physics-solutions-chapter-32-nuclear-physics-and-nuclear-radiation-92pp

Mastering Physics Solutions Chapter 5 Newton’s Laws Of Motion

May 21, 2018May 21, 2018 by Prasanna

Mastering Physics Solutions Chapter 5 Newton’s Laws Of Motion

Mastering Physics Solutions

Chapter 5 Newton’s Laws Of Motion Q.1CQ
Driving down the road, you hit the brakes suddenly. As a result, your body moves toward the front of the car. Explain, using Newton’s laws.
Solution:
When the brakes are applied, the car slows down. The body, however, keeps moving at the same speed. Thus, because of inertia, the body moves toward the front of the car.
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion1cqs

Chapter 5 Newton’s Laws Of Motion Q.1P
CE An object of mass m is initially at rest. After a force of magnitude F acts on it for a time T, the object has a speed v. Suppose the mass of the object is doubled, and the magnitude of the force acting on it is quadrupled. In terms of T, how long does it take for the object to accelerate from rest to a speed v now?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion1ps

Chapter 5 Newton’s Laws Of Motion Q.2CQ
You’ve probably seen pictures of someone pulling a tablecloth out from under glasses, plates, and silverware set out for a formal dinner. Perhaps you’ve even tried it yourself. Using Newton’s laws of motion, explain how this stunt works.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion2cqs

Chapter 5 Newton’s Laws Of Motion Q.2P
On a planet far, far away, an astronaut picks up a rock. The rock has a mass of 5.00 kg, and on this particular planet its weight is 40.0 N. Tf the astronaut exerts an upward force of 46.2 N on the rock, what is its acceleration?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion2ps

Chapter 5 Newton’s Laws Of Motion Q.3CQ
As you read this, you arc most likely sitting quietly in a chair. Can you conclude, therefore, that you are at rest? Explain.
Solution:
You are not at absolute rest, you are at rest with respect to the other objects in your surroundings. However when you are viewed from other planets, you are not at rest relative to the vantage point of the other planets.

Chapter 5 Newton’s Laws Of Motion Q.3P
In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 N. If the cart starts at rest, how far does it move in 2.50 s?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion3ps

Chapter 5 Newton’s Laws Of Motion Q.4CQ
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion4cq
Solution:
When a dog shakes its body, the water remains at rest because of inertia. The water drops fall away from the body when the position of the dog’s body is changed rapidly.

Chapter 5 Newton’s Laws Of Motion Q.4P
You are pulling your little sister on her sled across an icy (fric-tionless) surface. When you exert a constant horizontal force of 120 N, the sled has an acceleration of 2.5 m/s2. If the sled has a mass of 7.4 kg, what is the mass of your little sister?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion4ps

Chapter 5 Newton’s Laws Of Motion Q.5CQ
A young girl slides down a rope. As she slides faster and faster she tightens her grip, increasing the force exerted on her by the rope. What happens when this force is equal in magnitude to her weight? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion5cqs

Chapter 5 Newton’s Laws Of Motion Q.5P
· A 0.53-kg billiard ball initially at rest is given a speed of 12 m/s during a time interval of 4.0 ms. What average force acted on the ball during this time?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion5ps

Chapter 5 Newton’s Laws Of Motion Q.6CQ
A drag-racing car accelerates forward because of the force exerted on it by the road. Why, then, does it need an engine? Explain.
Solution:
The drag racer needs an engine to turn the wheels, which in turn makes them push against the ground. When the wheels push against the ground, the ground is able to exert a reaction force on the car to move it.

Chapter 5 Newton’s Laws Of Motion Q.6P
A 92-kg water skier floating in a lake is pulled from rest to a speed of 12 m/s in a distance of 25 m. What is the net force exerted on the skier, assuming his acceleration is constant?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion6ps

Chapter 5 Newton’s Laws Of Motion Q.7CQ
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion7cq
Solution:
(A) The upper string is exposed to two forces. One is the downward weight attached and the other is the force applied. So, in this case, the upper string will break.
(B) Because of the inertia of the block, the lower string will break.

Chapter 5 Newton’s Laws Of Motion Q.7P
CE Predict/Explain You drop two balls of equal diameter from the same height at the same time. Ball 1 is made of metal and has a greater mass than ball 2, which is made of wood. The upward force due to air resistance is the same for both balls, (a) Is the drop time of ball 1 greater than, less than, or equal to the drop time of ball 2? (b) Choose the best explanation from among the following:
I. The acceleration of gravity is the same for all objects, regardless of mass.
II. The more massive ball is harder to accelerate.
III. Air resistance has less effect on the more massive ball.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion7ps

Chapter 5 Newton’s Laws Of Motion Q.8CQ
An astronaut on a space walk discovers that his jet pack no longer works, leaving him stranded 50 m from the spacecraft. If the jet pack is removable, explain how the astronaut can still use it to return to the ship.
Solution:
The astronaut should push the jetpack away from him, in the opposite direction from the spaceship. As a result, the reaction force exerted on him by the pack will accelerate him toward the ship.

Chapter 5 Newton’s Laws Of Motion Q.8P
IP A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant acceleration over a distance of 0.750 m, what force does the ground exert on her? (b) If the parachutist comes to rest over a shorter distance, is the force exerted by the ground greater than, less than, or the same as in part (a)? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion8ps

Chapter 5 Newton’s Laws Of Motion Q.9CQ
Two untethered astronauts on a space walk decide to take a break and play catch with a baseball. Describe what happens as the game of catch progresses.
Solution:
Each time the astronauts catch or throw, an equal and opposite force acts on them. This causes the astronauts to move farther from each other, with increasing speed.

Chapter 5 Newton’s Laws Of Motion Q.9P
IP In baseball, a pitcher can accelerate a 0.15-kg ball from rest to 98 mi/h in a distance of 1.7 m. (a) What is the average force exerted on the ball during the pitch? (b) If the mass of the ball is increased, is the force required of the pitcher increased, decreased, or unchanged? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion9ps

Chapter 5 Newton’s Laws Of Motion Q.10CQ
What are the action-reaction forces when a baseball bat hits a fast ball? What is the effect of each force?
Solution:
Because of action-reaction forces, the ball will be returned with the same force that the ball applied to the bat.
The force exerted on the bat by the ball is action force, and the force exerted on the ball by the bat is reaction force.
The force exerted on the ball changes its direction.

Chapter 5 Newton’s Laws Of Motion Q.10P
A major-league catcher gloves a 92-mi/h pitch and brings it to rest in 0.15 m. If the force exerted by the catcher is 803 N, what is the mass of the ball?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion10ps

Chapter 5 Newton’s Laws Of Motion Q.11CQ
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion11cq
Solution:
Mr. Ed’s reasoning is incorrect because he is adding two action-reaction forces that act on different objects.
Wilbur should point out that the net force exerted on the cart is simply the force exerted on it by Mr. Ed, and so the cart will accelerate.
The equal and opposite reaction force acts on Mr. Ed, and does not cancel the force acting on the cart.

Chapter 5 Newton’s Laws Of Motion Q.11P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion11p
Solution:

Chapter 5 Newton’s Laws Of Motion Q.12CQ
A whole brick has more mass than half a brick, thus the whole brick is harder to accelerate. Why doesn’t a whole brick fall more slowly than half a brick? Explain.
Solution:
The whole brick also experiences twice the gravitational force.
As a result, these two effects (more inertia, more force) cancel each other out exactly. The free-fall acceleration is independent of mass.

Chapter 5 Newton’s Laws Of Motion Q.12P
Stopping a 747 A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.50 × 105 kg, its speed is 27.0 m/s, and the net braking force is 4.30 × 105 N, (a) what is its speed 7.50 s later? (b) How far has it traveled in this time?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion12ps

Chapter 5 Newton’s Laws Of Motion Q.13CQ
The force exerted by gravity on a whole brick is greater than the force exerted by gravity on half a brick. Why, then, doesn’t a whole brick fall faster than half a brick? Explain.
Solution:
Acceleration of the whole brick is the same as that of half the brick. This is because acceleration is directly proportional to force and inversely proportional to mass.

Chapter 5 Newton’s Laws Of Motion Q.13P
IP A drag racer crosses the finish line doing 202 mi/h and promptly deploys her drag chute (the small parachute used for braking), (a) What force must the drag chute exert on the 891-kg car to slow it to 45.0 mi/h in a distance of 185 m? (b) Describe the strategy you used to solve part (a).
SECTION 5-4 NEWTON’S THIRD LAW OF MOTION
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion13ps

Chapter 5 Newton’s Laws Of Motion Q.14CQ
Is it possible for an object at rest to have only a single force acting on it? If your answer is yes, provide an example. If your answer is no, explain why not.
Solution:
No.
If only a single force acts on the object, it will not stay at rest but will accelerate in the direction of the force.

Chapter 5 Newton’s Laws Of Motion Q.14P
CE Predict/Explain A small car collides with a large truck, (a) Is the magnitude of the force experienced by the car greater than, less than, or equal to the magnitude of the force experienced by the truck? (b) Choose the best explanation from among the following:
I. Action-reaction forces always have equal magnitude.
II. The truck has more mass, and hence the force exerted on it is greater.
III. The massive truck exerts a greater force on the lightweight car.
Solution:
(a) The magnitude of the force experienced by the car is equal to magnitude of the force experienced by the truck.
(b) This is because action reaction forces are always equal in magnitude according to ’s third law. So, option I is the best explanation.

Chapter 5 Newton’s Laws Of Motion Q.15CQ
Is it possible for an object to be in motion and yet have zero net force acting on it? Explain.
Solution:
Yes.
When the object is moving with a constant velocity, its acceleration is zero. Thus, the force acting on the object is also zero.

Chapter 5 Newton’s Laws Of Motion Q.15P
CE Predict/Explain A small car collides with a large truck, (a) Is the acceleration experienced by the car greater than, less than, or equal to the acceleration experienced by the truck? (b) Choose the best explanation from among the following:
I. The truck exerts a larger force on the car, giving it the greater acceleration.
II. Both vehicles experience the same magnitude of force, therefore the lightweight car experiences the greater acceleration.
III. The greater force exerted on the truck gives it the greater acceleration.
Solution:
(a) The acceleration experienced by the car is greater than the acceleration experienced by the truck.
(b) This is because of both cars experience same magnitude of force, therefore the lightweight car experiences greater acceleration. So, option II is the best explanation.

Chapter 5 Newton’s Laws Of Motion Q.16CQ
A bird cage, with a parrot inside, hangs from a scale. The parrot decides to hop to a higher perch. What can you say about the reading on the scale (a) when the parrot jumps, (b) when the parrot is in the air, and (c) when the parrot lands on the second perch? Assume that the scale responds rapidly so that it gives an accurate reading at all times.
Solution:
(a) The scale goes down at the moment the parrot jumps.
This happens because the parrot pushes down on its perch in order to jump, and as a result the scale reads a larger value.
(b) When the bird is in the air, the scale just reads the weight of the cage.
(c) When the parrot lands on a higher perch, the scale goes down and then reads a higher value again. This happens because the perch exerts an upward force on the bird in order to bring the bird to rest.

Chapter 5 Newton’s Laws Of Motion Q.16P
You hold a brick at rest in your hand. (a) How many forces act on the brick? (b) Identify these forces, (c) Are these forces equal in magnitude and opposite in direction? (d) Are these forces an action-reaction pair? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion16ps

Chapter 5 Newton’s Laws Of Motion Q.17CQ
Suppose you jump from the cliffs of Acapulco and perform a perfect swan dive. As you fall, you exert an upward force on the Earth equal in magnitude to the downward force the Earth exerts on you. Why, then, does it seem that you are the one doing all the accelerating? Since the forces are the same, why aren’t the accelerations?
Solution:
The acceleration of a body is inversely proportional to its mass. Earth has a huge mass compared to a man, and so has negligible acceleration towards the man. Thus, the forces are the same but not the accelerations.

Chapter 5 Newton’s Laws Of Motion Q.17P
Referring to Problem 16, you are now accelerating the brick upward, (a) How many forces act on the brick in this case? (b) Identify these forces, (c) Are these forces equal in magnitude and opposite in direction? (d) Are these forces an action-reaction pair? Explain.
Solution:
(A) There are two forces.
(B) Gravitational force and the upward force applied by your hand.
(C) No, these forces are not equal and opposite since the brick is moving up.
(D) No.

Chapter 5 Newton’s Laws Of Motion Q.18CQ
A friend tells you that since his car is at rest, there are no forces acting on it. How would you reply?
Solution:
The car is at rest, and the net force acting on the car is zero.
However, it is wrong to say that no force is acting on the car. When the car is at rest, gravitational force acts on the car, which is balanced by the normal force reaction of the ground upon the car.

Chapter 5 Newton’s Laws Of Motion Q.18P
On vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85 m/s2, (a) What is the net force exerted on the trailer? (b) What force does the traiter exert on the car? (c) What is the net force acting on the car?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion18ps

Chapter 5 Newton’s Laws Of Motion Q.19CQ
Since all objects are “weightless” in orbit, how is it possible for an orbiting astronaut to tell if one object has more mass than another object? Explain.
Solution:
If an astronaut pushes the object, the acceleration can indicate the object’s mass. The heavier mass will have lower acceleration compared to the lighter mass.

Chapter 5 Newton’s Laws Of Motion Q.19P
IP A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push, (a) Is the force experienced by the child more than, less than, or the same as the force experienced by the parent? (b) Is the acceleration of the child more than, less than, or the same as the acceleration of the parent? Explain, (c) If the acceleration of the child is 2.6 m/s2 in magnitude, what is the magnitude of the parent’s acceleration?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion19ps

Chapter 5 Newton’s Laws Of Motion Q.20CQ
To clean a rug, you can hang it from a clothesline and beat it with a tennis racket. Use Newton’s laws to explain why beating the rug should have a cleansing effect.
Solution:
As you hit the rug with the tennis racket you cause it to accelerate rapidly.
The dust from the rug, if it is not attached too firmly, will be left behind as the rug accelerates away from it.

Chapter 5 Newton’s Laws Of Motion Q.20P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion20p
Solution:

Chapter 5 Newton’s Laws Of Motion Q.21CQ
If you step off a high board and drop to the water below, you plunge into the water without injury. On the other hand, if you were to drop the same distance onto solid ground, you might break a leg. Use Newton’s laws to explain the difference.
Solution:
When we jump on the ground, we plunge a very small distance compared to jumping into water. The smaller the distance, the larger the acceleration, and thus the greater the force.

Chapter 5 Newton’s Laws Of Motion Q.21P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion21p
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Chapter 5 Newton’s Laws Of Motion Q.22CQ
A moving object is acted on by a net force. Give an example of a situation in which the object moves (a) in the same direction as the net force, (b) at right angles to the net force, or (c) in the opposite direction of the net force.
Solution:
(a) When a cart is pushed forward and set into motion, the direction of the force acting on it and the direction of its motion are equal.
(b) If a ball or any object is thrown up, then, at the top of its flight, its motion is in the horizontal direction. Additionally, the force of gravity acting on the ball (or any object) is in the vertical direction. Here, the direction of force and the direction of motion are perpendicular to each other. In this example, air resistance is neglected.
(c) While you are riding a bicycle, the frictional force acts in the direction opposite to the direction of motion of the bicycle.
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion22cqs

Chapter 5 Newton’s Laws Of Motion Q.22P
IP Two boxes sit side-by-side on a smooth horizontal surface. The lighter box has a mass of 5.2 kg; the heavier box has a mass of 7.4 kg. (a) Find the contact force between these boxes when a horizontal force of 5.0 N is applied to the light box. (b) If the 5.0-N force is applied to the heavy box instead, is the contact force between the boxes the same as, greater than, or less than the contact force in part (a)? Explain, (c) Verify your answer to part (b) by calculating the contact force in this case.
Solution:

Chapter 5 Newton’s Laws Of Motion Q.23CQ
Is it possible for an object to be moving in one direction while the net force acting on it is in another direction? If your answer is yes, provide an example. If your answer is no, explain why not.
Solution:
Yes, during the journey of a ball thrown upward, the direction of force is always downward. Thus, the force is in an opposite direction until the ball reaches its maximum height, and then it comes back down.

Chapter 5 Newton’s Laws Of Motion Q.23P
CE A skateboarder on a ramp is accelerated by a nonzero net force. For each of the following statements, state whether it is always true, never true, or sometimes true, (a) The skateboarder is moving in the direction of the net force, (b) The ac-celeration of the skateboarder is at right angles to the net force. (c) The acceleration of the skateboarder is in the same direction as the net force. (d) The skateboarder is instantaneously at rest.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion23ps

Chapter 5 Newton’s Laws Of Motion Q.24CQ
Since a bucket of water is “weightless” in space, would it hurt to kick the bucket? Explain.
Solution:
Answer: Yes
Explanation:
It is given that the bucket of water is weightless in space, but every object has a certain mass, and also has inertia. As inertia is the tendency of a body to resist any change in its state of motion or rest, so when you kick the bucket, it resists a change in its state of motion by exerting an equal and opposite force, which gives the hurt to your foot.

Chapter 5 Newton’s Laws Of Motion Q.24P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion24p
Solution:

Chapter 5 Newton’s Laws Of Motion Q.25CQ
In the movie The Rocketeer, a teenager discovers a jet-powered backpack in an old barn. The backpack allows him to fly at incredible speeds. In one scene, however, he uses the backpack to rapidly accelerate an old pickup truck that is being chased by “bad guys.” He does this by bracing his arms against the cab of the pickup and firing the backpack, giving the truck the acceleration of a drag racer. Is the physics of this scene “Good,” “Bad,” or “Ugly?” Explain.
Solution:
This scene is an example of bad physics. It is true that the jet-powered backpack produces enough force to push the truck, but this force is imparted to the truck through the arms of the teenager. This is a very unrealistic situation, because force that is great enough to accelerate an old truck would likely crush the teenager’s arms.

Chapter 5 Newton’s Laws Of Motion Q.25P
A farm tractor tows a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer? (Ignore friction.)
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion25ps

Chapter 5 Newton’s Laws Of Motion Q.26CQ
List three common objects that have a weight of approximately 1 N.
Solution:
Any object whose weight is approximately 100g has a weight equal to 1N
(i) An apple weighs approximately 1N
(ii) A chocolate bar
(iii) A pack of crayons

Chapter 5 Newton’s Laws Of Motion Q.26P
A surfer “hangs ten,” and accelerates down the sloping face of a wave. If the surfer’s acceleration is 3.25 m/s2 and friction can be ignored, what is the angle at which the face of the wave is inclined above the horizontal?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion26ps

Chapter 5 Newton’s Laws Of Motion Q.27P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion27p
Solution:

Chapter 5 Newton’s Laws Of Motion Q.28P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion28p
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Chapter 5 Newton’s Laws Of Motion Q.29P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion29p
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Chapter 5 Newton’s Laws Of Motion Q.30P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion30p
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Chapter 5 Newton’s Laws Of Motion Q.31P
IP Before practicing his routine on the rings, a 67-kg gymnast stands motionless, with one hand grasping each ring and his feet touching the ground. Both aims slope upward at an angle of 24° above the horizontal, (a) If the force exerted by the rings on each arm has a magnitude of 290 N, and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet? (b) If the angle his arms make with the horizontal is greater that 24°, and everything else remains the same, is the force exerted by the floor on his feet greater than, less than, or the same as the valne found in part (a)? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion31ps

Chapter 5 Newton’s Laws Of Motion Q.32P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion32p
Solution:

Chapter 5 Newton’s Laws Of Motion Q.33P
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 4.4 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion33ps

Chapter 5 Newton’s Laws Of Motion Q.34P
· A train is traveling up a 3.73° incline at a speed of 3.25 m/s when the last car breaks free and begins to coast without friction, (a) How long does it take for the last car to come to rest momentarily? (b) How far did the last car travel before mo-mentarily coming to rest?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion34ps

Chapter 5 Newton’s Laws Of Motion Q.35P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion35p
Solution:

Chapter 5 Newton’s Laws Of Motion Q.36P
You pull upward on a stuffed suitcase with a force of 105 N, and it accelerates upward at 0.705 m/s2. What are (a) the mass and (b) the weight of the suitcase?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion36ps

Chapter 5 Newton’s Laws Of Motion Q.37P
BIO Brain Growth A newborn baby’s brain grows rapidly. In fact, it has been found to increase in mass by about 1.6 mg per minute, (a) How much does the brain’s weight increase in one day? (b) How long does it take for the brain’s weight to increase by 0.15 N?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion37ps

Chapter 5 Newton’s Laws Of Motion Q.38P
Suppose a rocket launches with an acceleration of 30.5 m/s2. What is the apparent weight of an 92-kg astronaut aboard this rocket?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion38ps

Chapter 5 Newton’s Laws Of Motion Q.39P
At the bow of a ship on a stormy sea, a crewman conducts an experiment by standing on a bathroom scale. In calm waters, the scale reads 182 lb. During the storm, the crewman finds a maximum reading of 225 lb and a minimum reading of 138 lb. Find (a) the maximum upward acceleration and (b) the maximum downward acceleration experienced by the crewman.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion39ps

Chapter 5 Newton’s Laws Of Motion Q.40P
IP As part of a physics experiment, you stand on a bathroom scale in an elevator. Though your normal weight is 610 N, the scale at the moment reads 730 N. (a) Is the acceleration of the elevator upward, downward, or zero? Explain. (b) Calculate the magnitude of the elevator’s acceleration. (c) What, if anything, can you say about the velocity of the elevator? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion40ps

Chapter 5 Newton’s Laws Of Motion Q.41P
When you weigh yourself on good old terra firma (solid ground), your weight is 142 lb. In an elevator your apparent weight is 121 lb. What are the direction and magnitude of the elevator’s acceleration?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion41ps

Chapter 5 Newton’s Laws Of Motion Q.42P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion42p
Solution:

Chapter 5 Newton’s Laws Of Motion Q.43PWhen you lift a bowling ball with a force of 82 N, the ball accelerates upward with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion43ps

Chapter 5 Newton’s Laws Of Motion Q.44P
A 23-kg suitcase is being pulled with constant speed by a handle that is at an angle of 25° above the horizontal. If the normal force exerted on the suitcase is 180 N, what is the force F applied to the handle?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion44ps

Chapter 5 Newton’s Laws Of Motion Q.45P
(a) Draw a free-body diagram for the skier in Problem 32. (b) Determine the normal force acting on the skier.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion45ps

Chapter 5 Newton’s Laws Of Motion Q.46P
· A 9.3-kg child sits in a 3.7-kg high chair. (a) Draw a free-body diagram for the child, and find the normal force exerted by the chair on the child. (b) Draw a free-body diagram for the chair, and find the normal force exerted by the floor on the chair.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion46ps

Chapter 5 Newton’s Laws Of Motion Q.47P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion47p
Solution:

Chapter 5 Newton’s Laws Of Motion Q.48P
A 5.0-kg bag of potatoes sits on the bottom of a stationary shopping cart. (a) Sketch a free-body diagram for the bag of potatoes. (b) Now suppose the cart moves with a constant velocity. How does this affect your free-body diagram? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion48ps

Chapter 5 Newton’s Laws Of Motion Q.49P
IP (a) Find the normal force exerted on a 2.9-kg book resting on a surface inclined at 36° above the horizontal. (b) If the angle of the incline is reduced, do you expect the normal force to increase, decrease, or stay the same? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion49ps

Chapter 5 Newton’s Laws Of Motion Q.50P
IP A gardener mows a lawn with an old-fashioned push mower. The handle of the mower makes an angle of 35° with the surface of the lawn. (a) If a 219-N force is applied along the handle of the 19-kg mower, what is the normal force exerted by the lawn on the mower? (b) If the angle between the surface of the lawn and the handle of the mower is increased, does the normal force exerted by the lawn increase, decrease, or stay the same? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion50ps

Chapter 5 Newton’s Laws Of Motion Q.51P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion51p
Solution:

Chapter 5 Newton’s Laws Of Motion Q.52GP
·CE Predict/Explain Riding in an elevator moving upward with constant speed, you begin a game of darts. (a) Do you have to aim your darts higher than, lower than, or the same as when you play darts on solid ground? (b) Choose the best explanation from among the following:
I. The elevator rises during the time it takes for the dart to travel to the darfboard.
II. The elevator moves with constant velocity. Therefore, Newton’s laws apply within the elevator in the same way as on the ground.
III. You have to aim lower to compensate for the upward speed of the elevator.
Solution:
(a) You have to aim your darts same as when you play darts on solid ground.
(b) This is because the elevator is moving with constant velocity. Therefore ’s laws apply within the elevator in the same way as on earth.
Therefore option II is best explanation.

Chapter 5 Newton’s Laws Of Motion Q.53GP
CE Predict/Explain Riding in an elevator moving with a constant upward acceleration, you begin a game of darts. (a) Do you have to aim your darts higher than, lower than, or the same as when you play darts on solid ground? (b) Choose the best explanation from among the following:
I. The elevator accelerates upward, giving its passengers a greater “effective” acceleration of gravity.
II. You have to aim lower to compensate for the upward acceleration of the elevator.
III. Since the elevator moves with a constant acceleration, Newton’s laws apply within the elevator the same as on the ground.
Solution:
In the game of darts a player have to aim and throw the dart on the board to hit the bulls- eye.
(a)
When elevator is going upward with constant acceleration the board hanged on wall of the elevator will also move upward with same acceleration. When a person will throw a dart it will immediately start accelerating downward on the other hand board will keep accelerating upward. Therefore dart will not hit the aimed position. It will hit lower than the aimed position.
Hence in the accelerating elevator a person has to aim his darts higher than the normal.
(b)
When the elevator is accelerating upward then due to force exerted by the floor of the elevator effective acceleration of the passenger increases. Therefore first explanation is right.
As discussed in previous section passenger has aim higher to compensate for the upward acceleration. Therefor second explanation is wrong.
Newton’s law can be applied directly only for inertial reference frame (stationary reference point or moving with constant reference point). Since elevator is moving with constant acceleration therefore newton’s law applied within the elevator is not same as on the ground. Therefore third explanation is wrong
Hence option I is the best explanation.

Chapter 5 Newton’s Laws Of Motion Q.54GP
CE Give the direction of the net force acting on each of the following objects. If the net force is zero, state “zero.” (a) A car accelerating northward from a stoplight. (b) A car traveling southward and slowing down. (c) A car traveling westward with constant speed. (d) A skydiver parachuting downward with constant speed. (e) A baseball during its flight from pitcher to catcher (ignoring air resistance).
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion54gps

Chapter 5 Newton’s Laws Of Motion Q.55GP
·CE Predict/Explain You jump out of an airplane and open your parachute after an extended period of free fall. (a) To decelerate your fall, must the force exerted on you by the parachute be greater than, less than, or equal to your weight? (b) Choose the best explanation from among the following:
I. Parachutes can only exert forces that are less than the weight of the skydiver.
II. The parachute exerts a force exactly equal to the skydiver’s weight.
III. To decelerate after free fall, the net force acting on a skydiver must be upward.
Solution:
(a) To decelerate your fall the force exerted by the parachute must be greater than your weight.
(b) To decelerate after free fall, the net force acting on a sky diver must be upward. This can be achieved only when the force exerted by the parachute is greater than your weight
So option III is the best explanation.

Chapter 5 Newton’s Laws Of Motion Q.56GP
In a tennis serve, a 0.070-kg ball can be accelerated from rest to 36 m/s over a distance of 0.75 m. Find the magnitude of the average force exerted hy the racket on the ball during the serve.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion56gp

Chapter 5 Newton’s Laws Of Motion Q.57GP
A 51.5-kg swimmer with an initial speed of 1.25 m/s decides to coast until she comes to rest. If she slows with constant acceleration and stops after coasting 2.20 m, what was the force exerted on her by the water?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion57gps

Chapter 5 Newton’s Laws Of Motion Q.58GP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion58gp
Solution:
The three identical pucks are acted upon by a force of 3N.
F = ma
Since the mass and the force of the 3 pucks are the same, then the acceleration produced by this force must also be the same. The acceleration produced by a force does not depend on the speed of the object, so the acceleration of A, B, and C are all equal.

Chapter 5 Newton’s Laws Of Motion Q.59GP
IP The VASIMR Rocket NASA plans to use a new type of rocket, a Variable Specific Impulse Magnetoplasma Rocket (VASIMR), on future missions. A VASIMR can produce 1200 N of thrust (force) when in operation. If a VASIMR has a mass of 2.2 × 105 kg, (a) what acceleration will it experience? Assume that the only force acting on the rocket is its own thrust, and that the mass of the rocket is constant. (b) Over what distance must the rocket accelerate from rest to achieve a speed of 9500 m/s? (c) When the rocket has covered one-quarter the acceleration distance found in part (b), is its average speed 1/2, 1/3, or 1/4 its average speed during the final three-quarters of the acceleration distance? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion59gps

Chapter 5 Newton’s Laws Of Motion Q.60GP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion60gp
Solution:

Chapter 5 Newton’s Laws Of Motion Q.61GP
At the local grocery store, you push a 14.5-kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N, you now accelerate the cart from rest through a distance of 2.29 m in 3.00 s. What was the mass of the dog food?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion61gps

Chapter 5 Newton’s Laws Of Motion Q.62GP
· IP BIO The Force of Running Biomechanical research has shown that when a 67-kg person is running, the force exerted oh each foot as it strikes the ground can be as great as 2300 N. (a) What is the ratio of the force exerted on the foot by the ground to the person’s body weight? (b) If the only forces acting on the person are (i) the force exerted by the ground and (ii) the person’s weight, what are the magnitude and direction of the person’s acceleration? (c) If the acceleration found in part (b) acts for 10.0 ms, what is the resulting change in the vertical component of the person’s velocity?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion62gps

Chapter 5 Newton’s Laws Of Motion Q.63GP
IP BIO Grasshopper Liftoff To become airborne, a 2.0-g grasshopper requires a takeoff speed of 2.7 m/s. It acquires this speed by extending its hind legs through a distance of 3.7 cm. (a) What is the average acceleration of the grasshopper during takeoff? (b) Find the magnitude of the average net force exerted
on the grasshopper by its hind legs during takeoff. (c) If the mass of the grasshopper increases, does the takeoff acceleration increase, decrease, or stay the same? (d) If the mass of the grasshopper increases, does the required takeoff force increase, decrease, or stay the same? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion63gps

Chapter 5 Newton’s Laws Of Motion Q.64GP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion64gp
Solution:

Chapter 5 Newton’s Laws Of Motion Q.65GP
IP An archer shoots a 0.024-kg arrow at a target with a speed of 54 m/s. When it hits the target, it penetrates to a depth of 0.083 m. (a) What was the average force exerted by the target on the arrow? (b) If the mass of the arrow is doubled, and the force exerted by the target on the arrow remains the same, by what multiplicative factor does the penetration depth change? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion65gps

Chapter 5 Newton’s Laws Of Motion Q.66GP
An apple of mass m = 0.13 kg falls out of a tree from a height h = 3.2 m. (a) What is the magnitude of the force of gravity, mg, acting on the apple? (b) What is the apple’s speed, v, just before it lands? (c) Show that the force of gravity times the height, mgh, is equal to . (We shall investigate the significance of this result in Chapter 8.) Be sure to show that the dimensions are in agreement as well as the nu-merical values.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion66gps

Chapter 5 Newton’s Laws Of Motion Q.67GP
An apple of mass m = 0.22 kg falls from a tree and hits the ground with a speed of v = 14 m/s. (a) What is the magnitude of the force of gravity, mg, acting on the apple? (b) What is the time, t, required for the apple to reach the ground? (c) Show that the force of gravity times the time, mgt, is equal to mv. (We shall investigate the significance of this result in Chapter 9.) Be sure to show that the dimensions are in agreement as well as the numerical values.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion67gps

Chapter 5 Newton’s Laws Of Motion Q.68GP
BIO The Fall of T. rex Paleontologists estimate that if a Tyrnnnosaunis rex were to trip and fall, it would have experienced a force of approximately 260,000 N acting on its torso when it hit the ground. Assuming the torso has a mass of 3800 kg, (a) find the magnitude of the torso’s upward acceleration as it comes to rest. (For comparison, humans lose consciousness with an acceleration of about 7g.) (b) Assuming the torso is in free fall for a distance of 1.46 m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion68gps

Chapter 5 Newton’s Laws Of Motion Q.69GP
Deep Space I The NASA spacecraft Deep Space I was shut down on December 18, 2001, following a three-year journey to the asteroid Braille and the comet Borrelly. This spacecraft used a solar-powered ion engine to produce 0.064 ounces of thrust (force) by stripping electrons from neon atoms and accelerating the resulting ions to 70,000 mi/h. The thrust was only as much as the weight of a couple sheets of paper, but the engine operated continuously for 16,000 hours. As a result, the speed of the spacecraft increased by 7900 mi/h. What was the mass of Deep Space I? (Assume that the mass of the neon gas is negligible.)
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion69gps

Chapter 5 Newton’s Laws Of Motion Q.70GP
Your groceries are in a bag with paper handles. The handles will tear off if a force greater than 51.5 N is applied to them. What is the greatest mass of groceries that can be lifted safely with this bag, given that the bag is raised (a) with constant speed, or (b) with an acceleration of 1.25 m/s2?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion70gps

Chapter 5 Newton’s Laws Of Motion Q.71GP
IP While waiting at the airport for your flight to leave, you observe some of the jets as they take off. With your watch you find that it takes about 35 seconds for a plane to go from rest to takeoff speed. In addition, you estimate that the distance re-quired is about 1.5 km. (a) If the mass of a jet is 1.70 × 105 kg, what force is needed for takeoff? (b) Describe the strategy you used to solve part (a).
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion71gps

Chapter 5 Newton’s Laws Of Motion Q.72GP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion72gp
Solution:

Chapter 5 Newton’s Laws Of Motion Q.73GP
Your groceries are in a bag with paper handles. The handles will tear off if a force greater than 51.5 N is applied to them. What is the greatest mass of groceries that can be lifted safely with this bag, given that the bag is raised (a) with constant speed, or (b) with an acceleration of 1.25 m/s2?
Solution:
The Newton’s second law of motion defined the acceleration of an object produced by a net force is directly proportional to the magnitude of net force and inversely proportional to the mass of object.
The force acting on mass m moves with acceleration a is given as follows:
F=ma
Here, m is the mass and a is the acceleration.
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion73gps

Chapter 5 Newton’s Laws Of Motion Q.74GP
IP Responding to an alarm, a 102-kg fireman slides down a pole to the ground floor, 3.3 m below. The fireman starts at rest and lands with a speed of 4.2 m/s. (a) Find the average force exerted on the fireman by the pole. (b) If the landing speed is half that in part (a), is the average force exerted on the fireman by the pole doubled? Explain. (c) Find the average force exerted on the fireman by the pole when the landing speed is 2.1 m/s.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion74gps

Chapter 5 Newton’s Laws Of Motion Q.75GP
For a birthday gift, you and some friends take a hot-air balloon ride. One friend is late, so the balloon floats a couple of feet off the ground as you wait. Before this person arrives, the combined weight of the basket and people is 1220 kg, and the balloon is neutrally buoyant. When the late arrival climbs up into the basket, the balloon begins to accelerate downward at 0.56 m/s2. What was the mass of the last person to climb aboard?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion75gps

Chapter 5 Newton’s Laws Of Motion Q.76GP
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Solution:

Chapter 5 Newton’s Laws Of Motion Q.77GP
When two people push in the same direction on an object of mass m they cause an acceleration of magnitude a1. When the same people push in opposite directions, the acceleration of the object has a magnitude a2. Determine the magnitude of the force exerted by each of the two people in terms of m, a1 and a2.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion77gps

Chapter 5 Newton’s Laws Of Motion Q.78GP
· · · An air-track cart of mass m1 = 0.14 kg is moving with a speed v0 = 1.3 m/s to the right when it collides with a cart of mass m2 = 0.25 kg that is at rest. Each cart has a wad of putty on its bumper, and hence they stick together as a result of their collision. Suppose the average contact force between the carts is F = 1.5 N during the collision. (a) What is the acceleration of cart 1? Give direction and magnitude. (b) What is the acceleration of cart 2? Give direction and magnitude. (c) How long does it take for both carts to have the same speed? (Once the carts have the same speed the collision is over and the contact force vanishes.) (d) What is the final speed of the carts, vf? (e) Show that m1v0 is equal to (m1 + m2)vf. (We shall investigate the significance of this result in Chapter 9.)
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion78gps

Chapter 5 Newton’s Laws Of Motion Q.79PP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion79gp
Solution:

Chapter 5 Newton’s Laws Of Motion Q.80PP
A driver who does not wear a seatbelt continues to move forward with a speed of 18.0 m/s (due to inertia) until something solid like the steering wheel is encountered. The driver now comes to rest in a much shorter distance—perhaps only a few centimeters. Find the magnitude of the net force acting on a 65.0-kg driver who is decelerated from 18.0 m/s to rest in 5.00 cm.
A. 3240 N
B. 1.17 × 104 N
C. 2.11 × 105 N
D. 4.21 × 105 N
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion80pps

Chapter 5 Newton’s Laws Of Motion Q.81PP
Suppose the initial speed of the driver is doubled to 36.0 m/s. If the driver still has a mass of 65.0 kg, and comes to rest in 1.00 m, what is the magnitude of the force exerted on the driver during this collision?
A. 648 N
B. 1170 N
C. 2.11 × 104 N
D. 4.21 × 104 N
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion81pps

Chapter 5 Newton’s Laws Of Motion Q.82PP
If both the speed and stopping distance of a driver are doubled, by what factor does the force exerted on the driver change?
A. 0.5
B. 1
C. 2
D. 4
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion82pps

Chapter 5 Newton’s Laws Of Motion Q.83IP
IP Referring to Example 5-4 Suppose that we would like the contact force between the boxes to have a magnitude of 5.00 N, and that the only thing in the system we are allowed to change is the mass of box 2—the mass of box 1 is 10.0 kg and the applied force is 20.0 N. (a) Should the mass of box 2 be increased or decreased? Explain. (b) Find the mass of box 2 that results in a contact force of magnitude 5.00 N. (c) What is the acceleration of the boxes in this case?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion83ips

Chapter 5 Newton’s Laws Of Motion Q.84IP
Referring to Example 5-4 Suppose the force of 20.0 N pushes on two boxes of unknown mass. We know, however, that the acceleration of the boxes is 1.20 m/s2 and the contact force has a magnitude of 4.45 N. Find the mass of (a) box 1 and (b) box 2.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion84ips

Chapter 5 Newton’s Laws Of Motion Q.85IP
IP Referring to Figure 5-9 Suppose the magnitude of is increased from 41 N to 55 N, and that everything else in the system remains the same. (a) Do you expect the direction of the satellite’s acceleration to be greater than, less than, or equal to 32°? Explain. Find (b) the direction and (c) the magnitude of the satellite’s acceleration in this case.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion85ips

Chapter 5 Newton’s Laws Of Motion Q.86IP
IP Referring to Figure 5-9 Suppose we would like the acceleration of the satellite to be at an angle of 25°, and that the only quantity we can change in the system is the magnitude of . (a) Should the magnitude of be increased or decreased? Explain. (b) What is the magnitude of the satellite’s acceleration in this case?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion86ips