Content check — College Physics, Chapter 8: Linear Momentum and Collisions

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Chapter 8: Linear Momentum and Collisions

32 problems · 0 figures

(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of

7 . 50 m/s

. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of

600 m/s

. (c) What is the momentum of the 90.0-kg hunter running at

7 . 40 m/s

after missing the elephant?

Answer

(a) $1.50 \times 10^{4} kg \cdot m/s$

(b) 625 to 1

8-3 §8.1 source ↗

(a) At what speed would a

2 . 00 \times 10^{4} -kg

airplane have to fly to have a momentum of

1 . 60 \times 10^{9} kg \cdot m/s

(the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of

60.0 m/s

? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship.

Answer

(a) $8 . 00 \times 10^{4} m/s$

(b) $1 . 20 \times 10^{6} kg \cdot m/s$

(c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be $- 0 . 0100 m/s$ , which is probably not noticeable.

8-5 §8.1 source ↗

A runaway train car that has a mass of 15,000 kg travels at a speed of

5 .4 m/s

down a track. Compute the time required for a force of 1500 N to bring the car to rest.

Answer

54 s

8-7 §8.2 source ↗

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600.0 m/s in a time of 2.00 ms (milliseconds)?

Answer

9 . 00 \times 10^{3} N

8-9 §8.2 source ↗

A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s. (a) What is the average force exerted on the leg, taking the effective mass of the hand and forearm to be 1.50 kg? (b) Would the force be any different if the woman clapped her hands together at the same speed and brought them to rest in the same time? Explain why or why not.

Answer

a) $2 . 40 \times 10^{3} N$ toward the leg

b) The force on each hand would have the same magnitude as that found in part (a) (but in opposite directions by Newton’s third law) because the change in momentum and the time interval are the same.

8-11 §8.2 source ↗

Professional Application

Suppose a child drives a bumper car head on into the side rail, which exerts a force of 4000 N on the car for 0.200 s. (a) What impulse is imparted by this force? (b) Find the final velocity of the bumper car if its initial velocity was 2.80 m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor.

Answer

a) $800 kg \cdot m/s$ away from the wall

b) $1 . 20 m/s$ away from the wall

8-13 §8.2 source ↗

Professional Application

A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) Calculate the average force on the person if he is stopped by a padded dashboard that compresses an average of 1.00 cm. (b) Calculate the average force on the person if he is stopped by an air bag that compresses an average of 15.0 cm.

Answer

(a) $1 . 50 \times 10^{6} N$ away from the dashboard

(b) $1 . 00 \times 10^{5} N$ away from the dashboard

8-15 §8.2 source ↗

A cruise ship with a mass of

1 . 00 \times 10^{7} kg

strikes a pier at a speed of 0.750 m/s. It comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain’s finances. Calculate the average force exerted on the pier using the concept of impulse. (Hint: First calculate the time it took to bring the ship to rest.)

Answer

4 . 69 \times 10^{5} N

in the boat’s original direction of motion

8-17 §8.2 source ↗

Water from a fire hose is directed horizontally against a wall at a rate of 50.0 kg/s and a speed of 42.0 m/s. Calculate the magnitude of the force exerted on the wall, assuming the water’s horizontal momentum is reduced to zero.

Answer

2 . 10 \times 10^{3} N

away from the wall

8-19 §8.2 source ↗

Starting with the definitions of momentum and kinetic energy, derive an equation for the kinetic energy of a particle expressed as a function of its momentum.

Answer

\begin{array}{l} p = m v \Rightarrow p^{2} = m^{2} v^{2} \Rightarrow \frac{p^{2}}{m} = {mv}^{2} \\ \Rightarrow \frac{p^{2}}{2 m} = \frac{1}{2} {mv}^{2} = KE \\ KE = \frac{p^{2}}{2 m} \end{array}

8-21 §8.2 source ↗

When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racquet exerts a force of 540 N on the ball for 5.00 ms, giving it a final velocity of 45.0 m/s. Using these data, find the mass of the ball.

Answer

60.0 g

8-23 §8.3 source ↗

Professional Application

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second having a mass of 110,000 kg and a velocity of $- 0 . 120 m/s$ . (The minus indicates direction of motion.) What is their final velocity?

Answer

0.122 m/s

8-25 §8.3 source ↗

Professional Application

Consider the following question: A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seatbelt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70 kg. Would the answer to this question be different if the car with the 70-kg passenger had collided with a car that has a mass equal to and is traveling in the opposite direction and at the same speed? Explain your answer.

Answer

In a collision with an identical car, momentum is conserved. Afterwards

v_{f} = 0

for both cars. The change in momentum will be the same as in the crash with the tree. However, the force on the body is not determined since the time is not known. A padded stop will reduce injurious force on body.

8-27 §8.3 source ↗

A 1.80-kg falcon catches a 0.650-kg dove from behind in midair. What is their velocity after impact if the falcon’s velocity is initially 28.0 m/s and the dove’s velocity is 7.00 m/s in the same direction?

Answer

22.4 m/s in the same direction as the original motion

8-29 §8.4 source ↗

Professional Application

Two piloted satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of $4 . 00 \times 10^{3} kg$ , and the second a mass of $7 . 50 \times 10^{3} kg$ . If the two satellites collide elastically rather than dock, what is their final relative velocity?

Answer

0.250 m/s

8-31 §8.5 source ↗

A 0.240-kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts 0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How much kinetic energy in joules is lost during the collision? (c) What percent of the original energy is left?

Answer

(a) 86.4 N perpendicularly away from the bumper

(b) 0.389 J

8-33 §8.5 source ↗

Professional Application

Using mass and speed data from Example 8.1 and assuming that the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: (a) the final velocity if the ball and player are going in the same direction and (b) the loss of kinetic energy in this case. (c) Repeat parts (a) and (b) for the situation in which the ball and the player are going in opposite directions. Might the loss of kinetic energy be related to how much it hurts to catch the pass?

Example 8.1 View it in the book ↗

Answer

(a) 8.06 m/s

(b) -56.0 J

(c)(i) 7.88 m/s; (ii) -223 J

8-35 §8.5 source ↗

Professional Application

Two piloted satellites approaching one another, at a relative speed of 0.250 m/s, intending to dock. The first has a mass of $4 . 00 \times 10^{3} kg$ , and the second a mass of $7.50 \times 10^{3} kg$ . (a) Calculate the final velocity (after docking) by using the frame of reference in which the first satellite was originally at rest. (b) What is the loss of kinetic energy in this inelastic collision? (c) Repeat both parts by using the frame of reference in which the second satellite was originally at rest. Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.

Answer

(a) 0.163 m/s in the direction of motion of the more massive satellite

(b) 81.6 J

(c) $8 . 70 \times 10^{- 2} m/s$ in the direction of motion of the less massive satellite, 81.5 J. Because there are no external forces, the velocity of the center of mass of the two-satellite system is unchanged by the collision. The two velocities calculated above are the velocity of the center of mass in each of the two different individual reference frames. The loss in KE is the same in both reference frames because the KE lost to internal forces (heat, friction, etc.) is the same regardless of the coordinate system chosen.

8-37 §8.5 source ↗

Professional Application

Space probes may be separated from their launchers by exploding bolts. (They bolt away from one another.) Suppose a 4800-kg satellite uses this method to separate from the 1500-kg remains of its launcher, and that 5000 J of kinetic energy is supplied to the two parts. What are their subsequent velocities using the frame of reference in which they were at rest before separation?

Answer

0.704 m/s

–2.25 m/s

8-38 §8.5 source ↗

A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. (a) Calculate the recoil velocity of the rifle if it is held loosely away from the shoulder. (b) How much kinetic energy does the rifle gain? (c) What is the recoil velocity if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg? (d) How much kinetic energy is transferred to the rifle-shoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation. (e) Calculate the momentum of a 110-kg football player running at 8.00 m/s. Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Discuss its relationship to this problem.

Answer

(a) 4.58 m/s away from the bullet

(b) 31.5 J

(d) 3.38 J

8-40 §8.5 source ↗

Professional Application

The Moon’s craters are remnants of meteorite collisions. Suppose a fairly large asteroid that has a mass of $5 . 00 \times 10^{12} kg$ (about a kilometer across) strikes the Moon at a speed of 15.0 km/s. (a) At what speed does the Moon recoil after the perfectly inelastic collision (the mass of the Moon is $7 . 36 \times 10^{22} kg$ ) ? (b) How much kinetic energy is lost in the collision? Such an event may have been observed by medieval English monks who reported observing a red glow and subsequent haze about the Moon. (c) In October 2009, NASA crashed a rocket into the Moon, and analyzed the plume produced by the impact. (Significant amounts of water were detected.) Answer part (a) and (b) for this real-life experiment. The mass of the rocket was 2000 kg and its speed upon impact was 9000 km/h. How does the plume produced alter these results?

Answer

(a) $1 . 02 \times 10^{- 6} m/s$

(b) $5 . 63 \times 10^{20} J$ (almost all KE lost)

(c) Recoil speed is $6 . 79 \times 10^{- 17} m/s$ , energy lost is $6 . 25 \times 10^{9} J$ . The plume will not affect the momentum result because the plume is still part of the Moon system. The plume may affect the kinetic energy result because a significant part of the initial kinetic energy may be transferred to the kinetic energy of the plume particles.

8-42 §8.5 source ↗

What is the speed of a garbage truck that is

1 . 20 \times 10^{4} kg

and is initially moving at 25.0 m/s just after it hits and adheres to a trash can that is 80.0 kg and is initially at rest?

Answer

24.8 m/s

8-44 §8.5 source ↗

(a) During an ice skating performance, an initially motionless 80.0-kg clown throws a fake barbell away. The clown’s ice skates allow her to recoil frictionlessly. If the clown recoils with a velocity of 0.500 m/s and the barbell is thrown with a velocity of 10.0 m/s, what is the mass of the barbell? (b) How much kinetic energy is gained by this maneuver? (c) Where does the kinetic energy come from?

Answer

(a) 4.00 kg

(b) 210 J

(c) The clown does work to throw the barbell, so the kinetic energy comes from the muscles of the clown. The muscles convert the chemical potential energy of ATP into kinetic energy.

8-45 §8.6 source ↗

Two identical pucks collide on an air hockey table. One puck was originally at rest. (a) If the incoming puck has a speed of 6.00 m/s and scatters to an angle of

30 . 0º

,what is the velocity (magnitude and direction) of the second puck? (You may use the result that

θ_{1} - θ_{2} = 90º

for elastic collisions of objects that have identical masses.) (b) Confirm that the collision is elastic.

Answer

(a) 3.00 m/s, $60º$ below $x$ -axis

(b) Find speed of first puck after collision: $0 =$ $m v_{1}^{'} sin 30º -$ $m v_{2}^{'} sin 60º \Rightarrow v_{1}^{'} =$ $v_{2}^{'} \frac{sin 60º}{sin 30º} =$ $5.196 m/s$

Verify that ratio of initial to final KE equals one: $\begin{array}{l} KE = \frac{1}{2} {mv}_{1}^{2} = 18 m J \\ KE = \frac{1}{2} {mv'}_{1}^{2} + \frac{1}{2} {mv'}_{2}^{2} = 18 m J \end{array} \frac{KE}{KE′} = 1.00$

8-47 §8.6 source ↗

A 3000-kg cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity when it fires a 15.0-kg shell at 480 m/s at an angle of

20 . 0º

above the horizontal. (b) What is the kinetic energy of the cannon? This energy is dissipated as heat transfer in shock absorbers that stop its recoil. (c) What happens to the vertical component of momentum that is imparted to the cannon when it is fired?

Answer

(a) $- 2 . 26 m/s$

(b) $7 . 63 \times 10^{3} J$

(c) The ground will exert a normal force to oppose recoil of the cannon in the vertical direction. The momentum in the vertical direction is transferred to the earth. The energy is transferred into the ground, making a dent where the cannon is. After long barrages, cannon have erratic aim because the ground is full of divots.

8-49 §8.6 source ↗

Professional Application

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei $(^{4} He)$ from gold-197 nuclei $(^{197} Au)$ . The energy of the incoming helium nucleus was $8.00 \times 10^{- 13} J$ , and the masses of the helium and gold nuclei were $6.68 \times 10^{- 27} kg$ and $3.29 \times 10^{- 25} kg$ , respectively (note that their mass ratio is 4 to 197). (a) If a helium nucleus scatters to an angle of $120º$ during an elastic collision with a gold nucleus, calculate the helium nucleus’s final speed and the final velocity (magnitude and direction) of the gold nucleus. (b) What is the final kinetic energy of the helium nucleus?

Answer

(a) $5 . 36 \times 10^{5} m/s$ at $- 29.5º$

(b) $7 . 52 \times 10^{- 13} J$

8-51 §8.6 source ↗

Starting with equations $m_{1} v_{1} =$ $m_{1} v_{1}^{'} cos θ_{1} +$ $m_{2} v_{2}^{'} cos θ_{2}$ and $0 =$ $m_{1} v_{1}^{'} sin θ_{1} +$ $m_{2} v_{2}^{'} sin θ_{2}$ for conservation of momentum in the $x$ - and $y$ -directions and assuming that one object is originally stationary, prove that for an elastic collision of two objects of equal masses,

\begin{matrix} \frac{1}{2} {mv}_{1}^{2} = \\ \frac{1}{2} {mv'}_{1}^{2} + \frac{1}{2} {mv'}_{2}^{2} \\ + {mv'}_{1} {v'}_{2} cos (θ_{1} - θ_{2}) \end{matrix}

as discussed in the text.

Answer

We are given that $m_{1} = m_{2} \equiv m$ . The given equations then become:

v_{1} = v_{1} cos θ_{1} + v_{2} cos θ_{2}

and

0 = v_{1}^{'} sin θ_{1} + v_{2}^{'} sin θ_{2} .

Square each equation to get

\begin{array}{l} {v_{1}}^{2} & = & {v'}_{1}^{2} {cos}^{2} θ_{1} + {v'}_{2}^{2} {cos}^{2} θ_{2} + 2 {v'}_{1} {v'}_{2} cos θ_{1} cos θ_{2} \\ 0 & = & {v'}_{1}^{2} {sin}^{2} θ_{1} + {v'}_{2}^{2} {sin}^{2} θ_{2} + 2 {v'}_{1} {v'}_{2} sin θ_{1} sin θ_{2} . \end{array}

Add these two equations and simplify:

\begin{array}{l} v_{1}^{2} & = & {v'}_{1}^{2} + {v'}_{2}^{2} + 2 {v'}_{1} {v'}_{2} cos θ_{1} cos θ_{2} + sin θ_{1} sin θ_{2} \\ = & {v'}_{1}^{2} + {v'}_{2}^{2} + 2 {v'}_{1} {v'}_{2} \frac{1}{2} cos θ_{1} - θ_{2} + \frac{1}{2} cos θ_{1} + θ_{2} + \frac{1}{2} cos θ_{1} - θ_{2} - \frac{1}{2} cos θ_{1} + θ_{2} \\ = & {v'}_{1}^{2} + {v'}_{2}^{2} + 2 {v'}_{1} {v'}_{2} cos θ_{1} - θ_{2} . \end{array}

Multiply the entire equation by $\frac{1}{2} m$ to recover the kinetic energy:

\frac{1}{2} {mv}_{1}^{2} = \frac{1}{2} m {v'}_{1}^{2} + \frac{1}{2} m {v'}_{2}^{2} + m {v'}_{1} {v'}_{2} cos θ_{1} - θ_{2}

8-53 §8.7 source ↗

Professional Application

Antiballistic missiles (ABMs) are designed to have very large accelerations so that they may intercept fast-moving incoming missiles in the short time available. What is the takeoff acceleration of a 10,000-kg ABM that expels 196 kg of gas per second at an exhaust velocity of $2 . 50 \times 10^{3} m/s?$

Answer

39.2 {m/s}^{2}

8-55 §8.7 source ↗

Professional Application

Calculate the increase in velocity of a 4000-kg space probe that expels 3500 kg of its mass at an exhaust velocity of $2 . 00 \times 10^{3} m/s$ . You may assume the gravitational force is negligible at the probe’s location.

Answer

4 . 16 \times 10^{3} m/s

8-57 §8.7 source ↗

Derive the equation for the vertical acceleration of a rocket.

Answer

The force needed to give a small mass

Δ m

an acceleration

a_{Δ m}

F = Δ {ma}_{Δ m}

. To accelerate this mass in the small time interval

Δ t

requires

v_{e} = a_{Δ m} Δ t

, so

F = v_{e} \frac{Δ m}{Δ t}

. By Newton’s third law, this force is equal in magnitude to the thrust force acting on the rocket, so

F_{thrust} =

v_{e} \frac{Δ m}{Δ t}

, where all quantities are positive. Applying Newton’s second law to the rocket gives

F_{thrust} -

mg =

ma \Rightarrow a =

\frac{v_{e}}{m} \frac{Δ m}{Δ t} -

g

, where

m

is the mass of the rocket and unburnt fuel.

8-60 §8.7 source ↗

How much of a single-stage rocket that is 100,000 kg can be anything but fuel if the rocket is to have a final speed of

8.00 km/s

, given that it expels gases at an exhaust velocity of

2 . 20 \times 10^{3} m/s?

Answer

2 . 63 \times 10^{3} kg

8-61 §8.7 source ↗

Professional Application

(a) A 5.00-kg squid initially at rest ejects 0.250-kg of fluid with a velocity of 10.0 m/s. What is the recoil velocity of the squid if the ejection is done in 0.100 s and there is a 5.00-N frictional force opposing the squid’s movement. (b) How much energy is lost to work done against friction?

Answer

(a) 0.421 m/s away from the ejected fluid.

(b) $0 .237 J$ .

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Problems and answers are adapted from College Physics 2e by Paul Peter Urone, Roger Hinrichs, Kim Dirks and Manjula Sharma, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.