Homework #5: Momentum

Homework #5: Momentum

1.(I) What is the magnitude of the momentum of a 28-g sparrow flying with a speed of

2.(I) A constant friction force of 25 N acts on a 65-kg skier for 20 s. What is the skier’s change in velocity?

3.(II) A 0.145-kg baseball pitched at is hit on a horizontal line drive straight back toward the pitcher at If the contact time between bat and ball is calculate the average force between the ball and bat during contact.

4.(II) A child in a boat throws a 6.40-kg package out horizontally with a speed of . Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is 26.0 kg, and that of the boat is 45.0 kg. Ignore water resistance.

5.(II) A 12,600-kg railroad car travels alone on a level frictionless track with a constant speed of A 5350-kg load, initially at rest, is dropped onto the car. What will be the car’s new speed?

6.(II) A 9300-kg boxcar traveling at strikes a second boxcar at rest. The two stick together and move off with a speed of What is the mass of the second car?

7.(II) An atomic nucleus initially moving at emits an alpha particle in the direction of its velocity, and the remaining nucleus slows to If the alpha particle has a mass of 4.0 u and the original nucleus has a mass of 222 u, what speed does the alpha particle have when it is emitted?

8.(II) A golf ball of mass 0.045 kg is hit off the tee at a speed of The golf club was in contact with the ball for Find (a) the impulse imparted to the golf ball, and (b) the average force exerted on the ball by the golf club.

9.A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. What impulse was given to the ball by the floor?

10.A pitcher claims he can throw a 0.145-kg baseball with as much momentum as a 3.00-g bullet moving with a speed of 1.50 × 103 m/s. (a) What must the baseball’s speed be if the pitcher’s claim is valid? (b) Which has greater kinetic energy, the ball or the bullet?

11.A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.20 m/s in 0.832 s. What are the magnitudes of the linear impulse and the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates?

12.The front 1.20 m of a 1 400-kg car is designed as a “crumple zone” that collapses to absorb the shock of a collision. If a car traveling 25.0 m/s stops uniformly in 1.20 m, (a) how long does the collision last, (b) what is the magnitude of the average force on the car, and (c) what is the acceleration of the car? Express the acceleration as a multiple of the acceleration of gravity.

13.A 730-N man stands in the middle of a frozen pond of radius 5.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2-kg physics textbook horizontally toward the north shore at a speed of 5.0 m/s. How long does it take him to reach the south shore?

14.High-speed stroboscopic photographs show that the head of a 200-g golf club is traveling at 55 m/s just before it strikes a 46-g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.

15.A railroad car of mass 2.00 × 104 kg moving at 3.00 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s. (a) What is the speed of the three coupled cars after the collision? (b) How much kinetic energy is lost in the collision?

16.A 7.0-g bullet is fired into a 1.5-kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 12 cm. Find the initial speed of the bullet.

17.A 0.030-kg bullet is fired vertically at 200 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball?

Homework #6: Conservation of Momentum: Elastic Collisions

18.(II) A ball of mass 0.440 kg moving east ( direction) with a speed of collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

19.(II) Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If one ball’s initial speed was and the other’s was in the opposite direction, what will be their speeds after the collision?

20.(II) A 0.060-kg tennis ball, moving with a speed of collides head-on with a 0.090-kg ball initially moving away from it at a speed of Assuming a perfectly elastic collision, what are the speed and direction of each ball after the collision?

21.(II) A softball of mass 0.220 kg that is moving with a speed of collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of Calculate (a) the velocity of the target ball after the collision, and (b) the mass of the target ball.

22.(II) A 28-g rifle bullet traveling buries itself in a 3.6-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Determine the vertical and horizontal components of the pendulum’s displacement.

23.(II) A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.80, calculates the speed of the sports car at impact. What was that speed?

24.A 5.00-g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0-g object that is initially at rest. Find (a) the velocity of each object after the collision and (b) the fraction of the initial kinetic energy transferred to the 10.0-g object.

25.A 10.0-g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 15.0-g object moving in the opposite direction at 30.0 cm/s. Find the velocity of each object after the collision.

26.A 25.0-g object moving to the right at 20.0 cm/s overtakes and collides elastically with a 10.0-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision.

27.(II) An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other. If 7500 J were released in the explosion, how much kinetic energy did each piece acquire?

28.(II) A wooden block is cut into two pieces, one with three times the mass of the other. A depression is made in both faces of the cut, so that a firecracker can be placed in it with the block reassembled. The reassembled block is set on a rough-surfaced table, and the fuse is lit. When the firecracker explodes, the two blocks separate and slide apart. What is the ratio of distances each block travels?

Solutions #5 Momentum:

1.

2.From Newton’s second law, . For a constant mass object, . Equate the two expressions for .

.

If the skier moves to the right, then the speed will decrease, because the friction force is to the left.

The skier loses of speed.

3.Choose the direction from the batter to the pitcher to be the positive direction. Calculate the average force from the change in momentum of the ball.

4.The throwing of the package is a momentum-conserving action, if the water resistance is ignored. Let “A” represent the boat and child together, and let “B” represent the package. Choose the direction that the package is thrown as the positive direction. Apply conservation of momentum, with the initial velocity of both objects being 0.

The boat and child move in the opposite direction as the thrown package.

5.Consider the horizontal motion of the objects. The momentum in the horizontal direction will be conserved. Let “A” represent the car, and “B” represent the load. The positive direction is the direction of the original motion of the car.

6.Consider the motion in one dimension, with the positive direction being the direction of motion of the first car. Let “A” represent the first car, and “B” represent the second car. Momentum will be conserved in the collision. Note that .

7.Consider the motion in one dimension, with the positive direction being the direction of motion of the original nucleus. Let “A” represent the alpha particle, with a mass of 4 u, and “B” represent the new nucleus, with a mass of 218 u. Momentum conservation gives the following.

Note that the masses do not have to be converted to kg, since all masses are in the same units, and a ratio of masses is what is significant.

8.(a)The impulse is the change in momentum. The direction of travel of the struck ball is the positive direction.

(b)The average force is the impulse divided by the interaction time.

9.The velocity of the ball just before impact is found from as

and the rebound velocity with which it leaves the floor is

The impulse given the ball by the floor is then

10.(a)If ,
then

(b)The kinetic energy of the bullet is

while that of the baseball is
The by a factor of 48.4.

11.
Thus, , and

or

12.(a)

(b)

(c)

13.We shall choose southward as the positive direction.
The mass of the man is . Then, from conservation of momentum, we find
or
and
Therefore, the time required to travel the 5.0 m to shore is

14.Requiring that total momentum be conserved gives

or
and

15.(a)If M is the mass of a single car, conservation of momentum gives
, or

(b)The kinetic energy lost is , or

With , this yields

16.Let us apply conservation of energy to the block from the time just after the bullet has passed through until it reaches maximum height in order to find its speed V just after the collision.
becomes
or
Now use conservation of momentum from before until just after the collision in order to find the initial speed of the bullet, v.

from which

17.Let M = mass of ball, m = mass of bullet, v = velocity of bullet, and V = the initial velocity of the ball-bullet combination. Then, using conservation of momentum from just before to just after collision gives

Now, we use conservation of mechanical energy from just after the collision until the ball reaches maximum height to find

With the data values provided, this becomes

Homework #6 Solutions

18.Let A represent the 0.440-kg ball, and B represent the 0.220-kg ball. We have and . Use Eq. 7-7 to obtain a relationship between the velocities.

Substitute this relationship into the momentum conservation equation for the collision.

19.Let A represent the ball moving at 2.00 m/s, and call that direction the positive direction. Let B represent the ball moving at 3.00 m/s in the opposite direction. So and . Use Eq. 7-7 to obtain a relationship between the velocities.

Substitute this relationship into the momentum conservation equation for the collision, noting that .

The two balls have exchanged velocities. This will always be true for 1-D elastic collisions of objects of equal mass.

20.Let A represent the 0.060-kg tennis ball, and let B represent the 0.090-kg ball. The initial direction of the balls is the positive direction. We have and . Use Eq. 7-7 to obtain a relationship between the velocities.

Substitute this relationship into the momentum conservation equation for the collision.

Both balls move in the direction of the tennis ball’s initial motion.

21.Let A represent the moving softball, and let B represent the ball initially at rest. The initial direction of the softball is the positive direction. We have , , and .

(a)Use Eq. 7-7 to obtain a relationship between the velocities.

(b)Use momentum conservation to solve for the mass of the target ball.

22.From the analysis in the Example 7-10, we know that

From the diagram we see that

23Use conservation of momentum in one dimension. Call the direction of the sports car’s velocity the positive x direction. Let A represent the sports car, and B represent the SUV. We have and . Solve for .

The kinetic energy that the cars have immediately after the collision is lost due to negative work done by friction. The work done by friction can also be calculated using the definition of work. We assume the cars are on a level surface, so that the normal force is equal to the weight. The distance the cars slide forward is . Equate the two expressions for the work done by friction, solve for , and use that to find .

24(a)From conservation of momentum,
(1)
Also for an elastic, head-on, collision, we have , which becomes .(2)
Solving (1) and (2) simultaneously yields
, and

(b)
, so

25.Using conservation of momentum gives
(1)
For elastic, head on collisions, which becomes
(2)

Solving (1) and (2) simultaneously gives ,
and

26.Conservation of momentum gives
(1)
For head-on, elastic collisions, we know that .
Thus,(2)
Solving (1) and (2) simultaneously yields
, and

27.Use conservation of momentum in one dimension, since the particles will separate and travel in opposite directions. Call the direction of the heavier particle’s motion the positive direction. Let A represent the heavier particle, and B represent the lighter particle. We have , and

.

The negative sign indicates direction.

Since there was no mechanical energy before the explosion, the kinetic energy of the particles after the explosion must equal the energy added.

Thus

28.Let A represent the more massive piece, and B the less massive piece. Thus . In the explosion, momentum is conserved. We have .

For each block, the kinetic energy gained during the explosion is lost to negative work done by friction on the block.

But work is also calculated in terms of the force doing the work and the distance traveled.

Equate the two work expressions, solve for the distance traveled, and find the ratio of distances.

And so