Question 1

A 90-kg halfback running north with a speed of 10 m/s is tackled by a 120-kg opponent running south at 5.0 m/s.The collision is perfectly inelastic.Compute the velocity of the two players just after the tackle.
A) 1.4 m/s south
B) 2.8 m/s south
C) 2.8 m/s north
D) 1.4 m/s north

Accepted Answer

In a perfectly inelastic collision, momentum is conserved. The total momentum before the collision is (90 kg * 10 m/s) - (120 kg * 5 m/s) = 900 kg*m/s - 600 kg*m/s = 300 kg*m/s north. The combined mass after the collision is 90 kg + 120 kg = 210 kg. The velocity is 300 kg*m/s / 210 kg = 1.4 m/s north.

Question 2

A 6-kg object is moving to the right at 4 m/s and collides with a 4-kg object moving to the left at 6 m/s.The objects collide and stick together.After the collision,the combined object:
A) has the same kinetic energy that the system had before the collision.
B) has more kinetic energy than the system had before the collision.
C) has no kinetic energy.
D) has less momentum than the system had before the collision.

Accepted Answer

In an inelastic collision like this, the objects stick together and move as one. Kinetic energy is not conserved in such a collision, therefore the combined object has no kinetic energy after the collision. The momentum of the system, however, is conserved.

Question 3

A 20-g bullet moving at 1000 m/s is fired through a one-kg block of wood emerging at a speed of 200 m/s.What is the change in the kinetic energy of the bullet-block system as a result of the collision assuming the block is free to move?
A) 0 J
B) −9.5 kJ
C) 9.5 kJ
D) −18 J

Accepted Answer

The initial kinetic energy of the bullet is 10,000 J, and the final kinetic energy is 400 J. The block's final kinetic energy is 800 J. The change in kinetic energy is (400 + 800) - 10,000 = -9,500 J, or -9.5 kJ.

Question 4

A miniature spring-loaded,radio-controlled gun is mounted on an air puck.The gun's bullet has a mass of 5.00 g,and the gun and puck have a combined mass of 130g.With the system initially at rest,the radio controlled trigger releases the bullet causing the puck and empty gun to move with a speed of 0.500 m/s.What is the bullet's speed?
A) 4.80 m/s
B) 13.0 m/s
C) 48.0 m/s
D) 12.5 m/s

Accepted Answer

We can use the conservation of momentum to solve this problem. Initially, the momentum of the system is zero because it is at rest. After the gun is fired, the momentum of the system is:

(m_bullet)(v_bullet) + (m_system)(v_system) = 0

where v_bullet is the speed of the bullet, and v_system is the speed of the puck and empty gun. We can solve for v_bullet:

v_bullet = -(m_system/m_bullet) v_system 
       = -(130 g / 5.00 g) (0.500 m/s)
       = -13.0 m/s

The negative sign indicates that the bullet is moving in the opposite direction of the puck and gun. Taking the absolute value gives:

|v_bullet| = 13.0 m/s

Therefore, the bullet's speed is 13.0 m/s, which corresponds to choice B.

Question 5

A railroad freight car,mass 15,000 kg,is allowed to coast along a level track at a speed of 2.0 m/s.It collides and couples with a 60,000-kg loaded second car,initially at rest and with brakes released.What percentage of the initial kinetic energy of the 15,000-kg car is preserved in the two-coupled cars after collision?
A) 20%
B) 23%
C) 86%
D) 100%

Accepted Answer

The answer of A railroad freight car,mass 15,000 kg,is allowed...

Question 6

A machine gun is attached to a railroad flatcar that rolls with negligible friction.If the railroad car has a mass of 6.25 × 10⁴ kg,how many bullets of mass 25 g would have to be fired at 250 m/s off the back to give the railroad car a forward velocity of 0.3 m/s? A) 400 B) 2000 C) 3000 D) 5000

Accepted Answer

Using the principle of conservation of momentum, we can solve for the number of bullets:
m_car * V_car = m_bullet * V_bullet
where m_car and V_car represent the mass and velocity of the railroad car respectively, and m_bullet and V_bullet represent the mass and velocity of each bullet. 
Rearranging the equation, we get:
m_bullet = m_car * V_car / V_bullet
Substituting the given values, we get:
m_bullet = (6.25 × 10¹⁴ kg) * (0.3 m/s) / (0.025 kg * 250 m/s) 
m_bullet = 3,000
Therefore, 3,000 bullets of mass 25 g would have to be fired at 250 m/s off the back to give the railroad car a forward velocity of 0.3 m/s. 
Hence, the answer is C.

Question 7

A billiard ball (Ball #1)moving at 5.00 m/s strikes a stationary ball (Ball #2)of the same mass.After the elastic collision,Ball #1 moves at a speed of 3.50 m/s.Find the speed of Ball #2 after the collision.
A) 1.25 m/s
B) 1.50 m/s
C) 2.16 m/s
D) 3.57 m/s

Accepted Answer

In an elastic collision with equal masses, the relative speed of separation equals the relative speed of approach. Therefore, Ball #2 moves at 3.57 m/s after the collision.

Question 8

During a snowball fight,two balls with masses of 0.4 and 0.6 kg,respectively,are thrown in such a manner that they meet head-on and combine to form a single mass.The magnitude of initial velocity for each is 20 m/s.What is the speed of the 1.0-kg mass immediately after collision?
A) zero
B) 3 m/s
C) 4 m/s
D) 5 m/s

Accepted Answer

The answer of During a snowball fight,two balls with masses...

Question 9

Popeye,of mass 70 kg,has just downed a can of spinach.He accelerates quickly and stops Bluto,of mass 700 kg (Bluto is very dense),who is charging in at 5.0 m/s.What was Popeye's remarkable speed?
A) 10 m/s
B) 31 m/s
C) 50 m/s
D) 100 m/s

Accepted Answer

Using the conservation of momentum, the initial momentum of Bluto (700 kg * 5 m/s = 3500 kg*m/s) must equal the momentum of Popeye (70 kg * v) since he stops Bluto. Solving for v gives v = 3500 kg*m/s / 70 kg = 50 m/s.

Question 10

A 1000-kg vehicle is moving in the positive y-direction along the y-axis at 10 m/s.A 2000-kg vehicle is moving in the positive x-direction at along the x-axis at 20 m/s.The vehicles collide and stick together.They move as a single object in a direction ​θ as measured counterclockwise from the x-axis immediately after the collision.What is the value of tan ​θ?
A) 2.0
B) 4.0
C) 0.50
D) 0.25

Accepted Answer

Conservation of momentum in the x-direction:

2000(20) + 1000(0) = (2000+1000)(v)cosθ 
v = 13.3 m/s

Conservation of momentum in the y-direction:

1000(10) + 2000(0) = (2000+1000)(v)sinθ 
v = 3.3 m/s

tanθ = (3.3)/(13.3) ≈ 0.25

Therefore the answer is D) 0.25.

Question 11

A 1.00-kg duck is flying overhead at 1.50 m/s when a hunter fires straight up.The 0.0100-kg bullet is moving 200 m/s when it hits the duck and stays lodged in the duck's body.What is the speed of the duck and bullet immediately after the hit?
A) 1.49 m/s
B) 1.80 m/s
C) 2.48 m/s
D) 2.50 m/s

Accepted Answer

The answer of A 1.00-kg duck is flying overhead at...

Question 12

A 20-g bullet moving at 1000 m/s is fired through a one-kg block of wood emerging at a speed of 200 m/s.If the block had been originally at rest and is free to move,what is its resulting speed?
A) 800 m/s
B) 80 m/s
C) 16 m/s
D) 8 m/s

Accepted Answer

By conservation of momentum:
Initial momentum = final momentum 
(20 g) (1000 m/s) = (1 kg + 20 g) (v)
v = (20 g) (1000m/s) / (1 kg + 20 g) = 16.13 m/s
Therefore, the resulting speed of the block is approximately 16 m/s.

Question 13

In an inelastic collision between two objects with unequal masses:
A) the velocity of one will increase by the amount that the velocity of the other decreases.
B) the momentum of one will increase by the amount that the momentum of the other decreases.
C) the energy of one increases by the amount that the energy of the other decreases.
D) the total momentum of the system will decrease.

Accepted Answer

In an inelastic collision, the total momentum of the system is conserved, but not the kinetic energy. The kinetic energy is converted into other forms of energy, such as heat or sound. In a collision between two objects with unequal masses, the momentum of the lighter object will increase by the amount that the momentum of the heavier object decreases, because momentum is conserved. Therefore, choice B is correct. Choices A and C are incorrect because they refer to changes in velocity or energy, which are not conserved in an inelastic collision. Choice D is incorrect because the total momentum of the system is conserved, as stated above.

Question 14

Three masses are moving along the x-axis.Their masses and velocities are m₁ = m with v₁ = v,m₂ = 2m with v₂ = 2v,and m₃ = 3m with v₃ = -3v (the minus sign indicating this velocity is in the negative x-direction).The three masses collide and stick together in a completely inelastic collision.What is the resultant velocity of the combined mass? A) 0,since their combined momentum is zero. B) -2v/3 C) 14v/3 D) -v

Accepted Answer

Since the collision is completely inelastic, the total momentum before the collision must equal the total momentum after the collision.

Before the collision: 
total momentum = m*v + 2m*2v + 3m*(-3v) = -5mv

After the collision, the three masses combine to form a single mass of 6m. Let v' be the resultant velocity of this combined mass. Then: 
total momentum = (6m)(v')

Setting these two expressions equal to each other: 
-5mv = 6mv' 
v' = -5v/6

So the resultant velocity of the combined mass is -5v/6, which simplifies to -2v/3. Therefore, the best choice is B.

Question 15

A cannon of mass 1500 kg fires a 10-kg shell with a velocity of 200 m/s at an angle of 30° above the horizontal.Find the recoil velocity of the cannon across the level ground.
A) 1.15 m/s
B) 0.94 m/s
C) 2.41 m/s
D) 1.94 m/s

Accepted Answer

Momentum is conserved in the horizontal direction. The initial momentum of the system is zero. The final momentum of the system is the momentum of the shell. Therefore, the recoil velocity of the cannon is:$$v_r = \frac{m_sv_{sx}}{m_c}$$$$v_r = \frac{(10 	ext{ kg})(200 	ext{ m/s})\cos 30°}{1500 	ext{ kg}}$$$$v_r = 1.15 	ext{ m/s}$$

Question 16

A 70-kg man is standing in a 20-kg boat.Thinking he is stepping out onto the dock,the man steps to the right.However,the following will actually happen (ignore the friction of the water or air on the boat or the man):
A) The man only moves a short distance to the right while the boat moves a larger distance to the left.
B) The man actually stays still while the boat moves toward the left.
C) The boat doesn't move and the man moves to the right.
D) None of the above.

Accepted Answer

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. When the man steps to the right, he exerts a force on the boat toward the left. To maintain their center of mass, the boat also moves to the left, but since it has less mass, it moves a larger distance than the man. Therefore, the man only moves a short distance to the right while the boat moves a larger distance to the left.

Question 17

A miniature,spring-loaded,radio-controlled gun is mounted on an air puck.The gun's bullet has a mass of 5.00 g,and the gun and puck have a combined mass of 130g.With the system initially at rest,the radio-controlled trigger releases the bullet,causing the puck and empty gun to move with a speed of 0.500 m/s.Of the total kinetic energy of the gun-puck-bullet system,what percentage is in the bullet?
A) 50%
B) 96%
C) 3.4%
D) 100%

Accepted Answer

The total initial kinetic energy of the system is:
KE = (1/2)mv^2 = (1/2)(0.130 kg)(0.500 m/s)^2 = 0.01625 J
After the bullet is released, the puck and gun move with a slower speed, but since momentum is conserved in the absence of external forces, the total momentum is still zero and the bullet must have moved in the opposite direction with recoil speed v'.
Using conservation of momentum:
mv = m'v' + (m + m')V
where m' is the mass of the gun and puck, V is their common speed after the recoil, and v' is the speed of the bullet.
Solving for v', we get:
v' = (m/(m + m'))v = (0.005 kg/(0.135 kg))(0.500 m/s) = 0.0185 m/s
The kinetic energy of the bullet after recoil is therefore:
KE' = (1/2)mv'^2 = (1/2)(0.005 kg)(0.0185 m/s)^2 = 0.00000429 J
The percentage of the total kinetic energy that is in the bullet is:
(KE'/KE)(100%) = (0.00000429 J/0.01625 J)(100%) = 0.0264%
Therefore, the best choice is B) 96%.

Question 18

A baseball infielder,mass 75.0 kg,jumps up with velocity 2.00 m/s and catches a 0.150-kg baseball moving horizontally at 50.0 m/s.Of the following,which is closest to the final momentum of the system,infielder and baseball?
A) 158 kg
 m/s
B) 155 kg
 m/s
C) 152 kg
 m/s
D) 150 kg
 m/s

Accepted Answer

The answer of A baseball infielder,mass 75.0 kg,jumps up with...

Question 19

A 20-g bullet moving at 1000 m/s is fired through a one-kg block of wood emerging at a speed of 200 m/s.What is the kinetic energy of the block that results from the collision if the block had not been moving prior to the collision and was free to move?
A) 10 kJ
B) 9.8 kJ
C) 0.16 kJ
D) 0.13 kJ

Accepted Answer

The answer of A 20-g bullet moving at 1000 m/s...

Question 20

Two skaters,both of mass 75 kg,are initially at rest on skates on a frictionless ice pond.One skater throws a 0.3-kg ball at 10 m/s to his friend,who catches it and throws it back at 10 m/s.When the first skater has caught the returned ball,what is the velocity of each of the two skaters?
A) 0.08 m/s,moving apart
B) 0.04 m/s,moving apart
C) 0.08 m/s,moving towards each other
D) 0.04 m/s,moving towards each other

Accepted Answer

Explanation:Momentum is conserved, so the total momentum of the system (two skaters and ball) is zero before and after the ball is thrown. Initially, both skaters are at rest, so the total momentum is zero. After the first skater throws the ball, the ball has momentum of 0.3 kg * 10 m/s = 3 kg*m/s. To conserve momentum, the first skater must have equal and opposite momentum, so he will move in the opposite direction at a speed of 3 kg*m/s / 75 kg = 0.04 m/s. When the second skater catches the ball and throws it back, the ball again has momentum of 3 kg*m/s, but this time in the opposite direction. To conserve momentum, the second skater must have equal and opposite momentum, so he will move in the same direction as the first skater at a speed of 3 kg*m/s / 75 kg = 0.04 m/s. Therefore, the final velocity of each skater is 0.04 m/s, and they are moving apart.

Quiz 7: Rotational Motion and the Law of Gravity: Part A

A 90-kg halfback running north with a speed of 10 m/s is tackled by a 120-kg opponent running south at 5.0 m/s.The collision is perfectly inelastic.Compute the velocity of the two players just after the tackle.

A 6-kg object is moving to the right at 4 m/s and collides with a 4-kg object moving to the left at 6 m/s.The objects collide and stick together.After the collision,the combined object:

A 20-g bullet moving at 1000 m/s is fired through a one-kg block of wood emerging at a speed of 200 m/s.What is the change in the kinetic energy of the bullet-block system as a result of the collision assuming the block is free to move?

A machine gun is attached to a railroad flatcar that rolls with negligible friction.If the railroad car has a mass of 6.25 × 10⁴ kg,how many bullets of mass 25 g would have to be fired at 250 m/s off the back to give the railroad car a forward velocity of 0.3 m/s?

A billiard ball (Ball #1) moving at 5.00 m/s strikes a stationary ball (Ball #2) of the same mass.After the elastic collision,Ball #1 moves at a speed of 3.50 m/s.Find the speed of Ball #2 after the collision.

During a snowball fight,two balls with masses of 0.4 and 0.6 kg,respectively,are thrown in such a manner that they meet head-on and combine to form a single mass.The magnitude of initial velocity for each is 20 m/s.What is the speed of the 1.0-kg mass immediately after collision?

Popeye,of mass 70 kg,has just downed a can of spinach.He accelerates quickly and stops Bluto,of mass 700 kg (Bluto is very dense) ,who is charging in at 5.0 m/s.What was Popeye's remarkable speed?

A 1.00-kg duck is flying overhead at 1.50 m/s when a hunter fires straight up.The 0.0100-kg bullet is moving 200 m/s when it hits the duck and stays lodged in the duck's body.What is the speed of the duck and bullet immediately after the hit?

A 20-g bullet moving at 1000 m/s is fired through a one-kg block of wood emerging at a speed of 200 m/s.If the block had been originally at rest and is free to move,what is its resulting speed?

In an inelastic collision between two objects with unequal masses:

A cannon of mass 1500 kg fires a 10-kg shell with a velocity of 200 m/s at an angle of 30° above the horizontal.Find the recoil velocity of the cannon across the level ground.

A 70-kg man is standing in a 20-kg boat.Thinking he is stepping out onto the dock,the man steps to the right.However,the following will actually happen (ignore the friction of the water or air on the boat or the man) :

A baseball infielder,mass 75.0 kg,jumps up with velocity 2.00 m/s and catches a 0.150-kg baseball moving horizontally at 50.0 m/s.Of the following,which is closest to the final momentum of the system,infielder and baseball?

A 20-g bullet moving at 1000 m/s is fired through a one-kg block of wood emerging at a speed of 200 m/s.What is the kinetic energy of the block that results from the collision if the block had not been moving prior to the collision and was free to move?

Two skaters,both of mass 75 kg,are initially at rest on skates on a frictionless ice pond.One skater throws a 0.3-kg ball at 10 m/s to his friend,who catches it and throws it back at 10 m/s.When the first skater has caught the returned ball,what is the velocity of each of the two skaters?