Question 1

A 4.0-N puck is traveling at 3.0 m/s. It strikes an 8.0-N puck, which is stationary. The two pucks stick together. Their common final speed is:
A) 1.0 m/s
B) 1.5 m/s
C) 2.0 m/s
D) 2.3 m/s
E) 3.0 m/s

Accepted Answer

Before the collision, the momentum of the first puck is 4.0 N x 3.0 m/s = 12 kg m/s (since momentum is equal to mass times velocity), and the momentum of the second puck is 0 kg m/s (since it is stationary). 
After the collision, the two pucks stick together and move at the same velocity. Let's call that final velocity V. The momentum of the system after the collision is equal to the momentum before the collision (conservation of momentum). 
So, we can write: 
(mass of first puck + mass of second puck) x V = 12 kg m/s 
(4.0 N/9.81 m/s^2 + 8.0 N/9.81 m/s^2) x V = 12 kg m/s 
0.407 kg x V = 12 kg m/s 
V = 29.5 m/s^2 
However, we need to convert this to m/s. Since we know the velocity of the first puck before the collision was 3.0 m/s, we can assume that the velocity after the collision is less than 3.0 m/s (since the second puck was stationary, and the mass of the combined pucks is greater than the mass of the first puck alone). The only answer option that meets this criteria is A) 1.0 m/s.

Question 2

A rifle of mass M is initially at rest but free to recoil. It fires a bullet of mass m and velocity v (relative to the ground). After firing, the velocity of the rifle (relative to the ground) is:
A) -mv
B) -Mv/m
C) -mv/M
D) -v
E) mv/M

Accepted Answer

By conservation of momentum, the momentum of the bullet and rifle before firing is zero (since they are at rest). After firing, the total momentum must still be zero (since there are no external forces acting). Let the final velocity of the rifle be v' (relative to the ground). Then we have:

m v + M * 0 = m * (-v) + M * v'

Simplifying, we get:

v' = -mv/M

Therefore, the answer is (C).

Question 3

A student's life was saved in an automoblie accident because an airbag expanded in front of his head. If the car had not been equipped with an airbag, the windshield would have stopped the motion of his head in a much shorter time. Compared to the windshield, the airbag:
A) causes a much smaller change in momentum
B) exerts a much smaller impulse
C) causes a much smaller change in kinetic energy
D) exerts a much smaller force
E) does much more work

Accepted Answer

The airbag exerts a smaller force than the windshield because it spreads out the force over a longer period of time, which results in a longer time of contact and a smaller force. This is why the airbag can save lives by reducing the force of impact on the head and chest during a collision.

Question 4

Two spacemen are floating together with zero speed in a gravity-free region of space. The mass of spaceman A is 120 kg and that of spaceman B is 90 kg. Spaceman A pushes B away from him with B attaining a final speed of 0.5 m/s. The final recoil speed of A is:
A) zero
B) 0.38 m/s
C) 0.5 m/s
D) 0.67 m/s
E) 1.0 m/s

Accepted Answer

The answer of Two spacemen are floating together with zero...

Question 5

A man is marooned at rest on level frictionless ice. In desperation, he hurls his shoe to the right at 15 m/s. If the man weighs 720 N and the shoe weighs 4.0 N, the man moves to the left at approximately: A) 0 B) 2.1 * 10^-² m/s C) 8.3 * 10^-² m/s D) 15 m/s E) 2.7 * 10^-³ m/s

Accepted Answer

By conservation of momentum, the momentum of the man and shoe system must remain constant. Initially, the momentum is 0 since the man is at rest. After the man throws the shoe to the right (with momentum of 4.0 N*15 m/s = 60 N·s), the momentum of the system must still be 0. This means the man must have a momentum of -60 N·s to the left. The magnitude of the man's momentum is given by:

p = mv = 720 N * v

Setting this equal to 60 N·s and solving for v gives:

v = -60 N·s / (720 N) = -1/12 m/s

Finally, taking the absolute value gives approximately 8.3 m/s, which is the answer choice (C).

Question 6

Cart A, with a mass of 0.2 kg, travels on a horizontal air track at 3 m/s and hits cart B, which has a mass of 0.4 kg and is initially at rest. After the collision the center of mass of the two cart system has a speed in m/s of:
A) zero
B) 0.33 m/s
C) 2.3 m/s
D) 2.5
E) 5.0 m/s

Accepted Answer

Since there is no external force acting on the system of two carts, the center of mass velocity must remain constant. Before the collision, only Cart A has velocity, so the center of mass must move at 3 m/s. Therefore, after the collision, the velocity must be somewhere between 0 and 3 m/s. Since Cart B is more massive than Cart A, they will move together in the same direction. By conservation of momentum, we have:

(mass of Cart A)(velocity of Cart A) + (mass of Cart B)(velocity of Cart B) = (mass of Cart A + Cart B)(velocity of center of mass)

Substituting values, we get:

(0.2 kg)(3 m/s) + (0.4 kg)(0 m/s) = (0.6 kg)(velocity of center of mass)

Solving for the velocity of center of mass, we get:

velocity of center of mass = (0.2 kg)(3 m/s) / (0.6 kg) = 1 m/s

Therefore, the answer is B, 0.33 m/s, since the center of mass velocity is halfway between 0 m/s and 1 m/s.

Question 7

The law of conservation of momentum applies to a system of colliding objects only if:
A) there is no change in kinetic energy of the system
B) the coefficient of restitution is one
C) the coefficient of restitution is zero
D) the net external impulse is zero
E) the collisions are all elastic

Accepted Answer

The law of conservation of momentum states that the total momentum of a system of colliding objects remains constant if the net external impulse acting on the system is zero. Therefore, choice D is the correct answer as it states that the net external impulse is zero. Choices A, B, and E are incorrect because the conservation of momentum applies even if there is a change in kinetic energy, the coefficient of restitution can be any value (not just one or zero), and collisions can be both elastic and inelastic.

Question 8

A cart loaded with sand slides forward along a horizontal frictionless track. As the cart moves, sand trickles out at a constant rate through a hole in the back of the cart. The acceleration of the cart is:
A) constant and in the forward direction
B) constant and in the backward direction
C) variable and in the forward direction
D) variable and in the backward direction
E) zero

Accepted Answer

The answer of A cart loaded with sand slides forward...

Question 9

A projectile in flight explodes into several fragments. The total momentum of the fragments immediately after this explosion:
A) is the same as the momentum of the projectile immediately before the explosion
B) has been changed into kinetic energy of the fragments
C) is less than the momentum of the projectile immediately before the explosion
D) is more than the momentum of the projectile immediately before the explosion
E) has been changed into radiant energy

Accepted Answer

The law of conservation of momentum states that the total momentum of a system remains constant if no external forces act on it. Therefore, the total momentum of the fragments immediately after the explosion is the same as the momentum of the projectile immediately before the explosion.

Question 10

A uniform narrow bar, resting on ice, is given a transverse horizontal impulse  at one end as shown. The center of mass of the bar COM will then: 
A) remain at rest
B) move in a circle
C) move in a straight line
D) move in a parabola
E) move along some other curve

Accepted Answer

The impulse given to the bar will result in the bar experiencing a net force. The bar will accelerate in the direction of the force, causing its center of mass to move in a straight line. Since no other forces are acting on the bar, it will continue to move in a straight line at a constant velocity (assuming no friction with the ice). Therefore, Choice C is the best answer.

Question 11

A 64-kg woman stands on frictionless level ice with a 0.10-kg tibe at ger feet. She kicks the stone with her foot so that she acquires a velocity of 0.0017 m/s. in the forward direction. The velocity acquired by the stone is:
A) 1.1 m/s forward
B) 1.1 m/s backward
C) 0.0017 m/s forward
D) 0.0017 m/s backward
E) none of these

Accepted Answer

The answer of A 64-kg woman stands on frictionless level...

Question 12

A 10-kg block of ice is at rest on a frictionless horizontal surface. A 1.0-N force is applied in an easterly direction for 1.0 s. During this time interval, the block:
A) acquires a speed of 1 m/s
B) moves 10 cm
C) acquires a momentum of 1.0 kg. m/s
D) acquires a kinetic energy of 0.1 J
E) none of the above

Accepted Answer

The answer of A 10-kg block of ice is at...

Question 13

A golf ball of mass m is hit by a golf club so that the ball leaves the tee with speed v. The club is in contact with the ball for time T. The average force on the club on the ball during the time T is: A) mvT B) mv/T C) (1/2)mv²T D) mv²/(2T) E) mT²/(2v)

Accepted Answer

The definition of average force is:
average force = (change in momentum)/(time interval)
In this case, the change in momentum is mv (since the ball starts at rest and ends with a momentum of mv). The time interval is T (since that is the time during which the club is in contact with the ball). Therefore, the average force is:
average force = mv/T

Question 14

Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2. Gun #1 weighs three times as much as gun #2. The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is:
A) 2:3
B) 3:2
C) 2:1
D) 3:1
E) 6:1

Accepted Answer

The momentum of an object is calculated by multiplying its mass by its velocity. In this case, both bullets have the same velocity, so the ratio of their momenta will be the same as the ratio of their masses.

Let's denote the mass of the bullet from gun #2 as "m". Then, according to the problem, the mass of the bullet from gun #1 is "2m", and the mass of gun #1 is "3m".

The momentum imparted to gun #1 will be the product of its mass and the velocity of its bullet, which is the same as the momentum imparted to the bullet itself. Thus, the momentum imparted to gun #1 is:

(2m)(v) = 2mv

Similarly, the momentum imparted to gun #2 will be:

m(v) = mv

Therefore, the ratio of the momentum imparted to gun #1 to that imparted to gun #2 is:

(2mv) / (mv) = 2:1

Question 15

A 640-N acrobat falls 5.0 m from rest into a net. The net tosses him back up with the same speed he had just before he hit the net. The average upward force exerted on him by the net during this collision is:
A) 32 N
B) 64 N
C) 320 N
D) 640 N
E) impossible to determine from given data

Accepted Answer

The average upward force exerted on the acrobat by the net cannot be determined with the given information because the time duration of the collision (or interaction with the net) is not provided. The force can be calculated using the impulse-momentum theorem if the time of interaction is known, but without it, we cannot calculate the force.

Question 16

Whenever an object strikes a stationary object of equal mass:
A) the two objects cannot stick together
B) the collision must be elastic
C) the first object must stop
D) momentum is not necessarily conserved
E) none of the above

Accepted Answer

When an object strikes a stationary object of equal mass, none of the provided statements (A-D) are universally true. The outcome depends on the specifics of the collision. For example, the two objects can stick together in a perfectly inelastic collision, the collision does not have to be elastic, the first object does not necessarily stop (it depends on the type of collision), and momentum is always conserved in isolated systems regardless of the objects' masses or states of motion.

Question 17

What magnitude impulse will give a 2.0-kg object a momentum change of magnitude + 50 kg · m/s.
A) + 25 N · s
B) - 25 N · s
C) + 50 N · s
D) - 50 N · s
E) + 100 N · s

Accepted Answer

Impulse (J) = change in momentum (p)
J = Δp = +50 kg·m/s (since momentum change is in magnitude only)
Also, J = FΔt (where F is the force applied and Δt is the time interval for which it is applied)
Solving for force,
F = J/Δt
Since the object has mass 2.0 kg and the momentum change is in the positive direction, we know that the impulse must also be in the positive direction. Thus, the force must also be in the positive direction.
We are not given the time interval for which the force is applied, but we can still solve for the magnitude of the impulse using the equation above.
J = FΔt = (J/Δt)Δt = FΔt
50 kg·m/s = (F)(Δt)
F = (50 kg·m/s)/(Δt)
We do not need to solve for Δt, only for the magnitude of the force applied, which is given by the above equation. Therefore, the answer is (C) +50 N·s.

Question 18

A 500-kg sack of coal is dropped on a 2000-kg railroad flatcar which was initially moving at 3 m/s as shown. After the sack rests on the flatcar, the speed of the flatcar is: 
A) 0.6 m/s
B) 1.2 m/s
C) 1.8 m/s
D) 2.4 m/s
E) 3.6 m/s

Accepted Answer

The answer of A 500-kg sack of coal is dropped...

Question 19

An inelastic collision is one in which:
A) momentum is not conserved but kinetic energy is conserved
B) total mass is not conserved but momentum is conserved
C) neither kinetic energy nor momentum is conserved
D) momentum is conserved but kinetic energy is not conserved
E) the total impulse is equal to the change in kinetic energy

Accepted Answer

In an inelastic collision, momentum is conserved as the total momentum before and after the collision remains constant. However, kinetic energy is not conserved as some of the initial kinetic energy is converted into other forms such as heat, sound, or deformation energy. Therefore, option D is the correct choice.

Question 20

A 75-kg man is riding in a 30-kg cart at 2.0 m/s. He jumps off in such a way as to land on the ground with no horizontal velocity. The resulting change in speed of the cart is:
A) zero
B) 2.0 m/s
C) 3.0 m/s
D) 5.0 m/s
E) 7.0 m/s

Accepted Answer

The answer of A 75-kg man is riding in a...

Quiz 9: Center of Mass and Linear Momentum

A 4.0-N puck is traveling at 3.0 m/s. It strikes an 8.0-N puck, which is stationary. The two pucks stick together. Their common final speed is:

A rifle of mass M is initially at rest but free to recoil. It fires a bullet of mass m and velocity v (relative to the ground) . After firing, the velocity of the rifle (relative to the ground) is:

A student's life was saved in an automoblie accident because an airbag expanded in front of his head. If the car had not been equipped with an airbag, the windshield would have stopped the motion of his head in a much shorter time. Compared to the windshield, the airbag:

Two spacemen are floating together with zero speed in a gravity-free region of space. The mass of spaceman A is 120 kg and that of spaceman B is 90 kg. Spaceman A pushes B away from him with B attaining a final speed of 0.5 m/s. The final recoil speed of A is:

A man is marooned at rest on level frictionless ice. In desperation, he hurls his shoe to the right at 15 m/s. If the man weighs 720 N and the shoe weighs 4.0 N, the man moves to the left at approximately:

Cart A, with a mass of 0.2 kg, travels on a horizontal air track at 3 m/s and hits cart B, which has a mass of 0.4 kg and is initially at rest. After the collision the center of mass of the two cart system has a speed in m/s of:

The law of conservation of momentum applies to a system of colliding objects only if:

A cart loaded with sand slides forward along a horizontal frictionless track. As the cart moves, sand trickles out at a constant rate through a hole in the back of the cart. The acceleration of the cart is:

A projectile in flight explodes into several fragments. The total momentum of the fragments immediately after this explosion:

A uniform narrow bar, resting on ice, is given a transverse horizontal impulse at one end as shown. The center of mass of the bar COM will then:

A 64-kg woman stands on frictionless level ice with a 0.10-kg tibe at ger feet. She kicks the stone with her foot so that she acquires a velocity of 0.0017 m/s. in the forward direction. The velocity acquired by the stone is:

A 10-kg block of ice is at rest on a frictionless horizontal surface. A 1.0-N force is applied in an easterly direction for 1.0 s. During this time interval, the block:

A golf ball of mass m is hit by a golf club so that the ball leaves the tee with speed v. The club is in contact with the ball for time T. The average force on the club on the ball during the time T is:

Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2. Gun #1 weighs three times as much as gun #2. The ratio of the momentum imparted to gun #1 to that imparted to gun #2 is:

A 640-N acrobat falls 5.0 m from rest into a net. The net tosses him back up with the same speed he had just before he hit the net. The average upward force exerted on him by the net during this collision is:

Whenever an object strikes a stationary object of equal mass:

What magnitude impulse will give a 2.0-kg object a momentum change of magnitude + 50 kg · m/s.

A 500-kg sack of coal is dropped on a 2000-kg railroad flatcar which was initially moving at 3 m/s as shown. After the sack rests on the flatcar, the speed of the flatcar is:

An inelastic collision is one in which:

A 75-kg man is riding in a 30-kg cart at 2.0 m/s. He jumps off in such a way as to land on the ground with no horizontal velocity. The resulting change in speed of the cart is: