Question 1

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:

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 2

An inelastic collision is one in which:

Accepted Answer

In an inelastic collision, momentum is conserved (as in all collisions), but kinetic energy is not conserved (unlike in elastic collisions where both momentum and kinetic energy are conserved). This is because in an inelastic collision, some of the kinetic energy is converted to other forms of energy such as heat, sound, or deformation of the objects involved. Therefore, choice D is the best choice as it correctly states that momentum is conserved but kinetic energy is not conserved.

Question 3

Whenever an object strikes a stationary object of equal mass:

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 4

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:

Accepted Answer

According to Newton's first law of motion, an object in motion (or at rest) will continue to be in motion (or at rest) in a straight line with a constant speed unless acted upon by an external force. Since the impulse given to the bar is horizontal and transverse, it will cause the bar to move in a straight line in the direction of the impulse. The center of mass of the bar will also move in a straight line with a constant speed, assuming no other external forces act upon it. Therefore, the correct answer is (C) move in a straight line.

Question 5

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:

Accepted Answer

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

Question 6

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:

Accepted Answer

Let the mass of bullet from gun #2 be m, then the mass of bullet from gun #1 is 2m (as it is twice as heavy). Also, let the velocity of both bullets be v. 
The total momentum imparted to gun #1 and gun #2 will be equal and opposite (as per Newton's Third Law). Let the momentum imparted to gun #1 be P1 and that imparted to gun #2 be P2. 
Then, P1/P2 = (2m * v) / (m * v) = 2, as the velocity of both bullets is the same and cancels out. 
Also, we know that the gun #1 weighs three times as much as gun #2. Let the weight of gun #2 be W, then the weight of gun #1 is 3W. As the bullets from both guns are fired with the same velocity, the recoil velocity of the guns will be inversely proportional to their respective weights. 
Thus, the recoil velocity of gun #1 is (1/3) times that of gun #2. 
Now, momentum = mass * velocity, so the momentum imparted to each gun will be proportional to the mass of the bullet multiplied by the velocity of the bullet. Thus, the ratio of the momentum imparted to gun #1 to that imparted to gun #2 will be (2m * v) / (m * v) = 2:1, as found earlier.

Question 7

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:

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 8

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:

Accepted Answer

Before the collision, the momentum of cart A is: p1 = m1*v1 = 0.2 kg * 3 m/s = 0.6 kg*m/s
Cart B is at rest, so its momentum is p2 = m2*0 = 0
The initial total momentum of the system is p_initial = p1 + p2 = 0.6 kg*m/s
After the collision, the carts stick together and move with a common speed v_final. By conservation of momentum, we have:
p_initial = p_final
0.6 kg*m/s = (m1 + m2)*v_final
0.6 kg*m/s = 0.6 kg * v_final
v_final = 1 m/s
Therefore, the center of mass of the two cart system has a speed of 1 m/s after the collision. However, since the initial speed of the center of mass was zero (cart B was at rest), the change in center of mass velocity is:
delta_v_com = v_final - 0 = 1 m/s
Since the mass of the system is m1 + m2 = 0.2 kg + 0.4 kg = 0.6 kg, the velocity of the center of mass changes by delta_v_com / (0.6 kg), which gives us:
delta_v_com / (m1 + m2) = 1 m/s / 0.6 kg = 0.33 m/s
Therefore, the answer is B) 0.33 m/s.

Question 9

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

Accepted Answer

Impulse is defined as the change in momentum of an object. It is given by the formula:
Impulse = Force x Time = Δp (change in momentum)

From the problem, Δp = 50 kg·m/s and m (mass) = 2.0 kg. We need to find the magnitude of the force (F) or impulse (J) required.

Δp = F x t
50 kg·m/s = F x t
t = 50 kg·m/s ÷ F

To find the impulse (J), we need to multiply the force (F) with the time (t):
J = F x t
J = F x (50 kg·m/s ÷ F)
J = 50 kg·m/s

Therefore, the magnitude of the impulse required is +50 N·s, which is choice C.

Question 10

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:

Accepted Answer

According to conservation of momentum, the momentum of the man and the shoe before the throw is zero. After the throw, the total momentum of the system is still zero since there are no external forces acting on the system. Therefore, the momentum of the man and the shoe after the throw must be equal in magnitude and opposite in direction.

Let the velocity of the man after the throw be v. Then, using conservation of momentum:

(720 N)(0 m/s) + (4.0 N)(15 m/s) = (720 N + 4.0 N)(-v)

Solving for v, we get:

v = 8.3 m/s

Since the velocity is negative, the man moves to the left. Therefore, the answer is choice C.

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:

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:

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:

Accepted Answer

The average force on the ball can be calculated using Newton's second law, $F = \frac{\Delta p}{\Delta t}$, where $\Delta p$ is the change in momentum and $\Delta t$ is the time interval. Since the initial momentum of the ball is 0 (it starts from rest) and its final momentum is $mv$ (mass times velocity), the change in momentum $\Delta p = mv - 0 = mv$. Therefore, the average force $F = \frac{mv}{T}$.

Question 14

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:

Accepted Answer

Using conservation of momentum, the initial momentum of the system is (75 kg + 30 kg) * 2.0 m/s = 210 kg*m/s. After the man jumps off, the cart's momentum is 30 kg * v. Since the man's horizontal velocity is zero, the cart's velocity must be 7.0 m/s to conserve momentum, resulting in a change of 5.0 m/s.

Question 15

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

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 16

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:

Accepted Answer

By conservation of momentum, we have Mv = (M + m) v', where v' is the velocity of the rifle and bullet after firing. Solving for v', we get v' = -mv/M, which is choice C.

Question 17

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:

Accepted Answer

The airbag exerts a smaller force over a longer period of time, resulting in a smaller impulse and a slower change in momentum. This helps to reduce the risk of injury compared to the sudden and rapid deceleration that would occur if the head impacted the windshield directly.

Question 18

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:

Accepted Answer

Momentum is conserved, so the total momentum before the push is zero, and the total momentum after the push is also zero. Therefore, the momentum of spaceman A after the push must be equal in magnitude but opposite in direction to the momentum of spaceman B. The momentum of spaceman B is 90 kg * 0.5 m/s = 45 kg*m/s. Therefore, the momentum of spaceman A must be -45 kg*m/s. The mass of spaceman A is 120 kg, so his recoil speed must be -45 kg*m/s / 120 kg = -0.375 m/s. However, since the question asks for the final recoil speed, we must take the absolute value of this value, which is 0.38 m/s.

Question 19

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

Accepted Answer

The law of conservation of momentum states that the total momentum in a system remains constant if there is no net external impulse acting on the system. Therefore, choice D is the correct answer as it states that the net external impulse is zero. The other choices are not necessarily true for conservation of momentum; the coefficient of restitution can vary and the collisions can be inelastic and still follow conservation of momentum.

Question 20

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:

Accepted Answer

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

Quiz 9: Center of Mass and Linear Momentum