Question 1

A 3.00-g bullet traveling horizontally at 400 m/s hits a 3.00-kg wooden block, which is initially at rest on a smooth horizontal table.The bullet buries itself in the block without passing through.The speed of the block after the collision is:

Accepted Answer

The conservation of momentum principle applies here. The total momentum before the collision equals the total momentum after the collision. The bullet's mass is 0.003 kg, and its initial velocity is 400 m/s, so its initial momentum is $0.003 \, \text{kg} \times 400 \, \text{m/s} = 1.2 \, \text{kg} \cdot \text{m/s}$. The block's mass is 3.00 kg, and it is initially at rest, so its initial momentum is 0. After the collision, the bullet and block move together with a combined mass of $3.003 \, \text{kg}$. Let $v$ be their final velocity. The conservation of momentum gives $1.2 = (3.003) \cdot v$, solving for $v$ gives $v \approx 0.40 \, \text{m/s}$.

Question 2

For a completely inelastic two-body collision the kinetic energy retained by the objects is the same as:

Accepted Answer

In a completely inelastic collision, the objects stick together and move with the velocity of the center of mass. The kinetic energy retained is given by $1/2 M v_{com}^2$, where $M$ is the total mass and $v_{com}$ is the velocity of the center of mass.

Question 3

A 3.0-kg cart and a 2.0-kg cart approach each other on a horizontal air track.They collide and stick together.After the collision their total kinetic energy is 40 J.The speed of their center of mass is:

Accepted Answer

The total kinetic energy before the collision is equal to the total kinetic energy after the collision. The speed of the center of mass is the same before and after the collision. The speed of the center of mass is:$$v_{cm}=\frac{p}{m}=\frac{m_1v_1+m_2v_2}{m_1+m_2}=\frac{(3.0	ext{ kg})(0)+(2.0	ext{ kg})(0)}{3.0	ext{ kg}+2.0	ext{ kg}}=0	ext{ m/s}$$

Question 4

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...

Question 5

A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart.The carts stick together.The impulse exerted by one cart on the other has a magnitude of:

Accepted Answer

The answer of A 2-kg cart, traveling on a horizontal...

Question 6

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

According to the law of conservation of momentum, the total momentum before firing is zero since the rifle is at rest. After firing, the total momentum is mv (momentum of the bullet) and -MV (momentum of the rifle, since it moves in the opposite direction). Thus, MV = mv, and the velocity of the rifle is -mv/M, which is equivalent to -(1/M)(mv), which gives us choice C.

Question 7

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

In this scenario, the law of conservation of momentum applies, which states that the total momentum of an isolated system remains constant if no external forces act on it. Initially, the total momentum of the two spacemen is zero.

When spaceman A pushes B away from him with a certain force, there is an equal and opposite force acting on A (Newton's third law). As a result, the two spacemen will move in opposite directions such that the total momentum of the system remains zero.

Let's denote the final recoil speed of A by v. According to the law of conservation of momentum,

(mass of A) x (initial velocity of A) + (mass of B) x (initial velocity of B) = (mass of A) x (final velocity of A) + (mass of B) x (final velocity of B)

Plugging in the given values, we obtain:

(120 kg)(0 m/s) + (90 kg)(0 m/s) = (120 kg)(v) + (90 kg)(0.5 m/s)

Solving for v, we get:

v = 0.38 m/s

Therefore, the answer is B.

Question 8

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

The momentum imparted to each gun is equal in magnitude to the momentum of the bullet it fires. Since the bullet from gun #1 is twice as heavy as the bullet from gun #2, and both bullets are fired with the same velocity, the momentum of the bullet from gun #1 is twice that of gun #2. Therefore, the ratio of the momentum imparted to gun #1 to that imparted to gun #2 is 2:1.

Question 9

A 3-g bullet is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling.The block swings in an arc, rising 3 mm above its lowest position.The velocity of the bullet was:

Accepted Answer

The correct answer involves the conservation of momentum and energy principles. The bullet's velocity can be calculated using the information given and applying these principles, not requiring the heat generated in the collision.

Question 10

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 11

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

Using conservation of momentum, the man's velocity can be calculated as (mass of shoe * velocity of shoe) / mass of man, which gives approximately 8.3 * 10^-2 m/s.

Question 12

In a one-dimensional collision,

Accepted Answer

In a one-dimensional collision, all motion, both before and after the collision, must be along the same line. This ensures that the collision dynamics can be analyzed within a single dimension.

Question 13

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 14

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

The speed of the center of mass of the system can be calculated using the formula for the conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. Before the collision, only cart A has momentum, given by its mass times its velocity (0.2 kg * 3 m/s = 0.6 kg*m/s). Since cart B is at rest, its initial momentum is 0. The total mass of the system is 0.2 kg + 0.4 kg = 0.6 kg. The speed of the center of mass is the total momentum divided by the total mass, which is 0.6 kg*m/s / 0.6 kg = 1.0 m/s.

Question 15

An elastic collision is one in which:

Accepted Answer

In an elastic collision, both kinetic energy and momentum are conserved. This means that not only is the total momentum of the system conserved, but also the total kinetic energy of the system.

Question 16

An inelastic collision is one in which:

Accepted Answer

In an inelastic collision, momentum is conserved but kinetic energy is not conserved. This means that the total momentum of the system before and after the collision remains the same, but the total kinetic energy decreases. This happens because some of the kinetic energy is transformed into other forms of energy, such as heat or sound, during the collision.

Question 17

A 64-kg woman stands on frictionless level ice with a 0.10-kg stone at her 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 law of conservation of momentum states that the total momentum of a closed system is constant if no external forces are acting on it. When the woman kicks the stone, she exerts a force on it, and the stone exerts an equal and opposite force on her, according to Newton's third law. Since the system (woman + stone) is initially at rest, its total momentum is zero. After the kick, the total momentum must still be zero. If the woman moves forward with a small velocity, the stone must move backward with a velocity that keeps the total momentum of the system at zero. The correct answer is the stone moving backward, but the specific velocity isn't provided in the choices based on the information given, so the correct choice is B, indicating the direction of the stone's velocity.

Question 18

A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest.A small amount of gunpowder had been placed on one of the blocks.The explosion does not harm the blocks but it doubles their total kinetic energy.After the explosion the blocks move along the x axis and the incident block has a speed of:

Accepted Answer

The initial kinetic energy is (1/2)m(5^2) = 12.5m. After the explosion, the total kinetic energy is doubled to 25m. Using conservation of momentum and solving the equations, the incident block's speed is found to be 1.8 m/s.

Question 19

Blocks A and B are moving toward each other.A has a mass of 2.0 kg and a velocity of 50 m/s, while B has a mass of 4.0 kg and a velocity of -25 m/s.They suffer a completely inelastic collision.The kinetic energy lost during the collision is:

Accepted Answer

In a completely inelastic collision, the two objects stick together. The initial kinetic energy is 3750 J, and the final kinetic energy is 0 J, so the kinetic energy lost is 3750 J.

Question 20

Two identical carts travel at 1 m/s on a common surface.They collide head-on and are reported to rebound, each with a speed of 2 m/s.Then:

Accepted Answer

The answer of Two identical carts travel at 1 m/s...

Quiz 9: Center of Mass and Linear Momentum