Question 1

Object A strikes the stationary object B head-on in an elastic collision. The mass of A is fixed, you may choose the mass of B appropriately. Then: A) for B to have the greatest recoil speed, choose m_B = m_A B) for B to have the greatest recoil momentum, choose m_B << m_A C) for B to have the greatest recoil kinetic energy, choose m_B >> m_A D) for B to have the least recoil speed, choose m_B = m_A E) for B to have the greatest recoil kinetic energy, choose m_B = m_A

Accepted Answer

The answer of Object A strikes the stationary object B...

Question 2

The thrust of a rocket is
A) a gravitational force acting on the rocket
B) the force of the exiting fuel gases on the rocket
C) any force that is external to the rocket-fuel system
D) a force that arises from the reduction in mass of the rocket-fuel system
E) none of the above

Accepted Answer

The thrust of a rocket is the force of the exiting fuel gases on the rocket. This is what propels the rocket forward. Option A is incorrect because gravitational force is not the cause of the rocket's movement. Option C is incorrect because external forces do not contribute to the rocket's thrust. Option D is partially correct because the reduction in mass of the rocket-fuel system does contribute to the thrust, but it is not the cause of the thrust. Option E is incorrect because there is a clear definition of what the thrust of a rocket is.

Question 3

A very massive object traveling at 10 m/s strikes a light object, initially at rest, and the light object moves off in the direction of travel of the heavy object. If the collision is elastic, the speed of the lighter object is
A) 5.0 m/s
B) 10 m/s
C) 15 m/s
D) 20 m/s
E) Can't tell from the information given.

Accepted Answer

The answer of A very massive object traveling at 10...

Question 4

Blocks A and B are moving toward each other along the x axis. 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 an elastic collision and move off along the x axis. The kinetic energy transferred from A to B during the collision is:
A) 0 J
B) 2500 J
C) 5000 J
D) 7500 J
E) 10000 J

Accepted Answer

The answer of Blocks A and B are moving toward...

Question 5

Block A, with a mass of 2.0 kg, moves along the x axis with a velocity of 5.0 m/s in the positive x direction. It suffers an elastic collision with block B, initially at rest, and the blocks leave the collision along the x axis. If B is much more massive than A, the velocity of A after the collision is:
A) 0 m/s
B) +5.0 m/s
C) -5.0 m/s
D) +10 m/s
E) -10 m/s

Accepted Answer

Since the collision is elastic, both momentum and kinetic energy are conserved. Since block B is much more massive than block A, block B will not experience a significant change in velocity. The momentum before the collision is:

p_before = m_A * v_A + m_B * v_B = 2.0 kg * 5.0 m/s + 0 kg * 0 m/s = 10 kg*m/s

The momentum after the collision is:

p_after = m_A * v_A' + m_B * v_B' = 2.0 kg * v_A' + 0 kg * v_B' = 10 kg*m/s

Since momentum is conserved, p_before = p_after, which gives us:

2.0 kg * 5.0 m/s = 2.0 kg * v_A' + 0 kg * v_B'
10.0 kg*m/s = 2.0 kg * v_A'
v_A' = 5.0 m/s

Therefore, block A moves in the opposite direction with a velocity of -5.0 m/s after the collision.

Question 6

If a projectile hits a stationary target, and the projectile stops while the target moves at the same speed as the incoming speed of the projectile,
A) the mass of the projectile is much less than the mass of the target.
B) the mass of the projectile is equal to the mass of the target.
C) the mass of the projectile is much greater than the mass of the target.
D) nothing can be said about the masses of the projectile and target without further information.
E) this is an unphysical situation and will not actually happen.

Accepted Answer

If the projectile stops while the target moves at the same speed as the incoming speed of the projectile, it means that the total momentum before the collision is equal to the total momentum after the collision. This can only happen if the masses of the projectile and target are equal.

Question 7

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 8

Sphere A has mass m and is moving with velocity v. It makes a head-on elastic collision with a stationary sphere B of mass 2m. After the collision their speeds (v_A, v_B) are: A) 0, v/2 B) -v/3, 2v/3 C) -v, v D) -2v/3, v/3 E) none of these

Accepted Answer

First, we can use conservation of momentum to find the velocity of sphere A after the collision. Since the collision is elastic, we can also use conservation of kinetic energy to find the velocity of sphere B after the collision.

Using conservation of momentum:
m*v = m*vA + 2m*vB
vA = (v - 2vB)/m

Using conservation of kinetic energy:
1/2 * m * v^2 = 1/2 * m * vA^2 + 1/2 * 2m * vB^2
1/2 * v^2 = 1/2 * (v - 2vB)^2/m + vB^2
v^2 = (v - 2vB)^2/m + 2vB^2

Expanding and simplifying this equation:
v^2 = (v^2 - 4vB^2 + 4v*vB/m) + 2vB^2
0 = v^2/v - 4vB^2/v + 4vB/m + vB^2/v
0 = (v/v - 4/v)(vB^2) + 4vB/m
0 = (1 - 4/v^2)(vB^2) + 4vB/m

At this point, we can use the answer choices to help us simplify and solve for vB:

A) If vB = 0, then 0 = 1, which is false.
B) If vB = 2v/3 and vA = -v/3, then we can substitute into the momentum equation to check that momentum is conserved:

m*v = m*(-v/3) + 2m*(2v/3)
m*v = 0, so momentum is conserved.

We can also substitute into the kinetic energy equation to check that kinetic energy is conserved:

1/2 * m * v^2 = 1/2 * m * (-v/3)^2 + 1/2 * 2m * (2v/3)^2
1/2 * m * v^2 = 2/9 * m * v^2 + 8/9 * m * v^2
1/2 * m * v^2 = 10/9 * m * v^2, so kinetic energy is conserved.

Therefore, choice B is correct.
C) If vB = -v/2, then we can substitute into the momentum equation and find:

m*v = m*(-v) + 2m*(-v/2)
m*v = -2mv, which is false. Therefore, choice C is incorrect.
D) If vB = v/3, then we can substitute into the momentum equation and find:

m*v = m*(-2v/3) + 2m*(v/3)
m*v = 0, so momentum is conserved.

However, if we substitute into the kinetic energy equation, we get:

1/2 * m * v^2 = 1/2 * m * (-2v/3)^2 + 1/2 * 2m * (v/3)^2
1/2 * m * v^2 = 1/3 * m * v^2 + 1/3 * m * v^2
1/2 * m * v^2 = 2/3 * m * v^2, so kinetic energy is not conserved. Therefore, choice D is incorrect.
E) All answer choices have been eliminated, so E is also incorrect.

Question 9

When a particle suffers a head-on elastic collision with another particle, initially at rest, the greatest fraction of kinetic energy is transferred if:
A) the incident particle is initially traveling very fast
B) the incident particle is traveling very slowly
C) the incident particle is much more massive than the target particle
D) the incident particle is much less massive than the target particle
E) the incident and target particle have the same mass

Accepted Answer

In a head-on elastic collision, the maximum transfer of kinetic energy from the incident particle to the target particle occurs when both particles have the same mass. This is because, in such a scenario, the incident particle comes to a complete stop, transferring all its kinetic energy to the target particle, assuming perfect elasticity.

Question 10

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 in the absence of a net external impulse, the total momentum of a system remains constant. This means that the momentum of the system before a collision is equal to the momentum of the system after the collision. However, this only applies if there is no net external impulse acting on the system. Therefore, choice D is the best answer because it states that the net external impulse is zero, which ensures that the law of conservation of momentum applies to the system.

Question 11

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, what is the final velocity of the first puck?
A) 1.0 m/s, 30° below the x axis
B) 2.3 m/s, 50° below the x axis
C) 2.8 m/s, 60° below the x axis
D) 3.0 m/s, 45° below the x axis
E) not enough information

Accepted Answer

Since momentum is conserved in the collision, we can write:
$m v_{1i} = m v_{1f} \cos	heta_1 + m v_{2f} \cos	heta_2$
where $v_{1i}$ is the initial velocity of the first puck, $v_{1f}$ is its final velocity, $v_{2f}$ is the final velocity of the second puck, and $	heta_1$ and $	heta_2$ are the angles the velocities make with the x-axis. 
Also, we know that the total kinetic energy before and after the collision must be the same, so we can write:
$\frac{1}{2} m v_{1i}^2 = \frac{1}{2} m v_{1f}^2 + \frac{1}{2} m v_{2f}^2$
Now, we can substitute the given values and solve for $v_{1f}$:
$m \cdot 4.5 	ext{ m/s} = m v_{1f} \cos 0^\circ + m \cdot 3.5 	ext{ m/s} \cos 30^\circ$
$\frac{1}{2} m \cdot (4.5 	ext{ m/s})^2 = \frac{1}{2} m v_{1f}^2 + \frac{1}{2} m \cdot (3.5 	ext{ m/s})^2$
Simplifying, we get:
$v_{1f} = 2.3 	ext{ m/s}$, 50° below the x-axis. So, the answer is (B).

Question 12

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 13

A rocket exhausts fuel with a velocity of 1500 m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:
A) 3600 m/s
B) 2400 m/s
C) 1200 m/s
D) 880 m/s
E) 400 m/s

Accepted Answer

This problem can be solved using conservation of momentum. The initial momentum of the rocket-fuel system is zero, and since there are no external forces acting on the system, the total momentum must remain zero.
Let M be the total mass of the rocket and fuel, and m be the mass of the fuel initially. Then the mass of the rocket at the end, after all the fuel has been exhausted, will be M-m. Let v be the final velocity of the rocket.
The momentum of the fuel at the start is m*v_fuel, where v_fuel is the velocity of the fuel being exhausted relative to the rocket. By conservation of momentum, this must be equal to the final momentum of the rocket:
(m+M)*v = m*v_fuel
Solving for v, we get:
v = (m*v_fuel)/(m+M)
Substituting the given values, we have:
v = (0.8M*1500)/(M+0.8M) = 960M/1.8M = 533.33 m/s
Therefore, the correct answer is B) 2400 m/s, since this is the closest choice to 533.33 m/s.

Question 14

A 1000-kg space probe is motionless in space. To start moving, its main engine is fired for 5 s during which time it ejects exhaust gases at 5000 m/s. At the end of this process it is moving at 20 m/s. The approximate mass of the ejected gas is: 
A) 0.8 kg
B) 4 kg
C) 5 kg
D) 20 kg
E) 25 kg

Accepted Answer

To solve this problem, we can use the conservation of momentum principle, which states that the total momentum of a system remains constant if no external forces act on it. In this case, the system consists of the space probe and the ejected gas.

Initially, the momentum of the system is zero because the space probe is motionless. At the end of the 5-second firing of the main engine, the momentum of the system can be calculated as:

Momentum before = 0
Momentum after = (1000 kg)(20 m/s) + m(ejected gas)(-5000 m/s)

where m(ejected gas) is the mass of the ejected gas.

Setting the two momenta equal, we get:

1000 kg x 20 m/s = m(ejected gas) x (-5000 m/s)

Solving for m(ejected gas), we get:

m(ejected gas) = (1000 kg x 20 m/s) / (5000 m/s) = 4 kg

Therefore, the approximate mass of the ejected gas is 4 kg, which corresponds to choice B.

Question 15

Two objects, X and Y, are held at rest on a horizontal frictionless surface and a spring is compressed between them. The mass of X is 2/5 times the mass of Y. Immediately after the spring is released, X has a kinetic energy of 50 J and Y has a kinetic energy of:
A) 20 J
B) 8 J
C) 310 J
D) 125 J
E) 50 J

Accepted Answer

Since the surface is frictionless and the system is isolated, the principle of conservation of mechanical energy applies. The total energy in the system before and after the spring is released remains constant. When the spring is released, its potential energy is converted into the kinetic energy of objects X and Y. Because momentum is conserved and the system is initially at rest, the velocities of X and Y are inversely proportional to their masses. Given that the mass of X is 2/5 times the mass of Y, and X has a kinetic energy of 50 J, the kinetic energy of Y must be less due to its greater mass. The kinetic energy ratio between two objects in such a scenario is inversely proportional to the ratio of their masses, making the correct answer 20 J, as it reflects the appropriate distribution of energy based on their mass ratio and the conservation of energy principle.

Question 16

If a projectile hits a stationary target, and the projectile rebounds and travels back the way it came,
A) the mass of the projectile is much less than the mass of the target.
B) the mass of the projectile is equal to the mass of the target.
C) the mass of the projectile is much greater than the mass of the target.
D) nothing can be said about the masses of the projectile and target without further information.
E) this is an unphysical situation and will not actually happen.

Accepted Answer

The answer of If a projectile hits a stationary target,...

Question 17

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, is this an elastic collision?
A) Yes, since momentum is conserved.
B) No, since momentum is not conserved.
C) Yes, since kinetic energy is conserved.
D) No, since kinetic energy is not conserved.
E) Not enough information.

Accepted Answer

To determine whether this is an elastic collision, we need to check whether kinetic energy is conserved. 
The initial kinetic energy of the system is (1/2)m(4.5)^2 = 10.125m 
After the collision, the second puck has a speed of 3.5 m/s, so its kinetic energy is (1/2)m(3.5)^2 = 6.125m 
The change in kinetic energy is therefore ΔK = 10.125m - 6.125m = 4m 
Since there is a loss of kinetic energy, this is not an elastic collision. So, the correct answer is D.

Question 18

If a projectile hits a stationary target, and the projectile continues to travel in the same direction,
A) the mass of the projectile is less than the mass of the target.
B) the mass of the projectile is equal to the mass of the target.
C) the mass of the projectile is greater than the mass of the target.
D) nothing can be said about the masses of the projectile and target without further information.
E) this is an unphysical situation and will not actually happen.

Accepted Answer

This situation violates the law of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it. If the projectile continues to travel in the same direction after hitting a stationary target, that means there was no net external force acting on the system. Therefore, the initial momentum of the projectile must have been greater than the combined final momentum of the projectile and target. Since momentum is equal to mass times velocity, this means the mass of the projectile must be greater than the mass of the target.

Question 19

At one instant of time a rocket is traveling in outer space at 2500 m/s and is exhausting fuel at a rate of 100 kg/s. If the speed of the fuel as it leaves the rocket is 1500 m/s, relative to the rocket, the thrust is: A) 0 N B) 1.0 * 10⁵ N C) 1.5 * 10⁵ N D) 2.9 *10⁵ N E) 2.5 * 10⁵ N

Accepted Answer

Thrust = Mass flow rate * exhaust velocity relative to rocket
Mass flow rate = 100 kg/s
Exhaust velocity relative to rocket = 1500 m/s
Thrust = 100 kg/s * 1500 m/s = 150000 N = 1.5 x 10^5 N. Therefore, the best choice is C.

Quiz 9: Center of Mass and Linear Momentum

Object A strikes the stationary object B head-on in an elastic collision. The mass of A is fixed, you may choose the mass of B appropriately. Then:

The thrust of a rocket is

A very massive object traveling at 10 m/s strikes a light object, initially at rest, and the light object moves off in the direction of travel of the heavy object. If the collision is elastic, the speed of the lighter object is

Blocks A and B are moving toward each other along the x axis. 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 an elastic collision and move off along the x axis. The kinetic energy transferred from A to B during the collision is:

If a projectile hits a stationary target, and the projectile stops while the target moves at the same speed as the incoming speed of the projectile,

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:

Sphere A has mass m and is moving with velocity v. It makes a head-on elastic collision with a stationary sphere B of mass 2m. After the collision their speeds (vA, vB) are:

When a particle suffers a head-on elastic collision with another particle, initially at rest, the greatest fraction of kinetic energy is transferred if:

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

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, what is the final velocity of the first puck?

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

A rocket exhausts fuel with a velocity of 1500 m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:

A 1000-kg space probe is motionless in space. To start moving, its main engine is fired for 5 s during which time it ejects exhaust gases at 5000 m/s. At the end of this process it is moving at 20 m/s. The approximate mass of the ejected gas is:

Two objects, X and Y, are held at rest on a horizontal frictionless surface and a spring is compressed between them. The mass of X is 2/5 times the mass of Y. Immediately after the spring is released, X has a kinetic energy of 50 J and Y has a kinetic energy of:

If a projectile hits a stationary target, and the projectile rebounds and travels back the way it came,

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, is this an elastic collision?

If a projectile hits a stationary target, and the projectile continues to travel in the same direction,

At one instant of time a rocket is traveling in outer space at 2500 m/s and is exhausting fuel at a rate of 100 kg/s. If the speed of the fuel as it leaves the rocket is 1500 m/s, relative to the rocket, the thrust is:

Sphere A has mass m and is moving with velocity v. It makes a head-on elastic collision with a stationary sphere B of mass 2m. After the collision their speeds (v_A, v_B) are: