Practice Questions: Momentum

SECTION I: MULTIPLE CHOICE

Directions

Answer each of the following 20 questions. For each question, select the best answer from the four choices provided. You may use a calculator and the AP Physics 1 equation sheet.

Question 1

A 2.0 kg cart moving at 4.0 m/s to the right collides elastically with a stationary 2.0 kg cart on a frictionless track. After the collision, which of the following best describes the motion of the two carts?

1. Both carts move to the right at 2.0 m/s
2. The first cart stops and the second cart moves to the right at 4.0 m/s
3. The first cart moves to the left at 2.0 m/s and the second cart moves to the right at 2.0 m/s
4. Both carts move to the right at 4.0 m/s

Question 2

What is the final speed of the object after 3.0 seconds?

1. 2.0 m/s
2. 6.0 m/s
3. 10 m/s
4. 15 m/s

Question 3

Two objects, X and Y, collide and stick together. Object X has mass \(m\) and velocity \(v\), while object Y has mass \(2m\) and velocity \(-v\). What is the velocity of the combined system immediately after the collision?

1. \(\frac{v}{3}\) in the direction of object X's original motion
2. \(-\frac{v}{3}\) in the direction of object Y's original motion
3. \(\frac{v}{2}\) in the direction of object X's original motion
4. 0 m/s

Question 4

Experimental Data

A student performs three trials measuring the impulse delivered to carts of different masses. Which physics principle is best supported by this data?

1. Newton's second law: force is proportional to acceleration
2. The impulse-momentum theorem: impulse equals change in momentum
3. Conservation of energy: kinetic energy is conserved in collisions
4. Newton's third law: action and reaction forces are equal and opposite

Question 5

A ball of mass \(m\) moving horizontally at speed \(v_0\) strikes a vertical wall and rebounds horizontally at speed \(\frac{v_0}{2}\). What is the magnitude of the change in momentum of the ball?

1. \(\frac{1}{2}mv_0\)
2. \(mv_0\)
3. \(\frac{3}{2}mv_0\)
4. \(2mv_0\)

Question 6

A rocket in deep space fires its engine, expelling gas backward. Which statement best explains why the rocket accelerates forward?

1. The expelled gas pushes against space, creating a reaction force on the rocket
2. The momentum of the expelled gas must be balanced by an equal and opposite change in the rocket's momentum
3. The force of the engine creates an unbalanced force in the forward direction
4. The kinetic energy of the expelled gas is converted to kinetic energy of the rocket

Question 7

What is Bob's speed after the push?

1. 1.5 m/s
2. 2.0 m/s
3. 3.0 m/s
4. 4.5 m/s

Question 8

A 1200 kg car traveling at 20 m/s collides with a stationary 1200 kg car. The cars lock together and move as a single system. What type of collision is this, and what is the approximate kinetic energy lost?

1. Elastic collision; 0 J lost
2. Inelastic collision; 120,000 J lost
3. Perfectly inelastic collision; 120,000 J lost
4. Perfectly inelastic collision; 240,000 J lost

Question 9

What is the magnitude of the net force acting on the object during this time interval?

1. 2.0 N
2. 4.0 N
3. 10 N
4. 20 N

Question 10

A student drops a ball from rest. Just before hitting the ground, the ball has momentum \(p\) directed downward. During the collision with the ground, the ball experiences an average force \(F\) for time \(\Delta t\) and rebounds upward with momentum \(\frac{p}{2}\). Which expression correctly represents the magnitude of the average force?

1. \(F = \frac{p}{2\Delta t}\)
2. \(F = \frac{p}{\Delta t}\)
3. \(F = \frac{3p}{2\Delta t}\)
4. \(F = \frac{2p}{\Delta t}\)

Question 11

Three carts on a frictionless track are pushed together with a compressed spring between carts A and C. Cart B is in the middle. When released, cart A (mass \(m\)) moves left at speed \(v\), and cart C (mass \(m\)) moves right at speed \(v\). Cart B has mass \(2m\). What is the speed of cart B after release?

1. 0 m/s
2. \(\frac{v}{2}\)
3. \(v\)
4. \(2v\)

Question 12

What is the velocity of the 5.0 kg ball after the collision?

1. 0.4 m/s in its original direction
2. 0.4 m/s opposite to its original direction
3. 2.0 m/s opposite to its original direction
4. 7.6 m/s opposite to its original direction

Question 13

A system consists of two blocks connected by a light string over a frictionless pulley. Block A (mass 2 kg) rests on a frictionless table, and block B (mass 3 kg) hangs vertically. When released from rest, the system accelerates. What is the magnitude of the momentum of the two-block system 2.0 seconds after release? (Use g = 10 m/s²)

1. 12 kg·m/s
2. 24 kg·m/s
3. 30 kg·m/s
4. 60 kg·m/s

Question 14

During a collision between two objects on a horizontal frictionless surface, which of the following must be true?

1. The kinetic energy of the system is conserved
2. The momentum of each individual object is conserved
3. The total momentum of the system is conserved
4. The velocity of the center of mass increases

Question 15

What is the magnitude of the impulse delivered to the object during the first 2.0 seconds?

1. 4.0 N·s
2. 8.0 N·s
3. 16 N·s
4. 20 N·s

Question 16

A 60 kg person stands on a 40 kg platform initially at rest on a frictionless frozen lake. The person walks from one end of the 5.0 m long platform to the other. What is the displacement of the platform relative to the ice?

1. 2.0 m in the opposite direction of the person's walk
2. 2.5 m in the opposite direction of the person's walk
3. 3.0 m in the opposite direction of the person's walk
4. 5.0 m in the opposite direction of the person's walk

Question 17

In an inelastic collision between two objects, which of the following quantities is always conserved?

1. The kinetic energy of each individual object
2. The total kinetic energy of the system
3. The velocity of each individual object
4. The total momentum of the system

Question 18

If the cart was initially moving at 10 m/s toward the wall and rebounds elastically, what is the cart's velocity immediately after the collision?

1. 2.0 m/s away from the wall
2. 2.0 m/s toward the wall
3. 10 m/s away from the wall
4. 18 m/s away from the wall

Question 19

Two objects undergo a head-on collision. Object 1 has mass \(m_1\) and initial velocity \(v_1\), and object 2 has mass \(m_2\) and initial velocity \(v_2\). After the collision, object 1 has velocity \(v_1'\). Which expression gives the final velocity \(v_2'\) of object 2?

1. \(v_2' = \frac{m_1(v_1 - v_1')}{m_2} + v_2\)
2. \(v_2' = \frac{m_1(v_1 + v_1')}{m_2} - v_2\)
3. \(v_2' = \frac{m_2 v_2 - m_1 v_1}{m_1 + m_2}\)
4. \(v_2' = v_2 - \frac{m_1}{m_2}(v_1 - v_1')\)

Question 20

What is the velocity of the center of mass of this system?

1. 0.5 m/s to the right
2. 0.5 m/s to the left
3. 1.1 m/s to the right
4. 2.0 m/s to the right

SECTION II: FREE RESPONSE

Directions

Answer both of the following questions. Show all your work clearly, including equations, substitutions with units, and final answers with units. For derivations, begin with fundamental physics principles. For experimental questions, provide complete procedures and explanations.

FRQ 1: Mathematical Routines (12 minutes)

A ballistic pendulum is used to measure the speed of a projectile. A projectile of mass \(m\) is fired horizontally into a stationary pendulum bob of mass \(M\) hanging from a string of length \(L\). The projectile embeds itself in the bob, and the combined system swings upward to a maximum height \(h\) above the initial position.

1. Derive an expression for the speed \(v\) of the projectile just before impact. Begin your derivation by writing a fundamental physics principle or equation. Express your answer in terms of \(m\), \(M\), \(g\), and \(h\). Show all steps.
2. Justify why the collision between the projectile and bob cannot be elastic. Support your answer with a physics principle.
3. Suppose \(m = 0.050\) kg, \(M = 0.450\) kg, and the combined system swings to a height \(h = 0.15\) m. Calculate the initial speed of the projectile. Use g = 10 m/s².
4. A student proposes using a heavier pendulum bob to improve the accuracy of the measurement. Explain how increasing \(M\) while keeping \(m\) constant would affect the maximum height \(h\) reached for the same initial projectile speed \(v\).

FRQ 2: Experimental Design and Analysis (18 minutes)

A student investigates the relationship between impulse and momentum change using a dynamics cart on a horizontal track. The student uses a motion sensor to measure velocity and a force sensor attached to the cart to measure the force during collisions with a padded bumper.

1. Describe an experimental procedure the student could use to collect data on impulse and momentum change. In your description, include:
   • What measurements should be taken
   • How to calculate impulse from the force sensor data
   • How to calculate momentum change from the velocity data
   • What variable should be changed between trials
2. The student plans to graph the data to verify the impulse-momentum theorem. Describe what should be graphed on each axis and what the slope of the line represents. Include what value the slope should have if the impulse-momentum theorem is confirmed.
3. The student collects the following data:
   
   Complete the data table by calculating the momentum change for each trial. Show your calculation for Trial 1.
4. Using graph paper or a blank grid, plot the appropriate graph to verify the impulse-momentum theorem. Draw a best-fit line through your data points.
5. Calculate the slope of your best-fit line using two points that lie on the line (not necessarily data points). Show your work and include units.
6. Based on your experimental results, evaluate whether the data support the impulse-momentum theorem. Justify your answer using evidence from your graph and calculated slope. Identify one possible source of experimental uncertainty that could account for any deviation from the theoretical prediction.

ANSWER KEY

Part A: Multiple Choice Answer Table

Part B: Free Response Detailed Answers

FRQ 1 - Answer Key

Part A: Derive expression for projectile speed

Model derivation:

Step 1: Apply conservation of momentum to the collision
The collision between the projectile and bob is an inelastic collision. Before collision: projectile has momentum \(mv\), bob has momentum 0. After collision: combined system has velocity \(v'\).
\[mv + 0 = (m + M)v'\]
\[v' = \frac{mv}{m + M}\]

Step 2: Apply conservation of energy to the swing
After the collision, the combined system swings upward, converting kinetic energy to gravitational potential energy:
\[\frac{1}{2}(m + M)(v')^2 = (m + M)gh\]
\[v' = \sqrt{2gh}\]

Step 3: Combine equations to solve for v
Substitute the expression for \(v'\) from Step 2 into the equation from Step 1:
\[\frac{mv}{m + M} = \sqrt{2gh}\]
\[v = \frac{m + M}{m}\sqrt{2gh}\]

Final answer: \(\boxed{v = \frac{m + M}{m}\sqrt{2gh}}\)

Part B: Justify why collision cannot be elastic

The collision cannot be elastic because the projectile embeds itself in the bob, meaning the two objects stick together and move with the same final velocity. In an elastic collision, kinetic energy is conserved and the objects separate after collision. In this perfectly inelastic collision, kinetic energy is converted to other forms (deformation, heat, sound), so mechanical energy is not conserved during the collision. Only momentum is conserved.

Part C: Calculate initial speed

Given: m = 0.050 kg, M = 0.450 kg, h = 0.15 m, g = 10 m/s²

Solution:
\[v = \frac{m + M}{m}\sqrt{2gh}\]
\[v = \frac{0.050 + 0.450}{0.050}\sqrt{2(10)(0.15)}\]
\[v = \frac{0.500}{0.050}\sqrt{3.0}\]
\[v = 10 \times 1.73\]
\[v = 17.3 \text{ m/s}\]

Answer: The initial speed of the projectile is 17 m/s (or 17.3 m/s with more precision).

Part D: Effect of increasing M on maximum height h

From the derived equation \(v = \frac{m + M}{m}\sqrt{2gh}\), we can solve for h:
\[h = \frac{v^2 m^2}{2g(m + M)^2}\]

For constant projectile speed \(v\) and projectile mass \(m\), if the bob mass \(M\) increases, the denominator \((m + M)^2\) increases. This causes \(h\) to decrease.

Physical explanation: When the projectile embeds in a heavier bob, the combined system has more mass but the same momentum (from the collision). This results in a smaller velocity \(v'\) after collision. With less kinetic energy after collision, the system cannot swing as high, so \(h\) decreases. The heavier combined mass has more inertia and gains less speed from the same momentum transfer.

FRQ 2 - Answer Key

Part A: Experimental procedure

Model answer:

The student should follow this procedure:

1. Setup: Place a dynamics cart with attached force sensor on a horizontal track. Position a padded bumper at one end of the track and a motion sensor at the other end pointed at the cart.
2. Measurements to take:
   • Use the motion sensor to measure the cart's velocity immediately before and immediately after collision with the bumper
   • Use the force sensor to record force as a function of time during the collision
   • Record the mass of the cart (including force sensor)
3. Calculate impulse: The impulse is the area under the force vs. time curve. This can be calculated by:
   • Using software to integrate the force-time data, or
   • Counting squares under the curve and multiplying by scale factors
   The impulse \(J = \int F \, dt\) will be in N·s.
4. Calculate momentum change: Use the velocity measurements to calculate:
   \[\Delta p = m v_f - m v_i\]
   where \(v_i\) is initial velocity (before collision) and \(v_f\) is final velocity (after collision).
5. Variable to change: Conduct multiple trials with different initial velocities of the cart. This can be achieved by releasing the cart from different positions on an inclined ramp, or by giving it different pushes. The cart mass should remain constant to isolate one variable.
6. Repeat: Perform at least 4-5 trials with different initial speeds to collect sufficient data for graphing.

Part B: Graph description and slope interpretation

Model answer:

Axes:

• Vertical axis (y-axis): Change in momentum, Δp (in kg·m/s)
• Horizontal axis (x-axis): Impulse, J (in N·s)

Slope representation: The slope of the line represents the ratio \(\frac{\Delta p}{J}\).

Expected value: According to the impulse-momentum theorem, impulse equals change in momentum: \(J = \Delta p\). Therefore, the slope should equal 1 (dimensionless) if the theorem is confirmed. The graph should be linear and pass through the origin with a slope of 1.

Part C: Complete data table

Calculation for Trial 1:
Mass m = 0.50 kg
Initial velocity \(v_i = 1.2\) m/s
Final velocity \(v_f = -0.8\) m/s (negative indicates direction reversal)

\[\Delta p = m(v_f - v_i)\]
\[\Delta p = 0.50 \text{ kg} \times (-0.8 - 1.2) \text{ m/s}\]
\[\Delta p = 0.50 \times (-2.0)\]
\[\Delta p = -1.0 \text{ kg·m/s}\]

Completed data table:

Part D: Plot graph with best-fit line

Instructions for student:

Plot the four data points with Impulse (N·s) on the horizontal axis and Change in Momentum (kg·m/s) on the vertical axis. Both values are negative, so points should appear in the third quadrant if the origin is at the center, or in the main plotting area if axes are scaled to show negative values.

The points should be:
(-1.0, -1.0), (-1.4, -1.4), (-1.7, -1.7), (-2.1, -2.1)

Draw a best-fit straight line through these points. The line should pass through or very close to the origin (0, 0) and have a slope of approximately 1.

Part E: Calculate slope of best-fit line

Model calculation:

Using two points on the best-fit line (not necessarily data points):
Point 1: (-1.0 N·s, -1.0 kg·m/s)
Point 2: (-2.0 N·s, -2.0 kg·m/s)

\[\text{slope} = \frac{\Delta(\text{momentum change})}{\Delta(\text{impulse})}\]
\[\text{slope} = \frac{-2.0 - (-1.0)}{-2.0 - (-1.0)}\]
\[\text{slope} = \frac{-1.0 \text{ kg·m/s}}{-1.0 \text{ N·s}}\]
\[\text{slope} = 1.0\]

Answer: The slope is 1.0 (dimensionless, since N·s = kg·m/s).

Part F: Evaluate support for impulse-momentum theorem

Model answer:

The experimental data strongly support the impulse-momentum theorem. The graph shows a linear relationship between impulse and momentum change with a slope of 1.0, which matches the theoretical prediction that \(J = \Delta p\). All data points lie very close to the best-fit line, indicating good experimental consistency.

Evidence: The calculated slope of 1.0 means that for every 1 N·s of impulse delivered to the cart, the momentum changes by exactly 1 kg·m/s, confirming the direct proportionality predicted by the theorem. Additionally, the line passes through the origin, which is correct because zero impulse should produce zero momentum change.

Source of uncertainty: One possible source of experimental uncertainty is friction between the cart and track. Even though the track is designed to be low-friction, some friction force acts during the collision and motion, which would reduce the magnitude of the measured momentum change compared to the impulse. This could cause slight deviations from a perfect slope of 1.0. Other sources include: timing precision in the force sensor integration, motion sensor measurement error in velocity readings, or incomplete elastic rebound in the bumper causing energy dissipation not accounted for in the momentum calculation.