Unit 4 Practice Set: Linear Momentum

Unit overview · Games · Topics: 4.1 4.2 4.3 4.4

How to use this set

Part A: Multiple choice

1. A 0.50 kg ball moves at 8.0 m/s. A 2.0 kg ball has the same momentum. What is the speed of the 2.0 kg ball?

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p = (0.50)(8.0) = 4.0 kg·m/s. Then v = 4.0 / 2.0 = 2.0 m/s. Choice 4.0 m/s comes from matching kinetic energy instead of momentum (½·0.5·64 = 16 J gives v = 4.0 m/s for 2.0 kg).

2. A 4.0 kg cart moves at +3.0 m/s and a 2.0 kg cart moves at −3.0 m/s on the same track. What is the velocity of the centre of mass of the two carts?

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ptotal = (4.0)(3.0) + (2.0)(−3.0) = 6.0 kg·m/s. vcm = 6.0 / 6.0 = +1.0 m/s. "0 m/s" is tempting because the speeds are equal, but the masses are not, so the momenta do not cancel.

3. Two 2.0 kg pucks slide on frictionless ice. One moves east at 3.0 m/s, the other north at 3.0 m/s. What is the total momentum of the two pucks?

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Components: 6.0 east and 6.0 north (kg·m/s). Size √(6.0² + 6.0²) = 8.49 ≈ 8.5 kg·m/s at 45°, northeast. 12 adds sizes, which is only right for the same direction.

4. The graph shows the force on a tennis ball from a racket. What impulse does the racket give the ball?

01020t (ms) 6000F (N)
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Impulse = area under F-t = ½ × base × height = ½ × 0.020 s × 600 N = 6.0 N·s. 12 N·s treats the triangle as a rectangle. 6000 forgets to change ms to s.

5. A 0.15 kg baseball moving at 30 m/s is caught and brought to rest in 0.10 s. What is the size of the average force from the glove on the ball?

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Δp = 0.15 × (0 − 30) = −4.5 kg·m/s. Favg = Δp/Δt = 4.5 / 0.10 = 45 N (opposite the motion). 4.5 N forgets to divide by the time. (The ball's weight, 1.5 N, is small and the question asks for the glove force; the vertical part does not change this horizontal result.)

6. Two balls of equal mass hit a wall at the same speed. Ball X is clay and sticks. Ball Y is rubber and bounces straight back at the same speed. Which statement is correct?

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With toward-the-wall positive: X goes from +mv to 0, Δp = −mv. Y goes from +mv to −mv, Δp = −2mv. Impulse = Δp, so Y gets twice as much. The speed of Y is unchanged, but its velocity reverses, and momentum is a vector.

7. The graph shows the momentum of a cart moving on a straight track. What is the net force on the cart?

0123t (s) 280p (kg·m/s)
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Fnet = Δp/Δt = slope of p-t = (8 − 2)/3 = 2.0 N. No mass is needed: the slope of a momentum graph is already a force. 2.7 N uses 8/3 (final p over time) instead of the change.

8. In a crash, an airbag stops a driver who would otherwise hit the steering wheel. Compared with hitting the wheel, the airbag

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The driver goes from the same speed to zero either way, so Δp and the impulse are the same. Favg = Δp/Δt, so a longer Δt means a smaller force.

9. A 1.5 kg cart moves at +2.0 m/s. A constant force of −6.0 N acts on it for 0.75 s. What is the cart's velocity afterwards?

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J = FΔt = (−6.0)(0.75) = −4.5 N·s. pf = pi + J = (1.5)(2.0) − 4.5 = −1.5 kg·m/s. vf = −1.5/1.5 = −1.0 m/s. The cart stops and then moves backward.

10. A 50 kg skater stands at rest on frictionless ice and throws a 2.0 kg ball forward at 10 m/s. What is the skater's velocity just after the throw?

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Total momentum starts at zero and stays zero (no net external horizontal force). 0 = (2.0)(10) + (50)v, so v = −0.40 m/s, backward. 0.38 m/s comes from wrongly dividing by 52 kg (the ball and skater do not move together).

11. A ball falls freely toward the ground (ignore air). For which system is the total momentum constant while the ball falls?

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For the ball alone, gravity from Earth is an external force, so the ball's momentum grows downward. For ball + Earth, gravity is internal: the ball pulls Earth up as hard as Earth pulls the ball down, so the changes cancel and total momentum stays constant.

12. Two carts, 1.0 kg and 3.0 kg, are at rest with a compressed spring between them. The spring is released and the carts fly apart. Which statement is correct?

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Total momentum stays zero, so the momenta are equal in size and opposite. Then K = p²/(2m): same p, so the lighter cart has 3 times the KE. (Its speed is 3 times larger, and K grows with v², while m is 3 times smaller: 9 ÷ 3 = 3.)

13. A 2.0 kg cart moving at 3.0 m/s hits a 1.0 kg cart at rest. They stick together. What is their speed just after the collision?

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(2.0)(3.0) = (3.0)v → v = 2.0 m/s. 1.5 m/s halves the speed, which is only right for equal masses.

14. For the collision in question 13, how much kinetic energy is lost?

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Before: ½(2.0)(3.0)² = 9.0 J. After: ½(3.0)(2.0)² = 6.0 J. Lost: 3.0 J (turned into heat, sound and bending). 6.0 J is the KE left, not the KE lost.

15. A puck moving at speed v hits an identical puck at rest, head-on, in a perfectly elastic collision. What happens?

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Momentum: mv = mv₁ + mv₂. KE: ½mv² = ½mv₁² + ½mv₂². These have two solutions. v₁ = v, v₂ = 0 would mean the first puck passed straight through the second, so the real answer is v₁ = 0, v₂ = v: equal masses swap velocities. "Both at v/2" conserves momentum but loses half the KE, so it is the stick-together case, not elastic.

16. A 1.0 kg cart moving at +4.0 m/s hits a 1.0 kg cart at rest. Which pair of final velocities (cart 1, cart 2) is impossible?

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All four conserve momentum (sum = 4.0 kg·m/s). Check KE, which starts at 8.0 J: (a) 4.0 J, (b) 8.0 J, (c) 0.5 + 4.5 = 5.0 J, (d) 0.5 + 12.5 = 13 J. A collision cannot create kinetic energy from nothing (no spring or explosive here), so (d) is impossible.

17. The bar charts show the momentum of objects A and B before a collision and of A after it. The system is isolated. What is the momentum of B after the collision?

BeforeAfter +6−2+1? ABABp in kg·m/s (each grid step = 1)
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Total before = +6 + (−2) = +4. After: +1 + pB = +4, so pB = +3 kg·m/s. +5 comes from ignoring B's negative sign before (6 − 1). +7 comes from adding sizes.

18. A 0.010 kg dart moving at 400 m/s sticks into a 2.0 kg block at rest on a frictionless table. About how fast does the block move afterwards?

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(0.010)(400) = (2.010)v → v = 4.0/2.01 = 1.99 m/s ≈ 2.0 m/s. 28 m/s comes from wrongly using energy conservation (√(800 J × 2 / 2.01 kg) ≈ 28 m/s); KE is not conserved when objects stick: over 99% is lost.

19. A truck hits a fly head-on. During the collision, how does the size of the fly's change in momentum compare with the size of the truck's?

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By Newton's third law the forces are equal and opposite, and they act for the same time, so the impulses (and changes in momentum) are equal in size and opposite. The fly's change in velocity is far larger because its mass is tiny; that is the tempting mix-up.

Part B: Free response

FR 1: Mathematical routines AP (10 points)

Cart A (mass 0.60 kg) moves right at 1.2 m/s on a level track with negligible friction. It hits cart B (mass 0.40 kg), which is at rest. Just after the collision cart A moves right at 0.40 m/s. Take right as positive.

  1. Calculate the velocity of cart B just after the collision.
  2. Calculate the total kinetic energy before and after the collision. Classify the collision as elastic or inelastic and justify your answer.
  3. A force sensor shows that the force on B rises steadily from zero to a peak and falls steadily back to zero, a triangle lasting 0.050 s. Calculate the peak force on cart B.
  4. Calculate the velocity of the centre of mass of the two carts before and after the collision, and explain why the two values compare as they do.
Show worked answer and scoring guide
  1. Momentum conserved (no net external horizontal force): (0.60)(1.2) + 0 = (0.60)(0.40) + (0.40)vB → 0.72 = 0.24 + 0.40vB → vB = +1.2 m/s (right). [1 pt conservation of momentum written with correct terms; 1 pt correct answer with direction]
  2. Kbefore = ½(0.60)(1.2)² = 0.432 J. Kafter = ½(0.60)(0.40)² + ½(0.40)(1.2)² = 0.048 + 0.288 = 0.336 J. K decreased by 0.096 J (about 22%), so the collision is inelastic. [1 pt each KE; 1 pt classification consistent with the numbers]
  3. Impulse on B = ΔpB = (0.40)(1.2 − 0) = 0.48 N·s. Triangle area: ½ Fpeak(0.050) = 0.48 → Fpeak = 19.2 N ≈ 19 N. [1 pt impulse = Δp; 1 pt area of triangle; 1 pt answer]
  4. vcm = ptotal / M = 0.72 / 1.00 = 0.72 m/s before. After: (0.24 + 0.48)/1.00 = 0.72 m/s. They are equal because the forces between the carts are internal and there is no net external force, so the total momentum, and hence vcm, cannot change. [1 pt values; 1 pt reasoning naming internal forces / zero net external force]

FR 2: Translation between representations AP (8 points)

A 0.20 kg rubber ball is dropped. It hits the floor moving down at 5.0 m/s and leaves moving up at 3.0 m/s. It is in contact with the floor for 0.040 s. Take up as positive.

  1. Describe the free-body diagram of the ball while it touches the floor: name each force, its direction, and which is larger.
  2. Which graph below best shows the ball's momentum from just before it touches the floor to just after it leaves? Explain your choice using the values given.
  3. Calculate the average normal force from the floor on the ball.
  4. Explain how a feature of the graph you chose in (b) is consistent with your answer to (c).
Graph 1Graph 2Graph 3t (contact lasts 0.040 s); p in kg·m/s, up + −1.0+0.6+1.0−0.6−1.0+1.0
Show worked answer and scoring guide
  1. Two forces: the weight (gravity from Earth), down, mg = 1.96 N; and the normal force from the floor, up. The normal force is much larger, because the net force must be up to turn the ball's momentum from down to up. [1 pt both forces with directions; 1 pt normal larger, with reason]
  2. Graph 1. Before: p = (0.20)(−5.0) = −1.0 kg·m/s. After: p = (0.20)(+3.0) = +0.60 kg·m/s. Graph 2 has the signs reversed (it would be right if down were positive). Graph 3 shows the ball leaving at 5.0 m/s, which ignores the energy lost in the bounce. [1 pt choice; 1 pt both momenta computed with signs]
  3. Δp = +0.60 − (−1.0) = +1.6 kg·m/s. Fnet,avg = 1.6 / 0.040 = 40 N up. FN − mg = 40 → FN = 40 + 1.96 ≈ 42 N up. [1 pt Δp with signs; 1 pt net force; 1 pt adds the weight]
  4. The slope of the p-t graph during contact is Δp/Δt = 1.6/0.040 = 40 N, which is the net force, and it is positive (up), matching the net upward force in (c). [1 pt links slope to net force]

FR 3: Experimental design and analysis AP (10 points)

A student wants to test the claim that the impulse on an object equals its change in momentum. She has a 0.50 kg cart, a level track, a force sensor with a spring bumper fixed to the end of the track, a motion detector, and a computer that records force and velocity many times per second.

  1. Describe a procedure she could use. Say what she measures, how she measures it, and how she varies the conditions to get several trials.
  2. Her data are below (toward the bumper is positive). Complete the column for the size of the change in momentum.
  3. Which two quantities should she graph to test the claim with a straight line? What slope and intercept should she expect if the claim is true?
  4. The values of |Δp| come out slightly larger than the measured impulse in every trial. Give one physical reason for this.
Trialvbefore (m/s)vafter (m/s)Impulse from F-t area (N·s)|Δp| (kg·m/s)
1+0.40−0.300.34?
2+0.60−0.450.51?
3+0.80−0.620.69?
4+1.00−0.750.86?
5+1.20−0.881.02?
Show worked answer and scoring guide
  1. Place the motion detector at the far end of the track, pointing at the cart. Give the cart a push toward the bumper and let it roll freely. Record force vs time during the bounce, and velocity vs time before and after. Find the impulse as the area under the F-t graph (the software integrates it). Read v just before and just after contact from the v-t data. Repeat with different push strengths (different starting speeds) for at least five trials. [1 pt measures F vs t and finds area; 1 pt measures v before and after with detector; 1 pt varies initial speed over several trials]
  2. |Δp| = m|vafter − vbefore| = 0.50 × (0.70, 1.05, 1.42, 1.75, 2.08) = 0.35, 0.525 (≈0.53), 0.71, 0.875 (≈0.88), 1.04 kg·m/s. [1 pt uses the change including the sign flip (adds speeds); 1 pt correct values]
  3. Graph impulse (vertical) against |Δp| (horizontal), or the other way round. If impulse = Δp, the points lie on a straight line with slope 1 (no units) through the origin (intercept 0). [1 pt quantities; 1 pt slope 1; 1 pt intercept 0]
  4. The track is not quite level. If the bumper end is a little higher, part of the cart's weight pulls it away from the bumper during the whole contact. That extra impulse points the same way as the bumper's push, but the force sensor does not record it, so |Δp| comes out a bit larger than the measured impulse. (Also acceptable: the force sensor is slightly miscalibrated and reads low. Friction alone is a weak answer: it adds to the bumper's impulse while the cart moves in, but takes away while the cart moves out, so the two parts nearly cancel.) [2 pts: 1 for a physical cause, 1 for explaining the direction of the effect]

FR 4: Qualitative-quantitative translation AP (10 points)

Block A (mass m) slides at speed v₀ on a frictionless floor toward identical block B (mass m), which is at rest. Two cases are tested. Case 1: the blocks have sticky pads and stick together. Case 2: the blocks have magnets that repel, and the collision is perfectly elastic.

  1. Without using equations, explain in a clear paragraph which case gives the larger impulse to block B, or whether the impulses are equal.
  2. Derive an expression for the impulse on B in each case, in terms of m and v₀.
  3. Explain how your expressions in (b) support your reasoning in (a).
  4. In both cases the contact lasts the same time Δt. Write the ratio of the average force on B in case 2 to that in case 1.
Show worked answer and scoring guide
  1. The impulse on B equals B's change in momentum. In both cases the total momentum of the pair stays m v₀. When they stick, the momentum is shared, so B ends with only part of it. In the elastic case, the identical blocks swap velocities: A stops and B takes all the momentum. So B gains more momentum, and gets the larger impulse, in case 2. [1 pt impulse = change in B's momentum; 1 pt momentum shared vs fully transferred; 1 pt correct conclusion]
  2. Case 1: m v₀ = 2m v → v = v₀/2, so J₁ = m(v₀/2 − 0) = ½ m v₀. Case 2: momentum m v₀ = m vA + m vB and energy ½m v₀² = ½m vA² + ½m vB² give vA = 0, vB = v₀, so J₂ = m v₀. [1 pt case 1 velocity; 1 pt case 1 impulse; 1 pt uses both conservation laws in case 2; 1 pt case 2 impulse]
  3. J₂ = m v₀ is twice J₁ = ½ m v₀, matching the claim that B gets all the momentum in case 2 and half in case 1. [1 pt J₂ > J₁ agrees with the conclusion in (a); 1 pt connects the factor of 2 to the paragraph]
  4. Favg = J/Δt with the same Δt, so F₂ / F₁ = J₂ / J₁ = 2. [1 pt]

Score yourself

Part A: 1 point each, 19 points. Part B: 38 points (10 + 8 + 10 + 10). Total 57.

Most missed ideas: signs (momentum is a vector), impulse = change in momentum, and "momentum conserved but KE not" when objects stick.