Linear momentum practice problems AP Physics 1, Unit 4
Momentum, impulse, force-time graphs, conservation of momentum, and elastic and inelastic collisions. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.
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Key equations for linear momentum
| p = mv | momentum, a vector |
|---|---|
| J = FavgΔt = Δp | impulse equals the change in momentum |
| Σpbefore = Σpafter | for a system with no net external force |
| vcm = Σp / Σm | velocity of the centre of mass |
All questions use g = 9.8 m/s², as on the AP Physics 1 equation sheet. A calculator is allowed, as in the exam. The full sheet is in our AP Physics 1 equation sheet.
Warm-up questions
Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.
Hint
p = mv.
Worked solution
- p = 1500 × 20.
- p = 30,000 kg·m/s, in the direction of motion.
Hint
Impulse is force multiplied by the time it acts.
Worked solution
- J = FΔt = 400 × 0.050.
- J = 20 N·s, which equals the ball's change in momentum.
Hint
Is there a net external force on the two-cart system? Is the collision elastic?
Worked solution
- With no net external force, the total momentum of the system is always conserved.
- Carts that stick together have a perfectly inelastic collision: some kinetic energy becomes internal energy and sound.
- The answer is B.
Hint
Momentum is a vector. If the ball reverses, its velocity changes sign.
Worked solution
- Take the original direction as positive: pi = 0.20 × 10 = +2.0 and pf = 0.20 × (−8.0) = −1.6 kg·m/s.
- Δp = −1.6 − 2.0 = −3.6 kg·m/s.
- Magnitude: 3.6 kg·m/s. Subtracting the speeds to get 0.4 is the classic mistake.
Exam-style questions
Questions at the level of the AP Physics 1 exam. The ones marked Challenge combine more than one idea.
Hint
Total momentum before equals total momentum after.
Worked solution
- Momentum before: 2.0 × 3.0 + 0 = 6.0 kg·m/s.
- Afterwards they move together with mass 3.0 kg: 6.0 = 3.0v.
- v = 2.0 m/s.
Hint
Work out the total kinetic energy before and after.
Worked solution
- Before: ½ × 2.0 × 3.02 = 9.0 J.
- After: ½ × 3.0 × 2.02 = 6.0 J.
- Lost: 3.0 J, a third of the original kinetic energy.
Hint
The total momentum starts at zero, so it must stay zero.
Worked solution
- 0 = 50 × 1.4 + 70v, taking the 50 kg skater's direction as positive.
- v = −70 / 70 = −1.0 m/s.
- The 70 kg skater moves at 1.0 m/s in the opposite direction.
Hint
Both eggs stop from the same speed. What does that tell you about the impulse?
Worked solution
- Both eggs go from the same speed to rest, so the change in momentum, and the impulse, are the same.
- Impulse = FΔt. The pillow increases Δt, so the average force F is smaller.
- The answer is B, the same idea as crumple zones and airbags.
Hint
The area under a force-time graph is the impulse.
Worked solution
- Impulse = area of the triangle = ½ × 0.040 × 600 = 12 N·s.
- Impulse = Δp = mv − 0, so 12 = 0.40v.
- v = 30 m/s.
Hint
Write the momentum equation, and use the fact that in a head-on elastic collision the speed of approach equals the speed of separation.
Worked solution
- Momentum: 1.0(4.0) = 1.0v1 + 3.0v2.
- Elastic, head-on: v2 − v1 = 4.0, so v2 = v1 + 4.0.
- Substitute: 4.0 = v1 + 3v1 + 12, so 4v1 = −8.0.
- v1 = −2.0 m/s: the light cart bounces back. Check: v2 = 2.0 m/s, and kinetic energy 8 J before = 2 J + 6 J after.
Free-response practice
One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.
Experimental design and analysis style · 6 points · about 20 minutes
A student has two carts of different masses, a low-friction track, a balance and a motion sensor at each end of the track. The carts have Velcro on their bumpers so they stick together on contact.
- (a) Describe a procedure the student could use to test whether momentum is conserved when the carts collide and stick together. Say what is measured and how.
- (b) Explain how the student should use the data to decide whether momentum is conserved.
- (c) The student finds the momentum after is always slightly less than before. Suggest one physical reason.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
Measure each cart's mass with the balance. Place cart B at rest in the middle of the track. Push cart A towards it and use the motion sensor to record A's velocity just before the collision. Use a motion sensor to record the joint velocity of the two carts just after the collision. Repeat for several different starting speeds, and with extra mass added to one cart.
Part (b)
For each trial, calculate mAvA before and (mA + mB)vf after. If momentum is conserved, these are equal within experimental uncertainty. A graph of momentum after against momentum before should be a straight line through the origin with a slope of 1.
Part (c)
Friction on the track (or air resistance) is an external force. It acts during the time between the velocity measurements, slowing the carts and reducing the measured momentum.
Your mark: 0 of 6
Common mistakes in linear momentum
- Forgetting that momentum is a vector, so velocities in opposite directions need opposite signs.
- Assuming kinetic energy is conserved in every collision. Only elastic collisions conserve it.
- Using momentum conservation when there is a large external force, such as friction during a long time interval.
- Reading the height of a force-time graph as the impulse instead of the area.
Keep going
Revise the ideas behind these questions:
Then try real exam questions: AP Physics 1 past papers, with official scoring guidelines. · All AP Physics 1 units