AP Physics 1 / Topic

Linear Momentum

Unit 4 covers momentum and impulse, conservation of momentum in collisions and explosions, elastic versus inelastic collisions, and center of mass.

41 practice questions3 FRQs4 skill areas

What the exam asks

  • Relate impulse to momentum change and to force-time graphs
  • Apply conservation of momentum to 1D and simple 2D collisions
  • Classify collisions: elastic (KE conserved) versus perfectly inelastic (objects stick)
  • Explain why KE is lost in inelastic collisions while momentum is not
  • Locate and track the center of mass of a system

Key formulas and rules

Momentum and impulse

p = mv; J = FΔt = Δp; area under F-t graph

Conservation

m1v1 + m2v2 = m1v1' + m2v2' (isolated system)

Perfectly inelastic

m1v1 + m2v2 = (m1 + m2)v'

Elastic check

elastic collisions also conserve ½mv² totals

Center of mass

x(cm) = Σmixi / Σmi; no external force ⇒ v(cm) constant

Question bank breakdown

Multiple choice

41

Free response

3

Difficulty mix

10 easy · 22 medium · 9 hard

Skills covered

Change in Momentum and ImpulseConservation of Linear MomentumElastic and Inelastic CollisionsSystems and Center of Mass

Every question in the bank comes with a worked explanation, and most add a per-choice breakdown so you learn why each wrong answer is wrong - not just the key.

Sample question

From the free tier of the ScoreMint AP Physics 1 bank - try it, then check the answer.

A 0.50 kg0.50 \text{ kg} ball traveling at 8.0 m/s8.0 \text{ m/s} to the right strikes a wall and bounces back at 6.0 m/s6.0 \text{ m/s} to the left. The ball is in contact with the wall for 0.020 s0.020 \text{ s}. What is the magnitude of the average force exerted by the wall on the ball?

  • A. 50 N50 \text{ N}
  • B. 150 N150 \text{ N}
  • C. 350 N350 \text{ N}
  • D. 200 N200 \text{ N}
Show answer and explanation

Answer: C. Impulse equals the change in momentum: J=Δp=m(vfvi)J = \Delta p = m(v_f - v_i). Since momentum is a vector, direction matters critically. With rightward as positive: vi=+8.0v_i = +8.0 m/s and vf=6.0v_f = -6.0 m/s. Thus Δv=6.0(+8.0)=14\Delta v = -6.0 - (+8.0) = -14 m/s. The change in momentum is 0.50×(14)=7.00.50 \times (-14) = -7.0 kg m/s (the negative sign means the impulse is to the left, which makes sense). The average force magnitude is F=7.0/0.020=350|F| = 7.0/0.020 = 350 N. The most common error is subtracting speeds (86=28 - 6 = 2) instead of accounting for the reversal of direction.

Study tip

Momentum is a vector: pick a positive direction before plugging in, and keep the sign on every velocity. Most collision errors are sign errors, not physics errors.