Momentum · AP · 5 min
Rebounds double the momentum change if the speed reverses direction.
01 The trap
Wrong path
Students subtract speeds instead of velocities and lose the sign flip.
Why it feels right
The object may leave with the same speed it arrived with, which makes the change feel like zero.
02 Correct model
Impulse equals change in momentum: J = m(v_f - v_i). Choose a positive direction and keep it.
03 Mini-example
Prompt
A 0.20 kg ball hits a wall at 5 m/s and rebounds at 5 m/s. What is |J|?
Common wrong answer
Zero, because the speed is unchanged.
Correct reasoning
2.0 kg m/s, because velocity changes from -5 to +5 m/s.
Diagnostic cue
Any bounce or rebound is a sign-convention problem waiting to happen.
04 Guided practice
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
A 0.15 kg ball moving right at 8 m/s rebounds left at 8 m/s. Taking right as positive, find the impulse on the ball.
Use velocities with signs.
J = m(v_f - v_i) = 0.15(-8 - 8) = -2.4 N·s. The impulse points left.
Trap: Calling the impulse zero because the speed is unchanged.
A force pulse rises linearly to 120 N and returns to zero over 0.050 s. Find the impulse.
Impulse is area under F(t).
The area is ½(0.050)(120) = 3.0 N·s.
Trap: Multiplying peak force by total time as if the graph were a rectangle.
A ball hits a wall. Case A: it stops. Case B: it rebounds with the same speed. Which case has larger impulse magnitude?
Compare the change in velocity.
The rebound has larger impulse magnitude. Stopping changes velocity from v to 0; rebounding changes it from v to -v, a change of magnitude 2v.
Trap: Thinking rebound is gentler because the final speed is familiar.
A 0.10 kg ball moving right at 20 m/s rebounds left at 15 m/s after a 0.010 s contact. Find the average force on the ball, taking right as positive.
Use J = m(v_f - v_i) = F_avg Δt.
J = 0.10(-15 - 20) = -3.5 N·s. Thus F_avg = -3.5/0.010 = -350 N, so the average force is 350 N left.
Trap: Subtracting speeds as 20 - 15 and missing the direction reversal.
A 0.18 kg ball approaches a wall at 12 m/s and rebounds at 8.0 m/s. Take away from the wall as positive. What is the impulse on the ball?
The initial velocity is -12 m/s and the final velocity is +8.0 m/s.
J = 0.18[8.0 - (-12)] = +3.6 N·s, directed away from the wall.
Trap: Subtracting the speed magnitudes and obtaining -0.72 N·s.
05 Mock checks