CritABCD
A.2 · 6 · A.2 Forces and momentum · change in momentum

Momentum change: four cases

When the direction reverses, treat momentum as a vector. Use a number line.

SL+HLHarder16 minFree
Hook

A ball hits a wall at 5 m s⁻¹ and bounces back at 2 m s⁻¹. Is its change in momentum 3 units, or something else?

The four cases
1 / 3

A 0.2 kg ball, starting at 5 m s⁻¹

CaseΔp = m(v_{f} − v_{i})
1 web0.2 × (5 − 5) = 0 kg m s⁻¹no change
2 paper, out at 30.2 × (3 − 5) = −0.4 kg m s⁻¹decrease
3 stopped0.2 × (0 − 5) = −1.0 kg m s⁻¹
4 rebound at 20.2 × (−2 − 5) = −1.4 kg m s⁻¹the tricky one ★
Check 1 of 4

Case 2: 0.2 kg ball, 5 m s⁻¹ down to 3 m s⁻¹ (same direction). Δp is…

Check 2 of 4

Case 3: the same ball is brought to rest. Δp is…

Check 3 of 4

A ball’s velocity goes from +5 m s⁻¹ to −2 m s⁻¹. The change in velocity is…

Check 4 of 4

Which mistake gives the wrong answer for a rebound?

Apply

A 0.50 kg ball hits a wall at 4.0 m s⁻¹ and rebounds at 1.0 m s⁻¹. Find the change in momentum. Show your choice of positive direction.

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Remember earlier lessons
From lesson A.2 · 5

Momentum is a…

From lesson A.2 · 2

F_net = …

Summary card

Key points

  • • Δp = m(v_f − v_i), with signs.
  • • If the direction reverses, one velocity is negative.
  • • Use the number line: the change is the arrow from i to f.

Formulas

  • Δp = pf − pi = mΔv

Key terms

Change in momentum
:
final momentum minus initial momentum, a vector

Common errors

  • • Subtracting speeds instead of velocities.
  • • Dropping the sign of the final velocity after a rebound.
Checks solved: 0 of 4