impulse and momentum

Question

A 0.15 kg ball moving at 20 m/s is stopped by a catch in 0.05 s. What average force is exerted by the hand?

RankGuru · Accepted Answer

1. **Impulse-Momentum Theorem**: Impulse is defined as the change in momentum: $J = \Delta p = m(v_f - v_i)$.
2. **Step 1: Calculate Delta p**:
   - $m = 0.15$ kg, $u = 20$ m/s, $v = 0$ m/s.
   - $\Delta p = 0.15 	imes (-20) = -3$ kg·m/s.
3. **Step 2: Relate to Force**: Force is the rate of change of momentum (Newton's 2nd Law in derivative form).
   - $F_{avg} = \frac{\Delta p}{\Delta t}$
4. **Calculation**:
   - $|F_{avg}| = \frac{3 	ext{ kg·m/s}}{0.05 	ext{ s}} = \frac{300}{5} = 60$ N.
5. **Result**: The hand exerts an average force of **$60$ N**.

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