A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process be completed

Question

A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process be completed in 0.1 s, then the force of the blow exerted by the ball on the hands of the player is:

Accepted Answer

1. **Convert Units:** Mass $m = 150 	ext{ g} = 0.15 	ext{ kg}$.
2. **Analyze Momentum Change:** 
   - Initial velocity $u = 20 	ext{ m/s}$.
   - Final velocity $v = 0 	ext{ m/s}$ (the ball stops).
   - Change in momentum $\Delta p = m(v - u) = 0.15 	imes (0 - 20) = -3.0 	ext{ kg m/s}$.
   - Magnitude of momentum change $|\Delta p| = 3.0 	ext{ kg m/s}$ [Source 56].
3. **Apply Impulse-Momentum Theorem:** The average force is defined as the rate of change of momentum (or Impulse divided by time).
   - $F_{avg} = \frac{|\Delta p|}{\Delta t}$
   - Substituting the values: $F_{avg} = \frac{3.0 	ext{ kg m/s}}{0.1 	ext{ s}} = 30 	ext{ N}$ [Source 57, 66].
4. **Context:** This principle explains why a cricketer pulls his hands back while catching a ball: increasing the time ($Δt$) reduces the force ($F$) exerted on the hands [Source 56].

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