Solved: PHYSICS Question for NEET | Sushrut

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NEET PHYSICSMedium

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:

A

0.3 N

B

30 N

C

300 N

D

3000 N

Step-by-Step Solution

Convert Units: Mass $m = 150 \text{ g} = 0.15 \text{ kg}$ .
Analyze Momentum Change:

Initial velocity $u = 20 \text{ m/s}$ .
Final velocity $v = 0 \text{ m/s}$ (the ball stops).
Change in momentum $\Delta p = m(v - u) = 0.15 \times (0 - 20) = -3.0 \text{ kg m/s}$ .
Magnitude of momentum change $|\Delta p| = 3.0 \text{ kg m/s}$ [Source 56].

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: $F_{avg} = \frac{3.0 \text{ kg m/s}}{0.1 \text{ s}} = 30 \text{ N}$ [Source 57, 66].

Context: This principle explains why a cricketer pulls his hands back while catching a ball: increasing the time ( $\Delta t$ ) reduces the force ( $F$ ) exerted on the hands [Source 56].

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Solved: PHYSICS Question for NEET | Sushrut