A player caught a cricket ball of mass $150 \mathrm{~g}$ moving at a speed of $20 \mathrm{~m} / \mathrm{s}$.…

A player caught a cricket ball of mass $150 \mathrm{~g}$ moving at a speed of $20 \mathrm{~m} / \mathrm{s}$. If the catching process is completed in $0.1 \mathrm{~s}$, the magnitude of force exerted by the ball on the hand of the player is:
  1. $3 \mathrm{~N}$
  2. $300 \mathrm{~N}$
  3. $150 \mathrm{~N}$
  4. $30 \mathrm{~N}$

Solution

$\begin{aligned} & \mathrm{F}=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{\mathrm{mv}-0}{0.1} \\ & =\frac{150 \times 10^{-3} \times 20}{0.1}=30 \mathrm{~N}\end{aligned}$

Asked in: JEE Main 2024 (08 Apr Shift 1)

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