Q106 / 150 · JEE Main 2006 · Physics

A player caught a cricket ball of mass $150 \mathrm{~g}$ moving at a rate of $20 \mathrm{~m} / \mathrm{s}$. If the catching process is completed in $0.1 \mathrm{~s}$, the force of the blow exerted by the ball on the hand of the player is equal to

  • A
    $300 \mathrm{~N}$
  • B
    $150 \mathrm{~N}$
  • C
    $3 \mathrm{~N}$
  • Correct answer
    D
    $30 \mathrm{~N}$

Solution

$(m v-0) \Rightarrow 0.15 \times 20$ $\mathrm{F}=\frac{3}{0.1}=30 \mathrm{~N}$

Hence, the correct answer is option: 'D'.

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