JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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 _______.
- A \(150 \mathrm{~N}\)
- B \(3 \mathrm{~N}\)
- C \(30 \mathrm{~N}\)
- D \(300 \mathrm{~N}\)
Answer & Solution
Correct Answer
(C) \(30 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{\mathrm{mv}-0}{0.1}\) \(=\frac{150 \times 10^{-3} \times 20}{0.1}=30 \mathrm{~N}\)
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