A player caught a cricket ball of mass 150 ~g moving at a speed of 20 ~m / s . If the catching process is completed in 0.1 ~s , the magnitude of force exerted by the ball on the hand of the player is:

Laws of Motion PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q68 with Solution

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Correct answer: C

Given data
- Mass of ball: $m = 150\,\text{g} = 0.150\,\text{kg}$
- Initial speed: $u = 20\,\text{m/s}$
- Final speed after being caught: $v = 0$
- Time taken: $\Delta t = 0.1\,\text{s}$
Find change in momentum $\Delta p = m(v-u) = 0.150(0-20) = -3\,\text{kg m/s}$ Magnitude of change in momentum is: $|\Delta p| = 3\,\text{kg m/s}$
Use impulse-momentum theorem Average force magnitude is: $F = \frac{|\Delta p|}{\Delta t} = \frac{3}{0.1} = 30\,\text{N}$
Force exerted by ball on hand By Newton's third law, the magnitude of force exerted by the ball on the player's hand is equal to the magnitude of force exerted by the hand on the ball.
Check options
- A: $150\,\text{N}$ ❌
- B: $3\,\text{N}$ ❌
- C: $30\,\text{N}$ ✅
- D: $300\,\text{N}$ ❌

Therefore, the correct answer is Option C.