A cricket player catches a ball of mass 120 g moving with 25 m/s speed. If the catching process is completed in 0.1 s then the magnitude of force exerted by the ball on the hand of player will be (in SI unit):
- 24
- 12
- 25
- 30
The Correct Option is D
Solution and Explanation
Given: - Mass of the ball: \( m = 120 \, \text{g} = 0.12 \, \text{kg} \) - Initial speed of the ball: \( v = 25 \, \text{m/s} \) - Time taken to catch the ball: \( t = 0.1 \, \text{s} \) - Final speed of the ball: \( v_f = 0 \, \text{m/s} \) (since the ball is caught and comes to rest)
Step 1: Calculating the Change in Momentum
The change in momentum (\( \Delta p \)) of the ball is given by:
\[ \Delta p = m \cdot (v_f - v) \]
Substituting the given values:
\[ \Delta p = 0.12 \cdot (0 - 25) \, \text{kg} \cdot \text{m/s} \] \[ \Delta p = -3 \, \text{kg} \cdot \text{m/s} \]
The negative sign indicates a decrease in momentum.
Step 2: Calculating the Force Exerted
The force exerted by the ball on the hand of the player is given by Newton’s second law:
\[ F = \frac{\Delta p}{t} \]
Substituting the values:
\[ F = \frac{-3}{0.1} \, \text{N} \] \[ F = -30 \, \text{N} \]
The magnitude of the force is:
\[ |F| = 30 \, \text{N} \]
Conclusion:
The magnitude of the force exerted by the ball on the hand of the player is \( 30 \, \text{N} \).
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