Question

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):

• A. 24
• B. 12
• C. 25
• D. 30

Answer 1

The Correct Option is D

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 \( 30 \, \text{N} \).