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

Answer: (D)

30

Answer 2

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 $30 \, \text{N}$.