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

The impulse-momentum theorem can determine the force exerted by the ball on the player's hand. Impulse equals the change in momentum, represented by the formula: \(F \cdot \Delta t = \Delta p\).

• \(F\) signifies force.
• \(\Delta t\) is the time interval (0.1 s).
• \(\Delta p\) denotes the change in momentum.

The change in momentum is calculated as:

\(\Delta p = m(v_f - v_i)\)

• \(m\) is the ball's mass (0.12 kg).
• \(v_f\) is the final velocity (0 m/s).
• \(v_i\) is the initial velocity (25 m/s).

Plugging in the values yields:

\(\Delta p = 0.12 \cdot (0 - 25) = -3 \, \text{kg} \cdot \text{m/s}\)

Applying the impulse-momentum theorem:

\(F \cdot 0.1 = -3\)

Solving for \(F\):

\(F = \frac{-3}{0.1} = -30 \, \text{N}\)

The negative sign indicates the force opposes the initial motion. The magnitude of the force is 30 N. The final answer is:

30