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:
The force exerted by the cricket ball on the player's hand can be determined using the impulse-momentum theorem, which states that impulse equals the change in momentum. This is mathematically represented as:
\(F \cdot \Delta t = \Delta p\)
Where:
\(F\) represents the force.
\(\Delta t\) is the duration of the force application.
\(\Delta p\) is the momentum change.
First, calculate the initial and final momentum:
Initial momentum \(p_i = m \cdot v_i\), with \(m = 0.15 \text{ kg}\) (ball's mass in kg) and \(v_i = 20 \text{ m/s}\). Consequently, \(p_i = 0.15 \times 20 = 3 \text{ kg m/s}\).
Final momentum \(p_f = 0 \text{ kg m/s}\) as the ball is caught and stopped.
The momentum change is \(\Delta p = p_f - p_i = 0 - 3 = -3 \text{ kg m/s}\).
The magnitude of the impulse is 3 N s (ignoring the negative sign).
Next, calculate the force using the formula:
\(F = \frac{\Delta p}{\Delta t}\)
Substituting the values yields:
\(F = \frac{3}{0.1} = 30 \text{ N}\)
Therefore, the magnitude of the force the ball exerts on the player's hand is 30 N.
