Question

A player caught a cricket ball of mass \(150 \mathrm{~g}\) moving at the rate of \(20 \mathrm{~ms}^{-1}\). If the catching process be completed in $0.1 \mathrm{~s}$ the force of the blow exerted by the ball on the hands of player is (A) \(0.3 \mathrm{~N}\) (B) \(30 \mathrm{~N}\) (C) \(300 \mathrm{~N}\) (D) \(3000 \mathrm{~N}\)

Short answer

The magnitude of the force exerted by the ball on the player's hands is \(30 \mathrm{~N}\).

Step-by-step solution

Convert mass to kilograms.

First, we need to convert the mass of the ball from grams to kilograms, since all other units are in the SI system. To do this, just divide the given mass by 1000: \[ mass = \frac{150~g}{1000} = 0.15~kg \]

Calculate the initial momentum of the ball.

We'll call the initial velocity of the ball \(v_i =20~m/s\). Then, we can calculate the initial momentum of the ball as: \[ p_i = (mass)(v_i) = (0.15~kg)(20~m/s) = 3~kg \cdot m/s \]

Determine the final momentum of the ball.

Since the ball is caught and comes to a stop, its final velocity is zero (\(v_f = 0~m/s\)). Therefore, the final momentum of the ball is also zero: \[ p_f = (mass)(v_f) = (0.15~kg)(0~m/s) = 0~kg \cdot m/s \]

Calculate the change in momentum, or impulse.

The change in momentum, also known as impulse, can be calculated as the difference between the final and initial momenta: \[ \Delta p = p_f - p_i = 0~kg \cdot m/s - 3~kg \cdot m/s = -3~kg \cdot m/s \]

Use Newton's second law of motion to calculate the force.

Newton's second law of motion states that the force (\(F\)) acting on an object is equal to the change in momentum (\( \Delta p\)) divided by the time interval (\( \Delta t\)): \[ F = \frac{\Delta p}{\Delta t} \] Now we'll plug in the values we've calculated to find the force exerted by the ball on the hands of the player: \[ F = \frac{-3~kg \cdot m/s}{0.1~s} = -30~N \] As the force is negative, it means that the force exerted by the ball on the hands of the player is in the opposite direction of the ball's initial velocity. So the magnitude of the force: \[ |F| = 30 ~N \] Taking the magnitude of the force, the correct answer is (B) \(30 \mathrm{~N}\).