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:

Question

Accepted Answer

Correct answer: C. (c) The force exerted by the ball on the hand can be calculated using the formula derived from Newton's second law of motion, which is F= p t, where F is the force, p represents the change in momentum, and t is the time over which this change occurs. The change in momentum, p, can be calculated as the difference between the final momentum, p_f, and the initial momentum, p_i. In this scenario, because the ball comes to a stop in the player's hand, its final velocity (and hence, its final momentum) is 0 . Therefore, the change in momentum is equal to the initial momentum of the ball (since final momentum is zero). The initial momentum, p_i, of the ball can be calculated using the formula p= mv, where m is the mass of the ball and v is its velocity. Given that the mass of the ball is 150 g=0.15 kg (converting grams to kilograms) and its velocity is 20 m/s, we have: p_i= (0.15 kg ) (20 m/s )=3 kg m/s Since the change in momentum, p, equals the initial momentum ( p_i ) because the final momentum is 0, the force exerted can be found by substituting p and t into the first formula: F= 3 kg m/s0.1 s=30 N Therefore, the magnitude of force exerted by the ball on the hand of the player is 30 N, which corresponds to Option C.

$A player caught a cricket ball of mass 150 g moving at a speed of 20 m / s . If the catching process is completed in 0.1 s, the magnitude of force exerted by the ball on the hand of the player is:$

Solution

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