OnlineCBT
Physics · LAWS OF MOTION
हिन्दीA cricket ball of mass \( m = 150\text{ g} \) moving with a speed of \( v = 20\text{ m/s} \) is caught by a player. If the player stops the ball in \( \Delta t_1 = 0.1\text{ s} \), average force \( F_1 \) is exerted. If the player pulls his hands back extending the stopping time to \( \Delta t_2 = 0.4\text{ s} \), average force \( F_2 \) is exerted. Find \( F_2 \) and \( F_1 \) respectively.
Correct answer · Option 2
\( 7.5\text{ N} \) and \( 30\text{ N} \)
Explanation
Change in momentum \( |\Delta p| = m v = 0.15 \times 20 = 3\text{ N s} \). Force \( F_1 = \frac{3}{0.1} = 30\text{ N} \). Force \( F_2 = \frac{3}{0.4} = 7.5\text{ N} \). Thus \( F_2 \) and \( F_1 \) are \( 7.5\text{ N} \) and \( 30\text{ N} \) respectively.
