A cricket ball of mass 70 g moving with a velocity of 0.5 m/s is stopped by a player in 0.5 s. What is the force applied by the player to stop the ball?
Here, mass of the ball m=70 g=701000 kg.=7100 kg.m=70 g=701000 kg.=7100 kg. The initial velocity of the ball u=0.5 m/su=0.5 m/s The final velocity of the ball v=0v=0 Time taken t=0.5 st=0.5 s Therefore, acceleration a=v−uta=v−ut =0−0.50.5=−1 ms−2=0−0.50.5=−1 ms−2 Therefore, force applied by the player to stop the ball F=maF=ma =7100 kg.×−1 ms−2=7100 kg.×−1 ms−2 =−0.07 N=−0.07 N Therefore, the 0.07 N0.07 N force is applied by the player to stop the ball in the opposite direction which is indicated by −ve−ve sign.
