A Pakistani player shoaib Akhtar bowls a cricket ball of mass 250 g with a speed of 120 km/hr towards the batsman dhoni. (assume that the ball strikes the bat with the same speed). The time of impact of the bat with the ball is 0.1 sec. After striking, if the ball bounces back with half it's speed before striking, find the force which dhoni exerts on the ball
Answers
A proper run-up : A bowler must accelerate as he gets closer to the stumps, while keeping both arms close to the body (close levers ensure no wastage of energy).
A high jump: The momentum generated by the run-up is transferred to the jump. That's why most genuine quick bowlers - Imran Khan, Brett Lee, Malcolm Marshall - have a reasonably high jump
A proper landing: When the front foot hits the ground, the force generated is transferred to the hip before moving upwards. The bowler rotates the shoulder, which uses the force it receives from the movement in the hip. This force is then transferred to the wrist. The result: The force generated from the run-up, the landing, the hip movement, the shoulder rotation and the wrist movement is translated into the speed of the ball. The more aligned the movements, the better the outcome.
Given:
The mass of the ball m= 250 g
The speed of the ball u= -120 Km/hr
The time taken t= 0.1 sec
To find:
The force exerted by the dhoni on ball
Solution:
The velocity of the ball after striking the bat
v= u/2
v=120/2
v=-60 m/sec
Now calculate the change in momentum:
Force×time = change in momentum
F×0.1=(0.25)×(2000/60-1000/60)
F×0.1=0.25×50
F=125 N
So, the force exerting on the ball is F=125 N