Question

Can anyone please answer this question?

Please give a correct and useful answer, I'll mark your answer as brainliest.

Answer 1

Explanation:

Question

Answers

Related Questions

A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in constant for 0.001 s, the force that the batsman had to apply to hold the bat firmly at its place would be:

A. 10.510.5 N

B. 2121N

C. 1.05×1041.05×104 N

D. 2.1×1042.1×104N



Answer

Verified

79.5K+ Views

Hint: The force required to hold the ball at its place by the batsman is equal to the difference of final momentum and initial momentum divided by the time. The product of mass and velocity is called momentum.

Complete step by step answer:A cricket ball of mass m is 150g = 0.150 Kg(1g = 0.001Kg) moving with a speed u of 126 km/h hits the bat. The collision between them is completely elastic and the two remain in contact for time t=0.001s. The ball moves back with the same velocity as the collision is elastic but in an opposite direction i.e., v=-u.

We change the velocity from km/h to m/s, velocity = 126×518=35ms−1[1kmh−1=518ms−1]126×518=35ms−1[1kmh−1=518ms−1]

Force applied by the batsman (F)= Rate of change of momentum

F= (Final momentum −Initial momentum)timeF= (Final momentum −Initial momentum)time

F=(mv−mu)tF=(mv−mu)t

F=m(−u

hope it help u