Question

Brian Lara is a cricketer playing in the field on the second day of a cricket test-match. He exerts a forward force on the 0.145kg cricket ball, as he catches it, to bring it to rest from a speed of 38.2m/s. During the process, his hand recoils a distance of 0.135m. Determine the acceleration of the ball and the force which is applied to it by Brian Lara.

Answer 1

a = -3984.6 m/s²

F = 577.76 N

Step-by-step explanation:

The acceleration of the ball can be calculated by using the third equation of motion:

`2as = v_f^2 - v_i^2\\`

where,

a = acceleration of ball = ?

s = distance covered = recoil distance = 0.135 m

vf = final speed = 0 m/s

vi = initial speed = 38.2 m/s

Therefore,

`2(0.135\ m)a = (0\ m/s)^2-(38.2\ m/s)^2\\`

a = -3984.6 m/s²

here negative sign shows deceleration.

Now, for the force applied by Brian Lara will be equal in magnitude but opposite in direction of the force required to stop the ball:

`F = -ma\\F = -(0.145\ kg)(-3984.6\ m/s^2)\\`

F = 577.76 N