Question

A ball of mass 200 g moving with a velocity of 25 m/s is struck by the batsman to return along the same path in opposite direction with the same speed. The collision of the ball with the bat lasts for 0.5 s. Find the impulse of the force and the average force of impact.​

Answer 1

Explanation:

mass of the ball (m) =100g or 0.1kg

initial velocity (u) =25

s

m

time (t) = 0.025sec.

final velocity (v) = 0

Initial momentum = mu

= 0.1×25

= 2.5

s

kgm

Final momentum = mv (which is equal to 0, since v is zero)

Change in momentum = mv-mu

= 0 - 2.5 = -2.5

s

kgm

Average force =

time

Changeinmomentum

= -

0.025

(2.5)

= -100

s

kgm

= -100 N

The negative sign actually shows that the force was applied

opposite to the direction of the motion of the ball.

So, the average force applied by the player will be 100