A cricket ball of mass 120 g is moving with a velocity of $12 m s^{- 1}$ , it turns back and move with a velocity of $20 m s^{- 1}$ when hit by a bat. Find the impulse and the force, if the force acts for 0.02 s.

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Solution

Here, mass of the ball, $m = 0.12 k g$
initial velocity, $u = 12 m / s$
final velocity, $v = - 20 m / s$
Impulse = mass X change in velocity
$I = 0.12 (- 20 - 12) = 0.12 (- 32) = - 3.84 N s$
Force $= - \frac{I}{t} = \frac{3.84}{0.02} = - 192 N$

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A cricket ball of mass 120 g is moving with a velocity of 12ms−1, it turns back and move with a velocity of 20ms−1 when hit by a bat. Find the impulse and the force, if the force acts for 0.02 s.

Here, mass of the ball, m=0.12kginitial velocity,u=12m/sfinal velocity, v=−20m/sImpulse = mass X change in velocity I=0.12(−20−12)=0.12(−32)=−3.84NsForce =−It=3.840.02=−192N

A cricket ball of mass 120 g is moving with a velocity of $12 m s^{- 1}$ , it turns back and move with a velocity of $20 m s^{- 1}$ when hit by a bat. Find the impulse and the force, if the force acts for 0.02 s.

Here, mass of the ball, $m = 0.12 k g$
initial velocity, $u = 12 m / s$
final velocity, $v = - 20 m / s$
Impulse = mass X change in velocity
$I = 0.12 (- 20 - 12) = 0.12 (- 32) = - 3.84 N s$
Force $= - \frac{I}{t} = \frac{3.84}{0.02} = - 192 N$