A cricket ball of mass 100 g is moving with speed of 25 m/s and is brought to rest in 0.025 s. find the average force applied.

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Solution

mass of the ball (m) =100g or 0.1kg
initial velocity (u) =25 $\frac{m}{s}$
time (t) = 0.025sec.
final velocity (v) = 0
Initial momentum = mu
= $0.1 \times 25$
= 2.5 $\frac{k g m}{s}$
Final momentum = mv (which is equal to 0, since v is zero)
Change in momentum = mv-mu
= 0 - 2.5 = -2.5 $\frac{k g m}{s}$
Average force = $\frac{C h a n g e i n m o m e n t u m}{t i m e}$

= - $\frac{(2.5)}{0.025}$

= -100 $\frac{k g m}{s}$
= -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

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