A cricket ball of mass $200 g$ moving with a speed of $40 m s^{- 1}$ is brought to rest by a player in $0.04 s$ . Calculate the following :
(i) change in momentum of the ball,
(ii) average force applied by the player.

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Solution

$M a s s, m = 200 g = 0.2 k g$
$I n i t i a l v e l o c i t y, u = 40 m s^{- 1}$
$F i n a l v e l o c i t y, v = 0$
$T i m e, t = 0.04 s$

$I n i t i a l m o m e n t u m, p_{1} = m \times u = 0.2 \times 40 = 8 k g m s^{- 1}$
$F i n a l m o m e n t u m, p_{2} = m \times v = 0.2 \times 0 = 0$
$C h a n g e i n m o m e n t u m, p = p_{2} - p_{1} = 0 - 8 = - 8 k g m s^{- 1}$
$A v e r a g e f o r c e = \frac{m \times (v - u)}{t} = \frac{0.2 \times (0 - 40)}{0.04} - 200 N$

The negative sign shows that the force is applied in a direction opposite to the direction of motion of the ball.

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