A player caught a cricket ball of mass $150 g$ moving at a rate of $20 m / s$ . If the catching process is completed in $0.1 s$ , the average force of the blow exerted by the ball on the hand of the player is equal to :

300 N

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150 N

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3 N

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30 N

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Solution

The correct option is C $3 N$
$Step 1: Change in momentum$
After catching, the final velocity of the ball will be zero.

Magnitude of change in momentum is given by:
$| Δ \to P | = |_{f} -_{i} |$
$= | m \to v - m \to u |$
$= | 0.15 k g \times 0 - 0.15 k g \times 20 m / s |$ $= 3 k g m / s$

$Step 2: Average Force exerted by the ball$
$F_{A v .} = \frac{Δ P}{Δ t}$
$= \frac{3 k g - m / s}{0.1 s e c}$
$\Rightarrow | \to F | = 30 N$

Hence the magnitude of the force exerted by the ball on the hand of players is $30 N$ .
Hence, Option $C$ is correct.

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