Question

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

Answer 1

The force is given by the rate of change of linear momentum of the ball, which comes to rest on the player's hand. Thus,

$\displaystyle\longrightarrow\sf{F=-\dfrac{150\times10^{-3}(0-20)}{0.1}}$

$\displaystyle\longrightarrow\sf{F=-150\times10^{-3}\times-20\times10}$

$\displaystyle\longrightarrow\sf{\underline{\underline{F=30\ N}}}$

Answer 2

$\huge{\underline{\sf{Answer:-}}}$

$\boxed{\sf{F = 30 N}}$

$\huge{\underline{\sf{Explainationtion:-}}}$

★ Given:-

• Mass of cricket ball, m = 150 g = 0.15 kg

• Initial velocity of the ball, u = 0

• Final velocity of the ball, v = 20 m/s

• Time taken in the catching process, t = 0.1 s    

→ F = ma

→ m × (v - u)/t

→ F = 0.15 kg × 20 - 0 / 0.1

→ $\boxed{\sf{F = 30N}}$

The force of the blow exerted by the ball on the hand of the player is equal to 30N.

Hence, this is the required solution.

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