A cricket ball of mass 150 gm is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. The force of blow acts for 0.01s on the ball. The average force exerted by the bat on the ball is

480 N

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600 N

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500 N

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400 N

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Solution

The correct option is A
480 N

$v_{1} = - 12$ m\s and $v_{2} = + 20$ m/s [because direction is reversed]

$m = 150 g m = 0.15 k g, t = 0.01 s e c$

Force exerted by the bat on the ball $F = \frac{m [v_{2} - v_{1}]}{t} = \frac{0.15 [20 - (- 12)}{0.01}] = 480 N e w t o n$

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A cricket ball of mass 150 gm is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. The force of blow acts for 0.01s on the ball. The average force exerted by the bat on the ball is

The correct option is A 480 N v1=−12 m\s and v2=+20m/s [because direction is reversed] m=150 gm=0.15kg, t=0.01 sec Force exerted by the bat on the ball F=m[v2−v1]t=0.15[20−(−12)0.01]=480 Newton

The correct option is A
480 N

$v_{1} = - 12$ m\s and $v_{2} = + 20$ m/s [because direction is reversed]

$m = 150 g m = 0.15 k g, t = 0.01 s e c$

Force exerted by the bat on the ball $F = \frac{m [v_{2} - v_{1}]}{t} = \frac{0.15 [20 - (- 12)}{0.01}] = 480 N e w t o n$