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A ball reaches a racket at 60 m/s along + X direction and leaves the racket in the opposite direction with the same speed. Assuming that the mass of the ball as 50gm and the contact time is 0.02 second, the force exerted by the racket on the ball is

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$3, 00, 000 N along + X direction$

$300 N along + X direction$

$300 N along - X direction$

$3, 00, 000 N alng - X direction$

answer is B.

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Detailed Solution

change in momentum = - 2mu $= - 2 \times 50 \times 10^{- 3} \times 60 = - 6 {kgms}^{- 1}$

Then force, $F = \frac{- 6}{0.02} = - 300 N$

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