# A ball of mass 200g moving with a velocity of 25 m/s is struck by the batsman to return along the same path in opposite direction with the same speed . the collision of the ball with bat lasts for 0.5 sec find the impulse of the force and tge average force of impact

Initial velocity= u

final velocity= -u

Change in momentum= m(u-(-u)) = 2mu

$2\times 0.2\times 25\Rightarrow 10Kg-m/s(N-s)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Force=Rateofchangeinmomentum\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}F=\frac{0.2\times 2\times 25}{0.5}\Rightarrow 20N$

Regards.

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