Q. A ball of mass 200 gm, moving with a speed of 40 m/s, is deflected exactly with the same speed but at $ 90{}^\circ $ with its incident direction after striking with a bat. If the striking time is 2 s, the average force acts on the ball is
AMUAMU 1998
Solution:
: Resolve $ {{v}_{1}} $ and $ {{v}_{2}} $ along the bat and perpendicular to it along the normal ON Since $ {{v}_{1}}={{v}_{2}}, $ components along normal cancel out. components along bat add up. $ \therefore $ change in velocity $ =({{v}_{1}}+{{v}_{2}})\cos 45{}^\circ $ $ =\frac{(v+v)\times 1}{\sqrt{2}}=\frac{2v}{\sqrt{2}}=\sqrt{2}v $ $ \therefore $ change in momentum = mass $ \times \sqrt{2}v=\sqrt{2}\,mv $ time $ =t $
$ \therefore $ Force $ =\frac{\sqrt{2}mv}{t} $ $ =\sqrt{2}\times \left( \frac{200}{1000} \right)\times \frac{40}{2}=4\sqrt{2}N $
