question_answer

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

A)  4.0 N            

B)  \[4/\sqrt{2}N\]

C)  \[4\sqrt{2}N\]           

D)  zero

Correct Answer: C

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\[=\frac{\sqrt{2}mv}{t}\] \[=\sqrt{2}\times \left( \frac{200}{1000} \right)\times \frac{40}{2}=4\sqrt{2}N\]