Question

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

Answer 1

Answer: (C)

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