Question

A ball of mass ${}^{\prime }{m}^{\prime }$ moving horizontally which velocity ${}^{\prime }{u}^{\prime }$ hits a wedge of mass ${}^{\prime }{M}^{\prime }$. the wedge is situated on a smooth horizontal surface. If after striking wedge the ball starts moving in vertical direction and the wedge starts moving in horizontal plane. Calculate the coefficient of restitution $e$.

• A. $\frac{M}{m}+ta{n}^2\theta$
• B. $\frac{m}{M}+co{t}^2\theta$
• C. $\frac{m}{M}+ta{n}^2\theta$
• D. $\frac{M}{m}+co{t}^2\theta$

Answer 1

The correct option is C

$\frac{m}{M}+ta{n}^2\theta$

As no external force are acting on the system in horizontal direction the linear momentum should be constant conserved in horizontal direction.

Let velocity of wedge after collision be $V$.

Then $mu$ = $MV$ $\Rightarrow V$ = $\frac{mu}{M}$. . . . . . (i)

As there is no impulse on the ball in the direction parallel to sloping side hence the velocity of ball along the slope (tangent direction) should remain unchanged.

$\Rightarrow ucos\theta$ = $vsin\theta$ which gives $v$ = $ucot\theta$ . . . . . . (ii)

The coefficient of restitution is given as

$(v_2-v_1{)}_{normal}$ = $e(u_1-u_2{)}_{normal}$

$[Vsin\theta -(-vcos\theta \left)\right]$ = $e[usin\theta -0]$

$e$ = $\frac{vsin\theta +vcos\theta }{usin\theta }$

$e$ = $\frac{V}{u}+\frac{v}{u}cot\theta$ = $\frac{m}{M}+\left(cot\theta \right)cot\theta$ (from(i)&(ii))

$e$ = $\frac{m}{M}+co{t}^2\theta$