Question

A ball of mass '$m$' moving horizontally which velocity '$u$' hits a wedge of mass '$M$'. The wedge is situated on a smooth horizontal source. If after striking with wedge the ball starts moving in vertical direction and the wedge starts moving in horizontal plane. Calculated
a) The velocity of wedge $V$.
b) The velocity (v) at which the ball moves in vertical direction.
c) The impulse imparted by the ball on the wedge.
d) The coefficient of restitution $e=?$

Answer 1

a) As no external force is acting on the system in horizontal direction the linear momentum should be constant and conserved in horizontal direction.
Let velocity of wedge after collision be $V$.
Then $mu=MV⟹V=\frac{mu}{M}$ (i)
b) As there is no impulse on the ball in the direction parallel to slopping side hence the velocity of ball along the slope (tangent direction) should remain unchanged.
$⟹ucos\theta =\nu sin\theta$
which gives $\nu =ucot\theta$.
c) We can find the values $\nu$ and $V$ by impulse approach, when ball hits the wedge the impulse is generated between ball an wedge in the direction perpendicular to sloping surface. (normal direction)
For ball : $mu-Jsin\theta =0$
$Jsin\theta =mu$
$Jcos\theta =m\nu$
From Eqs. (i) and (ii), $\nu =ucot\theta$
From Eq. (iii), $J=mucosec\theta$
$Jsin\theta =MV⟹V=\frac{Jsin\theta }{M}$
which gives $V=\frac{mu}{M}$
d) the coefficient of restitution is given as
$({\nu }_2-{\nu }_1{)}_n=e(u_1-u_2{)}_n$
$[Vsin\theta -(-\nu cos\theta \left)\right]=e[usin\theta -0]$
$e=\frac{Vsin\theta +\nu cos\theta }{usin\theta }$
$e=\frac{V}{u}+\frac{\nu }{u}cot\theta =\frac{m}{M}+\left(cot\theta \right)cot\theta$
$e=\frac{m}{M}+co{t}^2\theta$