Question

A particle (a mud pallet, say) of mass $m$ strikes a smooth stationary wedge of mass $M$ with a velocity ${\nu }_0$, at an angle $\theta$ with horizontal. If the collision is perfectly inelastic, find the:
a) velocity of the wedge just after the collision.
b) change in $KE$ of the system $(M+m)$ in collision.

Answer 1

a) Let the system $(M+m)$ moves as a single mass with a velocity $\nu$.
Conserving the momentum of the system in horizontal, we have $m{\nu }_{0}cos\theta =(M+m)\nu$
This gives $\nu =\frac{m{\nu }_{0}cos\theta }{M+m}$
b) The change in $KE$ of the system is
$\Delta K=\frac{1}{2}(M+m){\nu }^2\frac{1}{2}m{\nu }_0^2$
where $\nu =\frac{m{\nu }_{0}cos\theta }{M+m}$
This gives $\Delta K=\frac{(M+m)}{2}{\left(\frac{m{\nu }_{0}cos\theta }{M+m}\right)}^2-\frac{1}{2}m{\nu }_0^2$
$=-\frac{m{\nu }_0^2}{2}(1-\frac{m}{M+m}{\cos }^2\theta )$
$=-\frac{(M+msi{n}^2\theta )m{\nu }_0^2}{2(M+m)}$