Question

A rod of mass m and length l is kept on a smooth wedge of mass M as shown in the figure. If the system is released, find the speed of the wedge when the rod hits the ground level, neglecting friction in all contacting surfaces.

• A. =
• B. =
• C. =
• D. None of these

Answer 1

The correct option is A

=

Let the speeds of M and m be $v_1$ and $v_2$ respectively when the rod descends through a vertical distance y. Hence the change in KE of the system.

$\Delta KE=\frac{1}{2}M{V}_1^2+\frac{1}{2}m{v}_2^2$
The change in gravitational potential energy of the system = $\Delta PE$ = -mg y (since the rod moves down).
Conservataion of energy yields,$\Delta PE+\Delta KE=0because{W}_N=0)$
$\Rightarrow -mgy+\frac{1}{2}M{v}_1^2+\frac{1}{2}m{v}_2^2=0$--------------------------(i)
Kinematics $y=xtan\theta \Rightarrow \frac{dy}{dt}=\frac{dx}{dt}tan\theta \Rightarrow v_2=v_{1}tan\theta$ --------------(ii)
Eliminating $v_2$ from (1) and (2),we obtain
$-mgy+\frac{1}{2}M{V}_1^2+\frac{1}{2}m(v_{1}tan\theta {)}^2=0\Rightarrow v_1=\sqrt{\frac{2mgy}{M+mta{n}^2\theta }}$
Putting y = H,we obtain $v_m$ =$\sqrt{\frac{2mgH}{M+mta{n}^2\theta }}$