Question

A rod of mass m and length l is kept on a smooth wedge of mass m

Answer 1

We can use conservation of energy to solve this... or by using forces (normal) between wedge M and the rod m.

Here on the system of rod m and wedge M, work is done only by the gravitational force on rod m.

The PE lost by the rod :   m g H

Wedge displaces horizontally by x. The rod displaces vertically by y.
=> y = x tanФ.        Differentiate wrt time, to get:
=> v1 = v2 tanФ
=> a1 = a2 tanФ

When the rod touches the ground, velocity of wedge = v2.
KE of Wedge = 1/2 M v2²
KE of the rod  = 1/2 m v1² = 1/2 m v2 tan²Ф

Using conservation of energy:  m g H = 1/2 v2² (M + m tan²Ф)

v2² = 2 m g H /(M + m tan²Ф)

v_{wedge}=\sqrt{\frac{2mgH}{M+mtan^\phi}}vwedge​=M+mtanϕ2mgH​​ 
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If we use the method of forces, then Normal reaction force N between the rod m and wedge M gives the acceleration to Wedge.

Draw the Free body diagram for wedge and the rod.

N SinФ = M a₂
m g - N CosФ = m a₁ = m a₂ tanФ
=>  m g - M a₂ CotФ = m a₂ TanФ
=>  a₂ = mg Tan Ф /(M+m tan²Ф )

Distance traveled by the wedge: H CotФ   (the base of wedge)
=>  v₂² = 2 a₂ H CotФ

Hence we get the same answer as above.