Question

A particle of mass 'm' strikes a smooth stationary wedge of mass M with a velocity $v_0$, at an angle $\theta$ with horizontal if the collision is perfectly inelastic, the impulse on the wedge is:

• A. $\frac{Mm{v}_0}{M+m}$
• B. $M{v}_0$
• C. $\frac{Mm{v}_{0}cos\theta }{M+m}$
• D. $\frac{Mm{v}_{0}sin\theta }{M+m}$

Answer 1

The correct option is C $\frac{Mm{v}_{0}cos\theta }{M+m}$

Initial momentum along x axis $=m{v}_0\cos \theta$

Final momentum $=(M+M)v$

$v=\frac{m{v}_0\cos \theta }{M+m}$

Impulse = Change in momentum

$=\frac{Mm{v}_0\cos \theta }{M+m}$