Question

A wedge of mass $M$ has one face making an angle $\alpha$ with horizontal and is resting on a smooth rigid floor. A particle of mass ${}^{\prime }{m}^{\prime }$ hits the inclined face of the wedge with horizontal velocity $v_0.$ It is observed that the particle rebounds in vertical direction after impact. Calculate the speed of particle after impact.
[Neglect friction between particle and the wedge and take $M=2m,v_0=10\text{m/s},\tan \alpha =2,g=10{\text{m/s}}^2.$]

• A. $5\text{m/s}$
• B. $0\text{m/s}$
• C. $20\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is B $0\text{m/s}$

After collision, assume speed of wedge is $v_1$ and speed of particle is $v_2$ as demonstrated in the below figure.

By momentum conservation along the wedge surface i.e. incline
$m{v}_0\cos \alpha =M{v}_1\cos \alpha +m{v}_2\sin \alpha$
$\Rightarrow m{v}_0\cos \alpha =2m{v}_1\cos \alpha +m{v}_2\sin \alpha (\because M=2m)$
(by dividing both sides with $\cos \alpha ,$ we get)
$v_0=2v_1+v_2\tan \alpha$
$\because (\tan \alpha =2)$
$\Rightarrow v_0=2v_1+2v_2$ ...(1)

Now, by momentum conservation perpendicular to the incline, we get
$m{v}_0\sin \alpha =M{v}_1\sin \alpha -m{v}_2\cos \alpha$
$\Rightarrow m{v}_0\sin \alpha =2m{v}_1\sin \alpha -m{v}_2\cos \alpha (\because 2m=M)$
(by dividing both sides with $\cos \alpha ,$ we get)
$v_0\tan \alpha =2v_1\tan \alpha -v_2$
$\because \tan \alpha =2$
$\Rightarrow 2v_0=4v_1-v_2$ ...(II)
Now, from eq. (I) and (II) we have
$v_0=2v_1$
and $v_2=0$
Hence, after the collision speed of particle along vertical direction is zero.