Question

A small particle of mass $m=2\text{kg}$ moving with constant horizontal velocity $u=10{\text{ms}}^{-1}$ strikes a wedge shaped block of mass $M=4\text{kg}$ placed on the smooth surface as shown in figure. After collision, the particle starts moving up the inclined plane. Find the velocity of the wedge immediately after collision-

• A. $1{\text{ms}}^{-1}$
• B. $3{\text{ms}}^{-1}$
• C. $2{\text{ms}}^{-1}$
• D. $4{\text{ms}}^{-1}$

Answer 1

The correct option is C $2{\text{ms}}^{-1}$


As given, after the collision particle starts moving up the inclined plane, it means particle velocity with respect to the wedge is along the incline i.e., $V_{rel}$.

Force on the wedge at the time of the collision will be perpendicular to the inclined plane. Hence, velocity will not change along the inclined plane.

$u\cos \alpha =V_{rel}+V\cos \alpha$

$\Rightarrow \frac{u}{\sqrt{2}}=V_{rel}+\frac{V}{\sqrt{2}}[\because \alpha ={45}^{\circ }]$

$V_{rel}=\frac{u-V}{\sqrt{2}}......\left(1\right)$

Horizontal component of velocity of particle after collision,

$V_x=V_{rel}\cos {45}^{\circ }+V=\frac{V_{rel}}{\sqrt{2}}+V$

Vertical component of velocity after collision,

$V_y=V_{rel}\sin {45}^{\circ }=\frac{V_{rel}}{\sqrt{2}}$

At the time of collision, there will be an external impulse force from the ground act on the wedge in vertical direction only. So, along horizontal direction, momentum will be conserved.

$mu=m{V}_x+MV$

$mu=m(\frac{V_{rel}}{\sqrt{2}}+V)+MV........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$mu=m(\frac{u-V}{2}+V)+MV$

$\Rightarrow 2\times 10=2(\frac{10-V}{2}+V)+4V$

$\Rightarrow 20=10+5V$

$\therefore V=2{\text{ms}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{C}\right)$ is the correct answer.

Why this question ? Caution: Since there will be impulsive force from the ground to wedge in vertical direction. Thus, momentum is not conserved in the vertical direction. Only horizontal direction momentum will be conserved.