Question

A body of mass $2\text{kg}$ is sliding on an inclined plane as shown


Mass of the wedge is $4\text{kg}$. Find the velocity of the wedge just before the body reaches the ground.

• A. $4\text{m/s}$
• B. $2\text{m/s}$
• C. $6\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is B $2\text{m/s}$

Let velocity of the wedge be $u$ and $v$ be the velocity of the body with respect to the wedge.


Velocity of body A w.r.t ground in horizontal direction is
$v_{Ax}=(v\cos {45}^{\circ }-u)=(\frac{v}{\sqrt{2}}-u)$
Now applying momentum conservation in horizontal direction
$0=2(v\cos {45}^{\circ }-u)-4u$
$\Rightarrow 2u=\frac{v}{\sqrt{2}}-u$
$\Rightarrow v=3\sqrt{2}u$
So, $v_{Ax}=\frac{v}{\sqrt{2}}-u$
$=\frac{3\sqrt{2}u}{\sqrt{2}}-u=2u$

Vertical velocity of block A w.r.t ground is
$v_{Ay}=v\sin {45}^{\circ }=\frac{v}{\sqrt{2}}=\frac{3\sqrt{2}u}{\sqrt{2}}=3u$
Now applying energy conservation :
$m_{A}gh=\frac{1}{2}\times m_B\times u^2+\frac{1}{2}{m}_A{v}_A^2$
$2\times 10\times 3=\frac{1}{2}\times 4\times u^2+\frac{1}{2}\times 2\left[\right(2u{)}^2+\left(3u{)}^2\right]$
$60=2u^2+13u^2$
$60=15u^2$
$\Rightarrow u=2\text{m/s}$ is the velocity of wedge.