Question

A ball of mass $m$ is released from $A$ inside a smooth wedge of mass $m$ as shown in the figure. What is the speed of the wedge when the ball reaches point $B$?

• A. ${\left(\frac{gR}{3\sqrt{2}}\right)}^{1/2}$
• B. $\sqrt{2gR}$
• C. ${\left(\frac{5gR}{2\sqrt{R}}\right)}^{1/2}$
• D. 0

Answer 1

The correct option is A ${\left(\frac{gR}{3\sqrt{2}}\right)}^{1/2}$

At point B,
let $u$ be velocity of wedge and $v$ be velocity of ball relative to the wedge,
Velocity of ball relative to ground is $=-v\cos {45}^{\circ }\hat{i}-v\sin {45}^{\circ }\hat{j}+u\hat{i}=-(v\cos {45}^{\circ }-u)\hat{i}-v\sin {45}^{\circ }\hat{j}$

From conservation of linear momentum
Initial momentum = Final momentum
$mu=m(v\cos {45}^{\circ }-u)$
$v\cos {45}^{\circ }=2u$
$v=2\sqrt{2}u$

By the law of conservation of energy
We know that –
Loss of P.E. = Gain in K.E.
P.E. at point B= K.E. before sliding + K.E. while sliding [Horizontal + vertical component]

$mgr\cos {45}^{\circ }=\frac{1}{2}m{u}^2+\frac{1}{2}m(v\cos {45}^{\circ }-u{)}^2$+$\frac{1}{2}m(v\sin {45}^{\circ }{)}^2$
$g\frac{r}{\sqrt{2}}=\frac{u^2}{2}+\frac{1}{2}(v\cos {45}^{\circ }-u{)}^2+\frac{1}{2}\frac{v^2}{2}$......(2)

Put the value of equation (1) in (2)
$\frac{gr}{\sqrt{2}}=\frac{u^2}{2}+\frac{1}{2}(2u-u{)}^2+\frac{1}{2}4u^2$
$\frac{gr}{\sqrt{2}}=\frac{u^2}{2}+\frac{u^2}{2}+2u^2$
$\frac{gr}{\sqrt{2}}=3u^2$
$u={\left(\frac{gR}{3\sqrt{2}}\right)}^{1/2}$