Question

Find the height by which the sphere rises when the wedge touches the wall. Let the initial distance between the wedge and the wall is $10\text{m}$ and the wedge is moved to the left until it touches the wall. (Assume all surfaces to be smooth)

• A. $10\text{m}$
• B. $10\sqrt{3}\text{m}$
• C. $\frac{10}{\sqrt{3}}\text{m}$
• D. $5\sqrt{3}\text{m}$

Answer 1

The correct option is C $\frac{10}{\sqrt{3}}\text{m}$

Let the velocity of wedge be $v_w$ and that of the sphere be $v_s$ . Then the velocities are shown below


Sphere will not have any horizontal velocity as it is also in contact with the wall.

Let the time taken by the wedge to touch the wall be $t$ and the rise in the height of the sphere be $h$.

As per the question given, the distance travelled by the wedge is $10m$

Now, from wedge constraints, we have
$v_s\cos {30}^{\circ }=v_w\cos {60}^{\circ }$

$\Rightarrow v_s=\frac{v_{w}cos{60}^{\circ }}{cos{30}^{\circ }}=\frac{v_w}{\sqrt{3}}m/s$

Also, time taken by the wedge to touch the wall will be equal to the time to which the sphere will rise
$\therefore$ $t=\frac{10}{v_w}=\frac{h}{v_s}\Rightarrow h=10\times \frac{v_s}{v_w}=\frac{10}{\sqrt{3}}m$