Question

A sphere of radius $R$ is in contact with a wedge. The point of contact is at a distance $\frac{R}{5}$ from the ground as shown in the figure. The wedge is moving with velocity $20\text{m/s}$ towards left, then the velocity $v$ of the sphere at this instant will be:

• A. $20\text{m/s}$
• B. $15\text{m/s}$
• C. $16\text{m/s}$
• D. $12\text{m/s}$

Answer 1

The correct option is B $15\text{m/s}$

Velocity of sphere and wedge at the given instant are shown in the figure given below. $\theta$ is the angle made by normal reaction $N$ on the sphere from the wedge, with the horizontal.


Normal reaction will act in a direction perpendicular to the inclined surface and passes through the centre of the sphere. Hence, from the geometry of the figure:
$\sin \theta =\frac{\left(\frac{4R}{5}\right)}{R}=\frac{4}{5}$
$\therefore \theta ={53}^{\circ }$

To maintain contact at the surface, the component of velocity of the wedge and sphere should be equal along the direction perpendicular to the inclined surface (or normal direction).
$v_2=v_1$ (from the diagram)
$\Rightarrow v\cos (90-\theta )=20\cos \theta$
$\Rightarrow v\cos {37}^{\circ }=20\cos {53}^{\circ }$
$\therefore v=\frac{(20\times \frac{3}{5})}{\frac{4}{5}}=15\text{m/s}$