Question

A sphere of radius $R$ is in contact with a wedge. The point of contact is $\frac{R}{5}$ from the ground as shown in the figure. Wedge is moving with velocity $20m{s}^{-1}$ towards left. The velocity of the sphere at this instant will be

• A. $20m{s}^{-1}$
• B. $15m{s}^{-1}$
• C. $16m{s}^{-1}$
• D. $12m{s}^{-1}$

Answer 1

The correct option is B $15m{s}^{-1}$

The given arrangement can be depicted as shown below.

So, from the figure we have,
$\cos \theta =\frac{R-\frac{R}{5}}{R}=\frac{4}{5}$
Similarly, $\sin \theta =\frac{3}{5}$

As given, velocity of wedge along $PQ=20m/s$
$\therefore$ Its Velocity along $PO=20\sin \theta$
$=20\times \frac{3}{5}=12m/s$
From the constrained condition, the components of the velocities of sphere and wedge along the common normal should be equal.
$\Rightarrow V_s\cos \theta =12$
$\Rightarrow V_s=\frac{12}{4/5}=15m/s$