Question

Find the velocity of separation of the spheres $1$ and $2$ if the sphere $3$ comes down with a velocity of $3m/s$. Assume all spheres to be identical and frictionless.

• A. $2\sqrt{3}m/s$
• B. $4\sqrt{3}m/s$
• C. $6\sqrt{3}m/s$
• D. $8\sqrt{3}m/s$

Answer 1

The correct option is C $6\sqrt{3}m/s$

The spheres are identical, the centre to centre distance between the spheres will be equal and thus, they will form an equilateral triangle.

So, the value of $\theta$ will be ${30}^{\circ }$ in the above figure shown.

By condition of wedge constraint, acceleration along normal to the common interface should be same.
So, $V_3\cos \theta =V_2\sin \theta$
$\Rightarrow V_2=V_3\cot \theta =3\cot {30}^{\circ }$
$\because \theta \text{is}{30}^{\circ }$
$\Rightarrow V_2=3\sqrt{3}$
Also, due to similarity $V_1$ will be $3\sqrt{3}$
Now, the spheres $1$ and $2$ will move apart in the opposite direction
$\therefore$ velocity of separation will be
$V_{12}=V_1-(-V_2)=3\sqrt{3}+3\sqrt{3}=6\sqrt{3}m/s$