Question

Two particles $A$ and $B$ each of mass $m$ are attached by a light inextensible string of length $2l$. The whole system lies on a smooth horizontal table with $B$ initially at a distance $l$ from $A$. The particle at end $B$ is projected across the table with speed $u=4\sqrt{3}m/s$ perpendicular to $AB$. Find velocity of ball $A$ just after the jerk?

Answer 1

Initial velocity along $AB=u\cos \theta$

And perpendicular to$AB=u\sin \theta$

When the string becomes, but the two bodies would acquire the same velocityalong $AB$

By conservation of momentum,

$2mv=mu\cos \theta v=\frac{u\cos \theta }{2}=\frac{u}{2}\times \frac{4l^2-l^2}{2l}=\frac{u}{2}\times \frac{\sqrt{3}l}{2l}=\frac{\sqrt{3}}{4}u$