Question

Two particles, each of mass $m$, are connected by a light inextensible string of length $2l$. Initially they lie on a smooth horizontal table at points $A$ and $B$ distance $l$ apart. the particle at $A$ is projected across the table with velocity $u$. Calculate (in terms of $m$ and $u$) the impulse of tension in the string.

If u is along BA and v be the velocity of the second particle, find u/v

Answer 1

$v_{com}=\frac{mu+m\left(0\right)}{m+m}=\frac{u}{2}\to \left(1\right)$

where $v_{com}$ = velocity of centre of mass of two block system

when string is taught both will move same velocity let

$v_{com}=\frac{mv+mv}{m+m}\frac{u}{2}=v\to \left(11\right)$

Hence $\frac{u}{v}=\frac{u}{\left(\frac{u}{2}\right)}=2$