Question

There spheres, each of mass m, can slide freely on a frictionless horizontal surface. Spheres A and B are attached to an inextensible inelastic cord of length l and are at rest in the position shown when sphere B is struck squarely by sphere C which is moving to the right with a velocity $v_0$. Knowing that the cord is taut when sphere B is struck by sphere C and assuming perfectly elastic impact between B and C, determine the velocity of each sphere immediately after impact.

Answer 1

After collision velocity of $ABC$ in common along $A\to B$ i.e. $u\cos 60°$.

Now according to momentum conservation in $x$-direction,

$m_c{v}_0=(m_B+m_c)u+m_{A}u{\cos }^{2}60m{v}_0=2mu+\frac{m}{4}u4v_0=9uu=\frac{4}{9}{v}_0$

Thus after collision, hence,

$v_B=v_C=\frac{4}{9}{v}_0{v}_A=v_B\cos 60°=v_C\cos 60°=\frac{4}{9}{v}_0\cos 60°v_A=\frac{4}{18}{v}_0$