Question

Steel ball of mass $0.5kg$ is fastened to a cord $20m$ long and fixed at the far end and is released when the cord is horizontal. At the bottom of its path the ball strikes a $2.5kg$ steel block initially at rest on a frictionless surface. The collision is elastic. Find the speed of the block, just after the collision.

Answer 1

Let the ball strike the block with a speed $u_1$. Since the initial speed (speed before collision) of the block $=u_2=0$ for the perfectly elastic collision, the speed of the block is given as
$v_2=\frac{2m_1}{m_1+m_2}{u}_1=\frac{2\frac{m_1}{m_2}}{\frac{m_1}{m_2}+1}{u}_1$
where $u_1$ can be found by conserving energy of the ball between the position $A$ and $B$ before collision of $m_1$
$\Rightarrow \frac{1}{2}{m}_1{u}_1^2-m_{1}gl=0$
$\Rightarrow u_1=\sqrt{2gl}=\sqrt{2\times 10\times 20}=20m/sec$
$\frac{m_1}{m_2}=\frac{0.5kg}{2.5kg}=\frac{1}{5}$
By putting $u_1=20m/sec$ and $\frac{m_1}{m_2}=\frac{1}{5}$ in (a) we obtain
$v_2=\left\{\frac{2}{\frac{1}{5}+1}\right\}\left\{\frac{1}{5}\right\}\left(20\right)=\frac{20}{3}m/s$