Question

A symmetric block of mass $m_1$ with groove of hemispherical shape of radius 'r' rest on a smooth horizontal surface in contact with the wall as shown in the figure. A small block of mass $m_2$ slides without friction from the initial position. Find the maximum velocity of the block $m_1$

Answer 1

The block will touch the wall until the washer comes to the lowest position. By this instant of time, the washer has acquired the velocity $v$ which can be determined from the energy law: $v^2=2gr$. During the subsequent motion of the system, the washer will " climb " the right-hand side of the block, accelerating it all the time in the rightward direction until the velocities of the washer and the block become equal. Then the washer will slide down the block, the block being accelerated until the washer passes through the lowest position. Thus, the block will have the maximum velocity at the instants at which the washer passes through the lowest position during its backward motion relative to the block. In order to calculate the maximum velocity of the block, we shall write the momentum conservation law for the instant at which the block is separated from the wall:
$m_2\sqrt{2gr}=m_1{v}_1+m_2{v}_2$, and the energy conservation law for the instants at which the washer passes through the lowest position:
$m_{2}gr=\frac{m_1{v}_1^2}{2}+\frac{m_2{v}_2^2}{2}$.
This system of equations has two solutions:
(1) $v_1=0,v_2=\sqrt{2gr}$,
(2) $v_1=\frac{2m_2}{m_1+m_2}\sqrt{2gr},v_2=\frac{m_2-m_1}{m_1+m_2}\sqrt{2gr}$.
Solution (1) corresponds to the instants at which the washer moves and the block is at rest. We are interested in solution (2) corresponding to the instants when the block has the maximum velocity :
$v_{1max}=\frac{2m_2\sqrt{2gr}}{m_1+m_2}$.