Question

A block of mass $m_1$ is projected with a velocity ${\nu }_0$ so that it climbs onto the smooth wedge of mass $m_2$. If the block does not leave the wedge find the maximum height attained by the block.

Answer 1

At the maximum height, both $m_1$ and $m_2$ ahve same velocity as the relative sliding between them stops. Since, no external force acts horizontally, kinear momentum is conserved in horizontal direction.
$({\overrightarrow{P}}_x{)}_f=({\overrightarrow{P}}_x{)}_i$
Where, ${\overrightarrow{P}}_{xi}=m_1{\nu }_0\hat{i}$ and ${\overrightarrow{P}}_{xf}=(m_1+m_2)\nu \hat{i}$
Then we have $(m_1+m_2)\nu =m_1{\nu }_0$
WE theorem: $W_{ext}+W_{int}=\Delta K$ (i)
Where $W_{ext}=-m_{1}gh,W_{int}=W_{contact}=0$.
and $\Delta K=\left[\frac{1}{2}\right(m_1+m_2){\nu }^2-\frac{m_1{\nu }_0^2}{2}]$
This gives $-m_{1}gh=\frac{1}{2}(m_1+m_2){\nu }^2-\frac{m_1{\nu }_0^2}{2}$ (ii)
Substituting $\nu$ from Eq. (i) in Eq. (ii), we have
$h=\frac{{\nu }_0^2}{2(1+\frac{m_1}{m_2})g}$