Solve the following :

Question:

A block of mass $\mathrm{m}$ is placed on a triangular block of mass $M$, which in turn is placed on a horizontal surface as shown in figure. Assuming frictionless surfaces between the velocities of the triangular block when the smaller block reaches the bottom end.

Solution:

Along x-direction ${ }^{a_{C O M}}=0$

$\Rightarrow \frac{m a_{x}-(M+m)\left(a_{2}\right)}{(M+m+m)}=0$

$\Rightarrow a_{2}=\frac{m a_{x}}{m+M}$

$m a=m g \sin \alpha$

$a_{x}=a \cos \alpha=g \sin \alpha \cos \alpha$

$a_{r}=$ resultant $=g \sin \alpha-a_{\alpha} \sin \alpha$

$=g \sin \alpha\left[\frac{M+m \sin ^{2} \alpha}{M+m}\right]$

Vertical distance travelled by $\frac{h}{\mathrm{~m}=\sin \alpha}=y$

Use law of kinematics,

$t=\sqrt{\frac{2 y}{a_{r}}}=\sqrt{\frac{2 h}{a_{r} \sin \alpha}}$

$\{$ velocity of $M$ block $\}$

$=\frac{m g \cos \alpha}{M+m} \sqrt{\frac{2 h \sin \alpha[M+m]}{g \sin \alpha \cdot\left[M+m \sin ^{2} \alpha\right]}}$

$=\sqrt{\frac{2 m^{2} g^{2} h \cos ^{2} \alpha}{(M+m)\left(M+m \sin ^{2} \alpha\right)}}$

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