Aditya's challenges in Mechanics 2

Since the only external forces acting on the two blocks are gravity and the normal reaction with the ground, the horizontal component of momentum is conserved. SInce the system is initially at rest, the horizontal component of the centre of mass of the system remains constant. As the small block moves down the surface, the large block will move away from the wall. When the small block reaches the lip of the semicircle, at the side nearest the wall, the system will be instantaneously at rest, and the same motion will repeat, but in the opposite direction, with the smaller block coming to rest at the lip of the semicircle, and the larger block coming to rest just touching the wall, as it was when it started. Thus the maximum speed of the small block on the return trip is the same as the maximum speed of the small block on the outward journey.

Suppose that $x$ is the horizontal distance of the centre of the semicircle on the big block from the wall, and let $\theta$ be the angle that the small block makes with the downward vertical. Thus $9x + (x + 10\sin\theta)$ remains constant, and so $\dot{x} = -\cos\theta\dot\theta$ Conservation of energy tells us that $\tfrac92\dot{x}^2 + \tfrac12\big[(\dot{x} + 10\cos\theta\dot{\theta})^2 + 100\sin^2\theta \dot\theta^2\big] \; = \; 100\cos\theta$ and hence, using the relationship between $\dot{x}$ and $\dot\theta$ , we deduce that $\dot\theta^2 \; = \; \frac{20\cos\theta}{9 + \sin^2\theta}$ If $V$ is the speed of the small block, then $V^2 \; = \; (\dot{x} + 10\cos\theta\dot\theta)^2 + 100\sin^2\theta\dot\theta^2 \; = \;\frac{20(81 + 19\sin^2\theta)\cos\theta}{9 + \sin^2\theta}$ Thus $V$ is maximized when $\theta=0$ , and so the maximum value of $V$ is $6\sqrt{5} = \boxed{13.4164}$ .

we can solve this problem with a very short method..... the max velocity of small block will be when it is at the bottom of the bigger block.... let this max velocity be v and the velocity of m2 at that moment be u...then using conservation of momentum.. we can write m1v = m2u also using energy conservation - m1(v^2)/2 + m2(u^2)/2 = m1gr hence solving both the equations .. we get (v^2) = 2gr(m2)/((m1)+(m2)) hence we get v = sqrt(180)