1. System Analysis

We analyze the system comprising the plank with the circular loop (mass $M = 0.8 \text{ kg}$) and the small ball (mass $m = 0.2 \text{ kg}$). Since the floor is frictionless, there is no external horizontal force, so linear momentum in the horizontal direction is conserved.

Initial State: Both are at rest. Ball is at height $h = 3R$.

Final State: Ball is at the topmost position of the loop (height $2R$).

2. Momentum Conservation

Let $V$ be the velocity of the plank to the right and $v_{rel}$ be the velocity of the ball relative to the plank to the left.

$$MV + m(V – v_{rel}) = 0 \implies V(M+m) = m v_{rel} \implies V = \frac{m v_{rel}}{M+m}$$

The velocity of the ball relative to ground is $v_b = V – v_{rel} = -v_{rel} \frac{M}{M+m}$.

3. Energy Conservation

Loss in PE = Gain in KE.

$$mg(3R) – mg(2R) = \frac{1}{2}MV^2 + \frac{1}{2}mv_b^2$$ $$mgR = \frac{1}{2}M \left(\frac{mv_{rel}}{M+m}\right)^2 + \frac{1}{2}m \left(\frac{Mv_{rel}}{M+m}\right)^2$$

Solving this yields: $\frac{v_{rel}^2}{R} = \frac{2g(M+m)}{M}$.

4. Force Analysis

At the top, equation of motion is: $N + mg = \frac{m v_{rel}^2}{R}$.

$$N = m \left[ \frac{2g(M+m)}{M} \right] – mg = mg \left( \frac{2(M+m)}{M} – 1 \right) = mg \left( 1 + \frac{2m}{M} \right)$$

Substituting values ($m=0.2, M=0.8, g=10$):

$$N = (0.2)(10) \left( 1 + \frac{0.4}{0.8} \right) = 2(1.5) = 3 \text{ N}$$