1. Define Masses and Variables

The problem contains a subtle definition regarding mass which is crucial for the solution.

• Let the mass of the small block be $m$.
• The problem states: “Total mass of the bowl and the block is $M$.”
• Therefore, the mass of the bowl alone is $M_{bowl} = M – m$.
• Let initial velocity given to the bowl be $v_0$.

2. Analyze Statement (a): Maximum Kinetic Energy

We need to determine the maximum kinetic energy of the bowl. The bowl starts with velocity $v_0$ and the block is initially at rest. As the block rises up the wall of the bowl, it gains potential energy at the expense of the system’s kinetic energy. The bowl slows down as it pushes the block.

Therefore, the Maximum Kinetic Energy of the bowl is at the initial instant (before it transfers any energy to the block).

To find this value in terms of $M, g, h$, we use the condition that the block rises to a maximum height $h$.

At maximum height $h$: The relative velocity between the block and bowl is zero. They move with common velocity $v_{com}$.

From Momentum Conservation:

$$(M_{bowl})v_0 = (M_{bowl} + m) v_{com} \implies (M-m)v_0 = M v_{com}$$

From Energy Conservation:

$$\frac{1}{2}(M-m)v_0^2 = \frac{1}{2}M v_{com}^2 + mgh$$

Substituting $v_{com} = \frac{M-m}{M}v_0$ into the energy equation:

$$\frac{1}{2}(M-m)v_0^2 – \frac{1}{2}M \left( \frac{M-m}{M} v_0 \right)^2 = mgh$$

$$\frac{1}{2}(M-m)v_0^2 \left[ 1 – \frac{M-m}{M} \right] = mgh$$

$$\frac{1}{2}(M-m)v_0^2 \left[ \frac{m}{M} \right] = mgh$$

Notice that the term $\frac{1}{2}(M-m)v_0^2$ is exactly the Initial Kinetic Energy of the Bowl ($K_{bowl,i}$).

$$K_{bowl,i} \cdot \frac{m}{M} = mgh \implies K_{bowl,i} = \frac{M}{m} (mgh) = Mgh$$

Thus, the maximum kinetic energy of the bowl is $Mgh$. Statement (a) is Correct.

3. Analyze Statement (b): Minimum Kinetic Energy

Statement (b): “If the bowl is lighter than the block, minimum kinetic energy of the bowl is zero.”

This condition implies $M_{bowl} < m$.

Consider the process after the block reaches max height and slides back down. When it reaches the bottom again, it’s equivalent to an elastic collision between the bowl (mass $M-m$) and the block (mass $m$).

If the bowl is lighter than the block ($M_{bowl} < m$), the collision logic dictates that the bowl will rebound (velocity reverses direction). Since the velocity changes from positive (initial) to negative (rebound), the velocity of the bowl must instantaneously be zero at some point.

Therefore, the minimum KE is zero. Statement (b) is Correct.

4. Analyze Statements (c) and (d)

Statement (c): To predict minimum KE, we must know the ratio of masses.

Yes. As seen in step 3, whether the velocity crosses zero (making min KE = 0) or simply drops to a non-zero minimum (like $v_{com}$) depends on whether $M_{bowl} < m$ or $M_{bowl} > m$. Statement (c) is Correct.

Statement (d): To predict maximum KE, we must know the ratio of masses.

No. We derived that Max KE $= Mgh$. This depends on the total mass $M$ and height $h$, but not the specific distribution (ratio) of $m$ and $M_{bowl}$. Statement (d) is Incorrect.