A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the ho

Q: Why is the horizontal momentum conserved but not the vertical?

In the horizontal direction, there are no external forces acting on the system (the table is frictionless). However, in the vertical direction, gravity acts downwards and the normal force from the table acts upwards. Because the center of mass accelerates downwards as the small mass slides, the net vertical force is not zero, so vertical momentum is not conserved.

Q: How did we get the absolute displacement of $m$ as $(d + R)$?

Displacement is a vector. The small mass $m$ moves a distance $R$ to the right *relative to the block*. Meanwhile, the block itself moves a distance $d$ (which will turn out to be negative, meaning it moves left) relative to the ground. The absolute displacement of $m$ relative to the ground is the sum of these two vectors: $d + R$.

Q: Why do we use the relation $P^2 / 2m$ for kinetic energy?

It's a mathematical shortcut! Since $P = mv$, we can write $P^2 = m^2 v^2$. If we divide this by $2m$, we get $\frac{m^2 v^2}{2m} = \frac{1}{2} m v^2$, which is exactly the kinetic energy. This form is incredibly useful when we know that the momenta of two objects are equal, as it allows us to instantly find the ratio of their kinetic energies.

Q: Does the normal force do any work here?

The normal force between the block and the table does no work because the block only moves horizontally, perpendicular to the normal force. The normal force between the small mass and the circular cut does no work *on the system* because it is an internal force, and the surfaces are frictionless, meaning the force is always perpendicular to the relative motion.

Q: Why is option (c) incorrect if it looks so similar to the displacement?

Option (c) includes an extra $\sqrt{2}$ factor. As we derived, the final position of the small mass is exactly equal to the displacement of the block, $d = -\frac{mR}{M + m}$. The $\sqrt{2}$ is a distractor designed to test if you carefully calculated the final absolute position rather than just guessing based on the formula's shape.

Question

A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $x = 0$, in a coordinate system fixed to the table. A point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is $x$ and the velocity is $v$. At that instant, which of the following option is/are correct?

Options:

(A) The velocity of the point mass $m$ is $v = \sqrt{\frac{2gR}{1 + \frac{m}{M}}}$

(B) The $x$ component of displacement of the centre of mass of the block $M$ is $-\frac{mR}{M + m}$

(C) The position of the point mass is $x = -\sqrt{2} \frac{mR}{M + m}$

(D) The velocity of the block $M$ is $V = -\frac{m}{M} \sqrt{2gR}$

Sigma Percentile · Accepted Answer

This problem is a beautiful symphony of two of the most powerful conservation laws in physics: the conservation of momentum and the conservation of mechanical energy. When you look at a system where parts are moving relative to each other, it can initially seem chaotic. But by anchoring our perspective to the center of mass, the chaos resolves into elegant mathematical relationships.

### Analyzing the Setup

Imagine a block of mass $M$ resting on a perfectly smooth, frictionless table. This block has a quarter-circular cut of radius $R$ on its top right corner. A small point mass $m$ is released from the very top of this circular cut. As the small mass slides down the curve, it pushes against the block. Because the table is frictionless, the block will start sliding in the opposite direction.

Here is the crucial insight: there are absolutely **no external horizontal forces** acting on the entire system (the block plus the small mass). Gravity acts downwards, and the normal force from the table acts upwards, but horizontally, the system is isolated. This means the horizontal position of the center of mass of the system must remain perfectly stationary throughout the motion.

### The Master Equation: Center of Mass

Let's define our coordinates. The right edge of the block is initially at $x = 0$. The small mass starts at the top of the cut, which is at a horizontal distance $R$ to the left of the right edge, so its initial position is $x_i = -R$.

Let the horizontal displacement of the block $M$ be $d$. Since the block is pushed to the left, we expect $d$ to be a negative value. As the small mass $m$ slides down to the bottom of the cut, it moves a horizontal distance $R$ *relative to the block*. Therefore, the absolute horizontal displacement of the small mass with respect to the ground is $(d + R)$.

Because the center of mass does not shift horizontally, the sum of the mass-weighted displacements must be zero:

$$M \Delta x_M + m \Delta x_m = 0$$

Substituting our expressions for the displacements:

$$M d + m(d + R) = 0$$

Expanding and solving for $d$:

$$Md + md + mR = 0$$
$$d(M + m) = -mR$$
$$d = -\frac{mR}{M + m}$$

This perfectly matches option (b). Now, what about the final position of the small mass $m$? Its final position $x$ is simply its initial position plus its absolute displacement:

$$x = x_i + (d + R) = -R + d + R = d$$

So, the final position of the small mass is also $x = -\frac{mR}{M + m}$. This reveals that option (c) is incorrect, as it contains an extra $\sqrt{2}$ factor.

### Final Calculation: Energy and Momentum

Now, let's determine the velocities of the two masses at the instant the small mass loses contact with the block (which happens at the bottom of the circular cut).

As the small mass descends a vertical distance $R$, the system loses gravitational potential energy equal to $mgR$. Since all surfaces are frictionless, this potential energy is entirely converted into the kinetic energy of the two masses:

$$K_m + K_M = mgR$$

Furthermore, because the initial horizontal momentum of the system was zero, the final horizontal momenta of the two masses must be equal in magnitude and opposite in direction:

$$P_m = P_M$$

We can express kinetic energy in terms of momentum using the relation $K = \frac{P^2}{2 	imes 	ext{mass}}$. Since their momenta are equal, we can write:

$$2m K_m = 2M K_M$$

This gives us the ratio of their kinetic energies:

$$\frac{K_m}{K_M} = \frac{M}{m}$$

This is a profound result: the kinetic energy is distributed inversely proportional to their masses. The lighter object gets a larger share of the energy. We can express the kinetic energy of the small mass $m$ as a fraction of the total energy:

$$K_m = \frac{M}{M + m} (mgR)$$

Now, we substitute the standard formula for kinetic energy, $K_m = \frac{1}{2} m v^2$:

$$\frac{1}{2} m v^2 = \frac{M}{M + m} (mgR)$$

Solving for $v$:

$$v = \sqrt{\frac{2MgR}{M + m}}$$

To match the format of option (a), we divide the numerator and denominator inside the square root by $M$:

$$v = \sqrt{\frac{2gR}{1 + \frac{m}{M}}}$$

This confirms that option (a) is correct.

Finally, let's check the velocity of the large block, $V$. Using the same logic, its kinetic energy is:

$$K_M = \frac{m}{M + m} (mgR)$$

Substituting $K_M = \frac{1}{2} M V^2$:

$$\frac{1}{2} M V^2 = \frac{m}{M + m} (mgR)$$

$$V = \sqrt{\frac{2m^2 gR}{M(M + m)}} = \frac{m}{M} \sqrt{\frac{2MgR}{M + m}}$$

Comparing this to option (d), we see that option (d) is missing the $(1 + \frac{m}{M})$ term inside the square root. Therefore, option (d) is incorrect.

By systematically applying conservation laws, we have successfully navigated through the relative motions and energy transformations of this elegant mechanics problem.

Select Answer:

Visualized Solution

Initial Setup and Constraints

Conservation of Center of Mass

Calculating Block's Displacement

Displacement of Block $M$

Final Position of Mass $m$

Conservation Laws for Velocity

Relating Kinetic Energies

Kinetic Energy of Mass $m$

Velocity of Mass $m$

Velocity of Block $M$

The Sigma Insight: Motion of Centre of Mass

Analyzing the Setup

The Master Equation

Final Calculation

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