A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m tra

Q: Why do we conserve angular momentum about the center of mass?

Conserving angular momentum about the center of mass is mathematically convenient because the torque due to pseudo forces (if we were in an accelerating frame) is zero. In the ground frame, the net external torque is zero about any point, but using the CM decouples the translational and rotational kinetic energies.

Q: Why is the final velocity of the particle taken as zero?

The problem explicitly states, "After collision, particle comes to rest." This simplifies our final momentum and energy equations significantly.

Q: How did we get the velocity of the point of contact as $v + \omega \frac{L}{2}$?

For a rigid body undergoing combined translation and rotation, the velocity of any point is the vector sum of the velocity of the CM and the velocity of that point relative to the CM. Since the rod rotates with $\omega$ and the end is at a distance $\frac{L}{2}$, the relative velocity is $\omega \frac{L}{2}$ in the same direction as $v$.

Q: Could we have used Conservation of Kinetic Energy instead of the Coefficient of Restitution?

Yes, absolutely! Since the collision is elastic, kinetic energy is conserved. However, using the coefficient of restitution equation ($e=1$) is often algebraically simpler because it avoids dealing with squared velocity terms.

Q: What would change if the surface had friction?

If the surface had friction, external impulsive forces would act during the collision, meaning linear momentum would not be conserved. The rod would also experience a torque from friction, complicating the angular momentum conservation.

Question

A rod of mass $M$ and length $L$ is lying on a horizontal frictionless surface. A particle of mass $m$ travelling along the surface hits at one end of the rod with a velocity $u$ in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $\left(\frac{m}{M}\right)$ is $\frac{1}{x}$. The value of $x$ will be ...... .

Sigma Percentile · Accepted Answer

### The Setup: A Dance of Particles and Rigid Bodies

Imagine a pristine, frictionless horizontal surface. On this surface lies a uniform rod of mass $M$ and length $L$, perfectly at rest.

Suddenly, a small particle of mass $m$ comes hurtling towards it with a velocity $u$. It doesn't just hit anywhere; it strikes the rod exactly at one of its ends.

The problem states that this collision is **perfectly elastic**, and immediately after the impact, the particle comes to a dead stop. Our mission is to find the ratio of their masses, $\frac{m}{M}$.

### The Three Pillars of Collision Mechanics

To solve this elegant problem, we need to rely on three fundamental pillars of mechanics.

First, because the surface is frictionless, there are no external horizontal forces acting on the system. This means **Linear Momentum is conserved**.

Second, there are no external torques acting on the system about the rod's center of mass. Therefore, **Angular Momentum is conserved**.

Third, the collision is perfectly elastic. This gives us our energy constraint, most easily expressed through the **Coefficient of Restitution ($e=1$)**.

### Pillar 1: Conservation of Linear Momentum

Let's look at the system before and after the collision. Initially, only the particle is moving. Its momentum is $mu$.

After the collision, the particle stops completely. Where did all that momentum go? It was transferred entirely to the rod. If the rod's center of mass moves with a velocity $v$, its momentum is $Mv$.

Equating the two, we get:
$$mu = Mv$$

From this, we can easily isolate the velocity of the rod's center of mass:
$$v = \frac{m}{M}u$$

### Pillar 2: Conservation of Angular Momentum

Now, let's analyze the rotational aspect. We will calculate the angular momentum about the rod's center of mass.

Before the collision, the particle is moving along a line that is at a perpendicular distance of $\frac{L}{2}$ from the rod's center. The initial angular momentum is simply the linear momentum multiplied by this lever arm:
$$L_{	ext{initial}} = mu \left(\frac{L}{2}ight)$$

After the collision, the rod starts spinning with an angular velocity $\omega$. The angular momentum of a rigid body rotating about its center of mass is $I\omega$. For a uniform rod, the moment of inertia $I$ is $\frac{ML^2}{12}$.
$$L_{	ext{final}} = \left(\frac{ML^2}{12}ight)\omega$$

Equating the initial and final angular momenta:
$$mu \frac{L}{2} = \frac{ML^2}{12} \omega$$

Solving for $\omega$, we find:
$$\omega = \frac{6mu}{ML}$$

### Pillar 3: The Elasticity Constraint

Because the collision is perfectly elastic, the **velocity of approach must equal the velocity of separation**.

Before the collision, the particle approaches the stationary rod with velocity $u$. So, $v_{	ext{approach}} = u$.

After the collision, the particle is at rest. But what about the point on the rod where the impact occurred? This point has two velocity components: the translational velocity of the center of mass ($v$) and the tangential velocity due to rotation ($\omega \frac{L}{2}$).

Since both components are in the same direction, the total velocity of the contact point is $v + \omega \frac{L}{2}$. This is our velocity of separation.

Setting them equal ($e=1$):
$$v + \frac{\omega L}{2} = u$$

### Bringing It All Together: The Final Calculation

We have our master equation. Now, we just need to substitute the expressions for $v$ and $\omega$ that we found earlier.

Substitute $v = \frac{m}{M}u$ and $\omega = \frac{6mu}{ML}$:
$$\frac{m}{M}u + \left(\frac{6mu}{ML}ight)\frac{L}{2} = u$$

Notice how the $L$ cancels out in the second term:
$$\frac{m}{M}u + \frac{3m}{M}u = u$$

Combine the terms on the left side:
$$\frac{4m}{M}u = u$$

The initial velocity $u$ cancels out from both sides, leaving us with a beautiful, clean ratio:
$$\frac{m}{M} = \frac{1}{4}$$

The problem defines this ratio as $\frac{1}{x}$. Therefore, comparing the two, we can confidently conclude that:
$$x = 4$$

Enter Numerical Value:

Visualized Solution

\text{Initial Setup}

\text{Governing Principles}

\text{Conservation of Linear Momentum}

\text{Conservation of Angular Momentum}

\text{Coefficient of Restitution}

\text{Substituting Values}

\text{Final Calculation}

\text{Food for Thought}

The Sigma Insight: Conservation of Angular Momentum

The Setup

The Three Pillars of Collision Mechanics

Pillar 1

Pillar 2

Pillar 3

Bringing It All Together

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