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

Q: Why is angular momentum conserved about the center of mass?

Because the net external force on the system in the horizontal plane is zero, and the collision forces are internal. The torque about the CM is zero.

Q: Why does the particle come to rest after the collision?

The problem states it comes to rest. This is a given condition that simplifies our conservation equations.

Q: How do we know the collision is elastic?

The problem explicitly states 'The collision is elastic', which means we can use the conservation of kinetic energy.

Q: What is the instantaneous center of rotation?

It's the point on a rigid body (or its extension) that has zero velocity at a specific instant. Here, it's point $P$.

Q: Why is the velocity of $P$ after time $t$ the vector sum of two perpendicular velocities?

After rotating by $90^\circ$, the rod is perpendicular to its initial orientation. The CM continues to move horizontally, but the rotational velocity of $P$ is now vertical.

Question

A rod $AB$ of mass $M$ and length $L$ is lying on a horizontal frictionless surface. A particle of mass $m$ travelling along the surface hits the end $A$ of the rod with a velocity $v_0$ in a direction perpendicular to $AB$. The collision is elastic. After the collision, the particle comes to rest.
(a) Find the ratio $m/M$.
(b) A point $P$ on the rod is at rest immediately after collision. Find the distance $AP$.
(c) Find the linear speed of the point $P$ a time $\pi L / 3v_0$ after the collision.

Sigma Percentile · Accepted Answer

The collision of a particle with a rigid rod is a classic problem that beautifully weaves together the three fundamental conservation laws of mechanics: linear momentum, angular momentum, and kinetic energy. In this epic journey, we will dissect the motion of the rod, locate its instantaneous center of rotation, and track its movement through time.

### Analyzing the Setup

Imagine a uniform rod $AB$ of mass $M$ and length $L$ lying peacefully on a smooth, frictionless horizontal surface. Suddenly, a small particle of mass $m$ strikes its end $A$ with a velocity $v_0$. The collision is perfectly elastic, and the particle stops dead in its tracks.

Because the particle stops, it transfers all its momentum and energy to the rod. The rod, now free from any external horizontal forces, will begin to translate and rotate simultaneously. Let's denote the velocity of its center of mass (CM) as $v$ and its angular velocity as $\omega$.

### The Master Equations

To unravel the kinematics of the rod, we must invoke our three conservation laws.

**1. Conservation of Linear Momentum:**
Since there are no external horizontal forces, the initial momentum of the particle must equal the final momentum of the rod.
$$mv_0 = Mv$$

**2. Conservation of Angular Momentum:**
The net torque about the center of mass is zero, meaning angular momentum about the CM is conserved. The particle strikes at a distance of $L/2$ from the CM.
$$mv_0 \left(\frac{L}{2}ight) = I\omega$$
Substituting the moment of inertia of a uniform rod about its CM, $I = \frac{ML^2}{12}$, we get:
$$mv_0 \frac{L}{2} = \frac{ML^2}{12} \omega \implies mv_0 = \frac{ML\omega}{6}$$

**3. Conservation of Kinetic Energy:**
The problem explicitly states the collision is elastic. Thus, the initial kinetic energy of the particle equals the sum of the translational and rotational kinetic energies of the rod.
$$\frac{1}{2}mv_0^2 = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$$
$$mv_0^2 = Mv^2 + \frac{ML^2}{12}\omega^2$$

### Solving for the Mass Ratio

Now comes the algebraic magic. We have three equations and three unknowns. Let's substitute $v$ and $\omega$ from the first two equations into the kinetic energy equation.

From our momentum equations, we know $v = \frac{m}{M}v_0$ and $\omega = \frac{6mv_0}{ML}$. Substituting these into the energy equation:
$$mv_0^2 = M\left(\frac{m}{M}v_0ight)^2 + \frac{ML^2}{12}\left(\frac{6mv_0}{ML}ight)^2$$

Notice how the $v_0^2$ terms cancel out beautifully! Simplifying the expression yields:
$$m = \frac{m^2}{M} + \frac{3m^2}{M}$$
$$m = \frac{4m^2}{M} \implies \mathbf{\frac{m}{M} = \frac{1}{4}}$$

### Locating the Instantaneous Center of Rotation

Next, we need to find a point $P$ on the rod that is momentarily at rest right after the collision. This point is known as the instantaneous center of rotation.

If $P$ is at a distance $x$ from the center of mass towards $B$, its net velocity is the vector sum of the CM's translational velocity $v$ and the rotational velocity $\omega x$. For $P$ to be at rest, these two must perfectly cancel each other out:
$$v = \omega x \implies x = \frac{v}{\omega}$$

Substituting the values of $v$ and $\omega$ we found earlier:
$$x = \frac{mv_0/M}{6mv_0/ML} = \frac{L}{6}$$

The question asks for the distance from $A$, which is $AP$. Since $A$ is $L/2$ away from the CM, we simply add the distances:
$$AP = \frac{L}{2} + \frac{L}{6} = \mathbf{\frac{2L}{3}}$$

### Tracking the Rod Through Time

Finally, we need the velocity of point $P$ after a specific time $t = \frac{\pi L}{3v_0}$. First, let's find out how much the rod has rotated in this time. The angle $	heta$ is simply $\omega$ multiplied by $t$:
$$	heta = \omega t = \left(\frac{6mv_0}{ML}ight) \left(\frac{\pi L}{3v_0}ight)$$

Using our mass ratio $m/M = 1/4$, we find:
$$	heta = 2\pi \left(\frac{m}{M}ight) = 2\pi \left(\frac{1}{4}ight) = \frac{\pi}{2}$$

This means the rod has rotated exactly $90^\circ$ and is now perpendicular to its initial position!

### The Final Velocity of Point P

Visualize the rod now. The center of mass is still moving horizontally with velocity $v$. However, because the rod has rotated by $90^\circ$, the rotational velocity of point $P$, which is $\omega x$, is now directed vertically.

Since we established earlier that $\omega x = v$, the total velocity of $P$ is the vector sum of two perpendicular velocities, both of magnitude $v$:
$$|\vec{v}_P| = \sqrt{v^2 + (\omega x)^2} = \sqrt{v^2 + v^2} = \sqrt{2}v$$

Substituting $v = \frac{m}{M}v_0 = \frac{1}{4}v_0$, we arrive at our final, elegant answer:
$$|\vec{v}_P| = \sqrt{2} \left(\frac{1}{4}v_0ight) = \mathbf{\frac{v_0}{2\sqrt{2}}}$$

What a spectacular demonstration of rotational dynamics!

Visualized Solution

$Collision Setup$

$Conservation of Linear Momentum$

$Conservation of Angular Momentum$

$Conservation of Kinetic Energy$

$Solving for \frac{m}{M}$

$Locating Point P$

$Calculating Distance A P$

$Rotation after time t$

$Velocity of P after time t$

The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Master Equations

Solving for the Mass Ratio

Locating the Instantaneous Center of Rotation

Tracking the Rod Through Time

The Final Velocity of Point P

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Collision Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Linear Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Angular Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Kinetic Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Mm​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Locating Point P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Distance AP

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rotation after time t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity of P after time t