A bar of mass M = 1.00 kg and length L = 0.20 m is lying on a horizontal frictionless surface. One e

Q: Why isn't linear momentum conserved in this collision?

Great question! In a typical free collision, linear momentum is always conserved. However, here the rod is attached to a fixed pivot. During the impact, the pivot exerts an impulsive reaction force on the rod to keep it in place. Because of this external force, the linear momentum of the system changes.

Q: Why is the final angular momentum of the particle negative?

Angular momentum is a vector quantity. If we take the initial direction of the particle's motion as positive, it creates a counter-clockwise angular momentum about the pivot. After the collision, it bounces back and moves in the opposite direction, creating a clockwise angular momentum. Hence, the negative sign!

Q: How do we calculate the velocity of separation correctly?

Velocity of separation is the relative speed at which the two objects move apart after the collision. The point on the rod moves forward with speed $\omega L/2$, while the particle moves backward with speed $v$. Since they are moving in opposite directions, their relative speed of separation is simply the sum of their speeds: $\omega L/2 + v$.

Q: Can we use conservation of kinetic energy instead of the restitution equation?

Yes, you absolutely can! Since the collision is perfectly elastic, kinetic energy is conserved. However, the kinetic energy equation involves squared terms ($v^2$ and $\omega^2$), which leads to a messy quadratic equation. The restitution equation is linear and much easier to solve algebraically.

Q: What is the moment of inertia of the rod, and why?

The rod is rotating about its end (the pivot), not its center of mass. The moment of inertia of a uniform rod about its center is $ML^2/12$. Using the parallel axis theorem, shifting the axis to the end adds $M(L/2)^2$, giving a total moment of inertia of $ML^2/3$.

Question

A bar of mass $M = 1.00	ext{ kg}$ and length $L = 0.20	ext{ m}$ is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass $m = 0.10	ext{ kg}$ is moving on the same horizontal surface with $5.00	ext{ m s}^{-1}$ speed on a path perpendicular to the bar. It hits the bar at a distance $L/2$ from the pivoted end and returns back on the same path with speed $v$. After this elastic collision, the bar rotates with an angular velocity $\omega$. Which of the following statement is correct ?

Options:

(A) $\omega = 6.98	ext{ rad s}^{-1}$ and $v = 4.30	ext{ m s}^{-1}$

(B) $\omega = 3.75	ext{ rad s}^{-1}$ and $v = 4.30	ext{ m s}^{-1}$

(C) $\omega = 3.75	ext{ rad s}^{-1}$ and $v = 10.0	ext{ m s}^{-1}$

(D) $\omega = 6.80	ext{ rad s}^{-1}$ and $v = 4.10	ext{ m s}^{-1}$

Sigma Percentile · Accepted Answer

### The Beauty of Collisions and Rotations

Imagine a pristine, horizontal frictionless surface. A uniform bar is resting peacefully, pivoted at one of its ends.

Suddenly, a small mass comes hurtling towards it, moving perpendicular to the bar. This isn't just a simple collision; it's a beautiful interplay of linear motion transforming into rotational motion.

When the mass strikes the bar exactly at its midpoint, it transfers energy and momentum. But because the bar is pinned at one end, it can't just slide away. It must rotate.

This problem is a classic test of your understanding of **Conservation of Angular Momentum** and the kinematics of **Elastic Collisions**. Let's break it down step by step.

### Analyzing the Setup

Before we dive into the equations, let's establish our physical parameters. We have a bar of mass $M = 1.00 	ext{ kg}$ and length $L = 0.20 	ext{ m}$.

The small mass $m = 0.10 	ext{ kg}$ approaches with an initial speed $u = 5.00 	ext{ m/s}$. It strikes the bar at a distance of $L/2$ from the pivot.

After the collision, the particle bounces back with a speed $v$, and the bar begins to swing with an angular velocity $\omega$.

Our goal is to find the exact values of $\omega$ and $v$. To do this, we need two independent equations.

### The Master Equation: Angular Momentum

Why can't we use linear momentum conservation here? The answer lies at the pivot.

During the impact, the pivot exerts a massive, unknown impulsive force on the rod to keep it anchored. This external force destroys the conservation of linear momentum for the system.

However, if we calculate the torque about the pivot point itself, the lever arm for this pivot force is zero. Thus, the **net external torque about the pivot is zero**.

This is our golden ticket: **Angular momentum about the pivot is conserved**.

Let's write down the initial angular momentum. It's entirely due to the incoming particle:

$$L_{	ext{initial}} = m u \left(\frac{L}{2}ight)$$

After the collision, the particle bounces back, so its angular momentum reverses direction. The rod also gains angular momentum as it rotates:

$$L_{	ext{final}} = -m v \left(\frac{L}{2}ight) + I \omega$$

Equating the two gives us our master equation:

$$m u \frac{L}{2} = -m v \frac{L}{2} + I \omega$$

### Calculating the Moment of Inertia

Before we plug in the numbers, we need the moment of inertia $I$ of the rod.

Since the rod is rotating about its end, we use the standard formula:

$$I = \frac{M L^2}{3}$$

Substituting the given values $M = 1.00 	ext{ kg}$ and $L = 0.20 	ext{ m}$:

$$I = \frac{1.00 	imes (0.20)^2}{3} = \frac{0.04}{3} 	ext{ kg m}^2$$

Now, let's substitute everything into our angular momentum equation:

$$0.10 	imes 5.00 	imes 0.10 = -0.10 	imes v 	imes 0.10 + \frac{0.04}{3} \omega$$

Simplifying this yields:

$$0.05 = -0.01 v + \frac{0.04}{3} \omega$$

Multiplying the entire equation by 100 to clear the decimals:

$$5 = -v + \frac{4}{3} \omega$$

And multiplying by 3 gives us a clean, linear equation:

$$15 = -3v + 4\omega$$

This is our first crucial relationship between $v$ and $\omega$.

### The Geometry of the Bounce: Coefficient of Restitution

We have two unknowns, so we need a second equation. The problem states that the collision is **elastic**.

This means kinetic energy is conserved, but using the energy equation leads to messy quadratics. Instead, we use the **Coefficient of Restitution ($e$)**, which is exactly $1$ for an elastic collision.

The restitution equation relates the relative velocities before and after the impact:

$$e = \frac{v_{	ext{separation}}}{v_{	ext{approach}}} = 1$$

The velocity of approach is simply the initial speed of the particle, $u = 5.00 	ext{ m/s}$.

The velocity of separation requires careful thought. After the hit, the point on the rod at $L/2$ moves forward with a tangential speed of $\omega (L/2)$.

Simultaneously, the particle bounces backward with speed $v$. Since they are moving in opposite directions, their relative speed of separation is the sum of their individual speeds:

$$v_{	ext{separation}} = \omega \frac{L}{2} - (-v) = \omega \frac{L}{2} + v$$

Setting up the restitution equation:

$$1 = \frac{\omega (0.10) + v}{5.00}$$

$$5.00 = 0.10 \omega + v$$

Rearranging for $v$, we get our second equation:

$$v = 5.00 - 0.10 \omega$$

### Final Calculation: Bringing It All Together

We now have a simple system of two linear equations. Let's substitute the expression for $v$ into our first equation:

$$15 = -3(5.00 - 0.10 \omega) + 4\omega$$

Expanding the bracket:

$$15 = -15 + 0.30 \omega + 4\omega$$

Moving the constants to one side:

$$30 = 4.30 \omega$$

Solving for $\omega$:

$$\omega = \frac{30}{4.30} \approx 6.9767 	ext{ rad/s}$$

Rounding to three significant figures, we get **$\omega = 6.98 	ext{ rad/s}$**.

Finally, let's find the rebound speed $v$ by plugging $\omega$ back into our second equation:

$$v = 5.00 - 0.10(6.9767)$$

$$v = 5.00 - 0.69767 \approx 4.3023 	ext{ m/s}$$

Rounding again, we get **$v = 4.30 	ext{ m/s}$**.

And there we have it! The physics perfectly predicts the aftermath of the collision, matching option (A).

Select Answer:

Visualized Solution

Visualizing the Setup

Conservation of Angular Momentum $L$

Applying Conservation Equation

Calculating Moment of Inertia $I$

Forming the First Equation

Coefficient of Restitution $e$

Velocity of Separation

Forming the Second Equation

Solving for $ω$ and $v$

What if $e < 1 ?$

The Sigma Insight: Conservation of Angular Momentum

The Beauty of Collisions and Rotations

Analyzing the Setup

The Master Equation

Calculating the Moment of Inertia

The Geometry of the Bounce

Final Calculation

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