A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and col

Q: Why isn't linear momentum conserved during the collision?

During the collision, the pivot point exerts an external impulsive force on the rod to keep it attached. Because there is a net external force acting on the system, linear momentum is not conserved.

Q: Why do we take the moment of inertia of the rod as $Ml^2/3$?

The rod is pivoted at one of its ends, not at its center of mass. The moment of inertia of a uniform rod about an axis passing through its end is $Ml^2/3$.

Q: How did we get the height $h = l(1 - \cos\theta)$?

Imagine the rod swinging. The vertical distance from the pivot to the bottom of the rod is initially $l$. When it swings by an angle $\theta$, the new vertical distance is $l\cos\theta$. The height it has risen is the difference: $l - l\cos\theta = l(1 - \cos\theta)$.

Q: Can we conserve kinetic energy during the collision?

No, the block sticks to the rod, making it a perfectly inelastic collision. In such collisions, maximum kinetic energy is lost to heat, sound, and deformation.

Q: What if the block hit the middle of the rod instead of the bottom?

The initial angular momentum would be $mv(l/2)$, and the block's contribution to the final moment of inertia would be $m(l/2)^2$. The rest of the logic remains exactly the same!

Question

A block of mass $m = 1	ext{ kg}$ slides with velocity $v = 6	ext{ m/s}$ on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about $O$ and swings as a result of the collision, making angle $	heta$ before momentarily coming to rest. If the rod has mass $M = 2	ext{ kg}$ and length $l = 1	ext{ m}$, then the value of $	heta$ is approximately (Take, $g = 10	ext{ m/s}^2$)

Options:

(A) $63^\circ$

(B) $55^\circ$

(C) $69^\circ$

(D) $49^\circ$

Sigma Percentile · Accepted Answer

### The Ballistic Pendulum: A Tale of Collision and Swing

Imagine a block sliding effortlessly across a frictionless horizontal surface, heading straight for a stationary, uniform vertical rod. Boom! The block collides with the rod and sticks to it. This is a classic example of a perfectly inelastic collision. But what happens next? Together, the rod and the block swing upwards like a pendulum. Our mission is to find the maximum angle $	heta$ they reach before momentarily coming to rest.

### The Collision Phase: Conserving Angular Momentum

During the collision, the pivot point $O$ exerts an external impulsive force on the rod. Because of this external force, the linear momentum of the system is **not** conserved. However, since this force acts exactly at the pivot, it creates zero torque about point $O$.

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

Let's write down the initial angular momentum. Only the block is moving, and it strikes the rod at a perpendicular distance $l$ from the pivot. So, the initial angular momentum is:
$$L_i = mvl$$

After the collision, the block and rod stick together and rotate as a single rigid body with an angular velocity $\omega$. The final angular momentum is:
$$L_f = I_{	ext{system}}\omega$$

The moment of inertia of the system is the sum of the moment of inertia of the rod (pivoted at its end) and the block (acting as a point mass at distance $l$):
$$I_{	ext{system}} = \frac{Ml^2}{3} + ml^2$$

Equating $L_i$ and $L_f$, we can solve for $\omega$:
$$\omega = \frac{mvl}{\frac{Ml^2}{3} + ml^2}$$

Plugging in the given values ($m = 1	ext{ kg}$, $v = 6	ext{ m/s}$, $M = 2	ext{ kg}$, $l = 1	ext{ m}$):
$$\omega = \frac{1 	imes 6 	imes 1}{\frac{2 	imes 1^2}{3} + 1 	imes 1^2} = \frac{6}{\frac{5}{3}} = \frac{18}{5}	ext{ rad/s}$$

### The Swing Phase: Conserving Mechanical Energy

Now that the collision is over, the system swings upwards. The only force doing work is gravity, which is a conservative force. Therefore, the mechanical energy of the system is conserved during the swing.

The rotational kinetic energy immediately after the collision will completely convert into gravitational potential energy at the highest point (where the system momentarily stops).

$$K_i = U_f$$
$$\frac{1}{2}I_{	ext{system}}\omega^2 = Mgh_1 + mgh_2$$

Here, $h_1$ is the height gained by the center of mass of the rod, and $h_2$ is the height gained by the block. Using simple trigonometry, we can express these heights in terms of the angle $	heta$:
$$h_1 = \frac{l}{2}(1 - \cos	heta)$$
$$h_2 = l(1 - \cos	heta)$$

### The Final Calculation

Let's substitute these heights back into our energy equation:
$$\frac{1}{2}\left(\frac{Ml^2}{3} + ml^2ight)\omega^2 = Mg\frac{l}{2}(1 - \cos	heta) + mgl(1 - \cos	heta)$$

Now, we carefully plug in the numbers:
$$\frac{1}{2}\left(\frac{2}{3} + 1ight)\left(\frac{18}{5}ight)^2 = 2 	imes 10 	imes \frac{1}{2}(1 - \cos	heta) + 1 	imes 10 	imes 1(1 - \cos	heta)$$

$$\frac{1}{2} 	imes \frac{5}{3} 	imes \frac{324}{25} = 10(1 - \cos	heta) + 10(1 - \cos	heta)$$

$$\frac{54}{5} = 20(1 - \cos	heta)$$

Dividing both sides by 20:
$$1 - \cos	heta = \frac{54}{100} = 0.54$$

$$\cos	heta = 1 - 0.54 = 0.46$$

Finally, taking the inverse cosine, we find the maximum angle:
$$	heta = \cos^{-1}(0.46) \approx 63^\circ$$

And there we have it! The system swings up to an impressive $63^\circ$ before gravity pulls it back down.

Select Answer:

Visualized Solution

Visualizing the Setup

Conservation of Angular Momentum

Setting up the Equation

Calculating Angular Velocity

Conservation of Mechanical Energy

Analyzing the Heights

Solving for $cos θ$

Final Answer

The Way Forward

The Sigma Insight: Conservation of Angular Momentum

The Ballistic Pendulum

The Collision Phase

The Swing Phase

The Final Calculation

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