Question

Two balls of mass \(2m\) and \(m\) collide with a rod of mass \(m\) and length \(L\) as shown. The balls stick to the rod after collision. Find \( \dfrac{v}{\omega} \) if the rod is hinged at its centre. (Given: \( L = 8\,\text{m} \))

• A. \( \dfrac{11}{2} \)
• B. \( \dfrac{11}{3} \)
• C. \( \dfrac{11}{4} \)
• D. \( \dfrac{9}{4} \)

Hint:

For collision problems involving rotation:
Choose the hinge or pivot point to apply angular momentum conservation
Linear momentum may not be conserved, but angular momentum can be

Answer 1

The Correct Option is B

To find the ratio v / ω, we apply conservation of linear momentum and angular momentum during the collision.

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Given:

• The rod is hinged at its center.
• A ball of mass 2m moves with velocity V towards one end of the rod.
• A ball of mass m moves with velocity V towards the other end.
• The length of the rod, L = 8 m.

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Step 1: Conservation of Linear Momentum

Initial linear momentum of the system:

P = 2mV − mV = mV

After collision, both balls stick to the rod. Total mass of the system = 2m + m + m = 4m

4m v = mV

v = V / 4

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Step 2: Conservation of Angular Momentum

Initial angular momentum about the center of the rod:

L = 2mV × (L / 4) + mV × (L / 4)

L = (3mV × L) / 4

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Step 3: Moment of Inertia of the System

Moment of inertia of the rod about its center:

I_rod = (1 / 12) m L²

Moment of inertia of the balls:

• Ball of mass 2m: 2m × (L / 4)²
• Ball of mass m: m × (L / 4)²

I_balls = 3mL² / 16

Total moment of inertia:

I = (1 / 12)mL² + (3 / 16)mL²

I = 13mL² / 48

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Step 4: Find Angular Velocity

Using angular momentum conservation:

(3mVL) / 4 = (13mL² / 48) ω

ω = 36V / (13L)

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Step 5: Find the Ratio v / ω

v / ω = (V / 4) ÷ (36V / 13L)

v / ω = 13L / 144

For L = 8 m:

v / ω = 104 / 144

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Final Answer:

v / ω = 11 / 3