Question

Two objects of masses \(m\) and \(2m\) are moving at speeds \(v\) and \(v/2\), respectively. After undergoing a completely inelastic collision, they move together with a speed of \(v/3\). The angle between the initial velocity vectors of the two objects is:

• A. \(60^\circ\)
• B. \(120^\circ\)
• C. \(45^\circ\)
• D. \(90^\circ\)

Hint:

For perfectly inelastic collisions, always apply vector momentum conservation. Use cosine law when directions differ.

Answer 1

The Correct Option is B

Step 1: Conservation of momentum. 
In a completely inelastic collision, momentum is conserved: \[ \vec{p}_{\text{initial}} = \vec{p}_{\text{final}} \] Final momentum \(= (m + 2m) \cdot \frac{v}{3} = m v\). 
 

Step 2: Initial momentum components. 
Magnitude of resultant momentum before collision: \[ |\vec{p}_i| = \sqrt{(mv)^2 + (2m \cdot \frac{v}{2})^2 + 2(mv)(mv)\cos\theta} \] Simplify: \[ m v = m v \sqrt{1 + 1 + 2\cos\theta} \Rightarrow 1 = \sqrt{2(1 + \cos\theta)} \] \[ \Rightarrow 1 = \sqrt{4\cos^2(\theta/2)} \Rightarrow \cos(\theta/2) = \frac{1}{2} \Rightarrow \theta = 120^\circ \]

Step 3: Conclusion. 
The angle between their initial velocity vectors is \(120^\circ.\)