Two particles of equal masses moving with same momentum collide perfectly inelastically. After the collision the combined mass moves with half of the speed of the individual masses. The angle between the initial momenta of individual particle is

Options

A. `60^(@)`
B. `90^(@)`
C. `120^(@)`
D. `45^(@)`

Explanation

To solve the problem, we need to analyze the situation involving two particles of equal mass that collide perfectly inelastically. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
We have two particles of equal mass \( m \) moving with the same momentum \( p \). After a perfectly inelastic collision, they stick together and move with a speed that is half of the speed of the individual masses.

### Step 2: Define the Initial Momentum
Let the momentum of each particle be represented as:
- For particle 1: \( \vec{p_1} = m \vec{v_1} \)
- For particle 2: \( \vec{p_2} = m \vec{v_2} \)

Since both particles have the same momentum, we can write:
\[ |\vec{p_1}| = |\vec{p_2}| = p \]

### Step 3: Analyze the Collision
After the collision, the two particles combine into a single mass of \( 2m \) and move with a velocity \( \vec{V} \). According to the problem, the speed of the combined mass is half that of the individual masses:
\[ |\vec{V}| = \frac{1}{2} |\vec{v_1}| \]

### Step 4: Apply Conservation of Momentum
The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision:
\[ \vec{p_1} + \vec{p_2} = \vec{P} \]
Where \( \vec{P} \) is the momentum of the combined mass after the collision:
\[ \vec{P} = (2m) \vec{V} \]

### Step 5: Express the Momentum in Terms of Magnitude
Since \( |\vec{p_1}| = |\vec{p_2}| = p \), we can write:
\[ p_1 + p_2 = 2m \cdot \frac{1}{2} |\vec{v_1}| \]
This simplifies to:
\[ p + p = m |\vec{v_1}| \]
Thus:
\[ 2p = m |\vec{v_1}| \]

### Step 6: Relate the Magnitudes
Using the law of cosines for the momenta:
\[ |\vec{P}|^2 = |\vec{p_1}|^2 + |\vec{p_2}|^2 + 2 |\vec{p_1}| |\vec{p_2}| \cos \theta \]
Substituting \( |\vec{p_1}| = |\vec{p_2}| = p \):
\[ |\vec{P}|^2 = p^2 + p^2 + 2p^2 \cos \theta \]
This simplifies to:
\[ |\vec{P}|^2 = 2p^2(1 + \cos \theta) \]

### Step 7: Substitute for \( |\vec{P}| \)
From the earlier step, we know:
\[ |\vec{P}| = 2m \cdot \frac{1}{2} |\vec{v_1}| = m |\vec{v_1}| \]
Thus:
\[ |\vec{P}|^2 = (m |\vec{v_1}|)^2 = m^2 v_1^2 \]

### Step 8: Set the Equations Equal
Equating the two expressions for \( |\vec{P}|^2 \):
\[ m^2 v_1^2 = 2p^2(1 + \cos \theta) \]

### Step 9: Substitute for \( p \)
Since \( p = m v_1 \):
\[ m^2 v_1^2 = 2(m v_1)^2(1 + \cos \theta) \]
This simplifies to:
\[ 1 = 2(1 + \cos \theta) \]
Thus:
\[ 1 = 2 + 2 \cos \theta \]
Rearranging gives:
\[ 2 \cos \theta = -1 \]
So:
\[ \cos \theta = -\frac{1}{2} \]

### Step 10: Find the Angle
The angle \( \theta \) that corresponds to \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \]

### Final Answer
The angle between the initial momenta of the individual particles is \( 120^\circ \).

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