Two particles of masses m and 3m approach each other with different velocities. After collision, the particle of mass m has velocity `vec( v) ` in their centre of mass frame. Velocity of particle of mass 3m in the centre of mass frame is

Options

A. (a)`- 2 vec( v ) `
B. (b)` - ( vec(v))/(2)`
C. (c)`-( vec( v ))/(3)`
D. (d)`( vec( v ))/(4)`

Answer

To solve the problem, we need to find the velocity of the particle of mass 3m in the center of mass frame after the collision, given that the particle of mass m has a velocity \( \vec{v} \) in that frame.

### Step 1: Understand the System
We have two particles:
- Particle 1 with mass \( m \) and velocity \( \vec{v_1} \) (approaching).
- Particle 2 with mass \( 3m \) and velocity \( \vec{v_2} \) (approaching).

After the collision, we know that the velocity of particle 1 (mass \( m \)) in the center of mass frame is \( \vec{v} \).

### Step 2: Center of Mass Frame
In the center of mass frame, the total momentum is zero. Therefore, we can write the equation for momentum conservation:
\[ m \vec{v_1} + 3m \vec{v_2} = 0 \]
This implies:
\[ \vec{v_1} + 3 \vec{v_2} = 0 \]
or
\[ \vec{v_1} = -3 \vec{v_2} \]

### Step 3: Relate Velocities in the Center of Mass Frame
After the collision, we need to express the velocity of the second particle (mass \( 3m \)) in terms of the known velocity \( \vec{v} \) of the first particle (mass \( m \)):
Let the velocity of the particle of mass \( 3m \) in the center of mass frame be \( \vec{v'} \).

Using the conservation of momentum in the center of mass frame:
\[ m \vec{v} + 3m \vec{v'} = 0 \]
This leads to:
\[ \vec{v} + 3 \vec{v'} = 0 \]
or
\[ 3 \vec{v'} = -\vec{v} \]

### Step 4: Solve for \( \vec{v'} \)
Now, we can solve for \( \vec{v'} \):
\[ \vec{v'} = -\frac{\vec{v}}{3} \]

### Conclusion
The velocity of the particle of mass \( 3m \) in the center of mass frame is:
\[ \vec{v'} = -\frac{\vec{v}}{3} \]

### Final Answer
The velocity of the particle of mass \( 3m \) in the center of mass frame is \( -\frac{\vec{v}}{3} \).