Two particles of mass m and 2 m moving in opposite directions collide elastically with velocities v and 2v. Find their velocities after collision.

Options

A. 0, 3v
B. 1, 3v
C. 0, 1v
D. 0, 4v

Explanation

To solve the problem of two particles colliding elastically, we will use the principles of conservation of momentum and conservation of kinetic energy.

### Step 1: Define the system
Let:
- Mass of particle 1, \( m_1 = m \)
- Mass of particle 2, \( m_2 = 2m \)
- Initial velocity of particle 1, \( u_1 = v \) (moving in the positive direction)
- Initial velocity of particle 2, \( u_2 = -2v \) (moving in the negative direction)

### Step 2: Apply conservation of momentum
The total momentum before the collision must equal the total momentum after the collision. The equation for momentum conservation is:

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

Substituting the known values:

\[ m(v) + 2m(-2v) = m(v_1) + 2m(v_2) \]

This simplifies to:

\[ mv - 4mv = mv_1 + 2mv_2 \]

\[ -3mv = mv_1 + 2mv_2 \quad \text{(1)} \]

### Step 3: Apply conservation of kinetic energy
For an elastic collision, the total kinetic energy before the collision equals the total kinetic energy after the collision:

\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

Substituting the known values:

\[ \frac{1}{2} m(v^2) + \frac{1}{2} (2m)(-2v)^2 = \frac{1}{2} m(v_1^2) + \frac{1}{2} (2m)(v_2^2) \]

This simplifies to:

\[ \frac{1}{2} mv^2 + \frac{1}{2} (2m)(4v^2) = \frac{1}{2} mv_1^2 + \frac{1}{2} (2m)v_2^2 \]

\[ \frac{1}{2} mv^2 + 4mv^2 = \frac{1}{2} mv_1^2 + mv_2^2 \]

\[ \frac{9}{2} mv^2 = \frac{1}{2} mv_1^2 + mv_2^2 \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (1):

\[ -3 = v_1 + 2v_2 \quad \text{(3)} \]

From equation (2):

\[ 9 = v_1^2 + 2v_2^2 \quad \text{(4)} \]

Now, we can express \( v_1 \) in terms of \( v_2 \) using equation (3):

\[ v_1 = -3 - 2v_2 \]

Substituting \( v_1 \) into equation (4):

\[ 9 = (-3 - 2v_2)^2 + 2v_2^2 \]

Expanding:

\[ 9 = (9 + 12v_2 + 4v_2^2) + 2v_2^2 \]

Combining like terms:

\[ 9 = 9 + 12v_2 + 6v_2^2 \]

Subtracting 9 from both sides:

\[ 0 = 12v_2 + 6v_2^2 \]

Factoring out \( 6v_2 \):

\[ 0 = 6v_2(2 + v_2) \]

This gives us two solutions:

1. \( v_2 = 0 \)
2. \( v_2 = -2 \)

### Step 5: Find \( v_1 \)
Using \( v_2 = -2 \):

Substituting back into equation (3):

\[ v_1 = -3 - 2(-2) = -3 + 4 = 1 \]

### Final Velocities
Thus, the final velocities after the collision are:

- \( v_1 = 1 \)
- \( v_2 = -2 \)

### Summary
The final velocities of the two particles after the elastic collision are:
- Particle of mass \( m \) has a velocity of \( 1 \).
- Particle of mass \( 2m \) has a velocity of \( -2 \).