A particle of mass m moving with speed V collides eleastically with another particle of mass 2mFind speed of smaller mass after head on collision

Options

A. `v/3`
B. `-v/3`
C. `(2v)/3`
D. `(-2v)/3`

Answer

To solve the problem of a particle of mass \( m \) moving with speed \( V \) colliding elastically with another particle of mass \( 2m \), we will use the principles of conservation of momentum and conservation of kinetic energy.

### Step 1: Understand the scenario
We have two particles:
- Particle 1 (mass \( m \)) moving with speed \( V \).
- Particle 2 (mass \( 2m \)) initially at rest (speed = 0).

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

**Before collision:**
- Momentum of Particle 1 = \( mV \)
- Momentum of Particle 2 = \( 2m \cdot 0 = 0 \)

Total initial momentum \( P_i = mV + 0 = mV \).

**After collision:**
Let \( V_1 \) be the speed of Particle 1 after the collision and \( V_2 \) be the speed of Particle 2 after the collision.

Total final momentum \( P_f = mV_1 + 2mV_2 \).

Setting initial momentum equal to final momentum:
\[ mV = mV_1 + 2mV_2 \]
Dividing through by \( m \):
\[ V = V_1 + 2V_2 \quad \text{(Equation 1)} \]

### Step 3: Apply conservation of kinetic energy
In an elastic collision, the total kinetic energy before the collision must equal the total kinetic energy after the collision.

**Before collision:**
Kinetic energy of Particle 1 = \( \frac{1}{2} m V^2 \)
Kinetic energy of Particle 2 = \( 0 \)

Total initial kinetic energy \( KE_i = \frac{1}{2} m V^2 + 0 = \frac{1}{2} m V^2 \).

**After collision:**
Total final kinetic energy \( KE_f = \frac{1}{2} m V_1^2 + \frac{1}{2} (2m) V_2^2 = \frac{1}{2} m V_1^2 + m V_2^2 \).

Setting initial kinetic energy equal to final kinetic energy:
\[ \frac{1}{2} m V^2 = \frac{1}{2} m V_1^2 + m V_2^2 \]
Dividing through by \( \frac{1}{2} m \):
\[ V^2 = V_1^2 + 2V_2^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( V = V_1 + 2V_2 \) (Equation 1)
2. \( V^2 = V_1^2 + 2V_2^2 \) (Equation 2)

From Equation 1, we can express \( V_1 \) in terms of \( V_2 \):
\[ V_1 = V - 2V_2 \]

Substituting \( V_1 \) into Equation 2:
\[ V^2 = (V - 2V_2)^2 + 2V_2^2 \]
Expanding the left side:
\[ V^2 = V^2 - 4VV_2 + 4V_2^2 + 2V_2^2 \]
Combining like terms:
\[ V^2 = V^2 - 4VV_2 + 6V_2^2 \]
Subtracting \( V^2 \) from both sides:
\[ 0 = -4VV_2 + 6V_2^2 \]
Factoring out \( V_2 \):
\[ 0 = V_2(-4V + 6V_2) \]
This gives us two solutions:
1. \( V_2 = 0 \) (not valid since it implies no collision)
2. \( -4V + 6V_2 = 0 \) leading to:
\[ 6V_2 = 4V \implies V_2 = \frac{2V}{3} \]

Now substituting \( V_2 \) back into Equation 1 to find \( V_1 \):
\[ V_1 = V - 2\left(\frac{2V}{3}\right) = V - \frac{4V}{3} = -\frac{V}{3} \]

### Conclusion
The speed of the smaller mass \( m \) after the collision is:
\[ V_1 = -\frac{V}{3} \]
This indicates that it moves in the opposite direction with a speed of \( \frac{V}{3} \).