Two identical balls A and B collide head on elastically. If velocities of A and B, before the collision are `+0.5 m/s and -0.3 m/s` respectively, then their velocities, after the collision, are respectively

Options

A. `-0.5 m/s and + 0.3 m/s`
B. `+0.5 m/s and + 0.3 m/s`
C. `+0.3 m/s and - 0.5m/s`
D. `-0.3 m/s and + 0.5 m/s`

Explanation

To solve the problem of two identical balls A and B colliding elastically, we will follow these steps:

### Step 1: Understand the initial conditions
- The initial velocity of ball A, \( V_A = +0.5 \, \text{m/s} \)
- The initial velocity of ball B, \( V_B = -0.3 \, \text{m/s} \)

### Step 2: Use the conservation of momentum
For an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since both balls have the same mass \( m \), we can write:

\[ m V_A + m V_B = m V_1 + m V_2 \]

Dividing through by \( m \) (mass cancels out), we get:

\[ V_A + V_B = V_1 + V_2 \]

Substituting the known values:

\[ 0.5 - 0.3 = V_1 + V_2 \]

This simplifies to:

\[ 0.2 = V_1 + V_2 \quad \text{(Equation 1)} \]

### Step 3: Use the conservation of kinetic energy
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Thus, we have:

\[ \frac{1}{2} m V_A^2 + \frac{1}{2} m V_B^2 = \frac{1}{2} m V_1^2 + \frac{1}{2} m V_2^2 \]

Again, dividing through by \( \frac{1}{2} m \):

\[ V_A^2 + V_B^2 = V_1^2 + V_2^2 \]

Substituting the known values:

\[ (0.5)^2 + (-0.3)^2 = V_1^2 + V_2^2 \]

Calculating the left side:

\[ 0.25 + 0.09 = V_1^2 + V_2^2 \]

This simplifies to:

\[ 0.34 = V_1^2 + V_2^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
We have two equations:
1. \( V_1 + V_2 = 0.2 \)
2. \( V_1^2 + V_2^2 = 0.34 \)

From Equation 1, we can express \( V_2 \) in terms of \( V_1 \):

\[ V_2 = 0.2 - V_1 \]

Substituting this into Equation 2:

\[ V_1^2 + (0.2 - V_1)^2 = 0.34 \]

Expanding the equation:

\[ V_1^2 + (0.04 - 0.4 V_1 + V_1^2) = 0.34 \]

Combining like terms:

\[ 2V_1^2 - 0.4V_1 + 0.04 = 0.34 \]

Rearranging gives:

\[ 2V_1^2 - 0.4V_1 - 0.3 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( V_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2, b = -0.4, c = -0.3 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-0.4)^2 - 4 \cdot 2 \cdot (-0.3) = 0.16 + 2.4 = 2.56 \]

Now, applying the quadratic formula:

\[ V_1 = \frac{0.4 \pm \sqrt{2.56}}{4} = \frac{0.4 \pm 1.6}{4} \]

Calculating the two possible values:

1. \( V_1 = \frac{2.0}{4} = 0.5 \, \text{m/s} \)
2. \( V_1 = \frac{-1.2}{4} = -0.3 \, \text{m/s} \)

### Step 6: Find \( V_2 \)
Using \( V_2 = 0.2 - V_1 \):

1. If \( V_1 = 0.5 \, \text{m/s} \), then \( V_2 = 0.2 - 0.5 = -0.3 \, \text{m/s} \)
2. If \( V_1 = -0.3 \, \text{m/s} \), then \( V_2 = 0.2 - (-0.3) = 0.5 \, \text{m/s} \)

### Conclusion
Thus, the velocities after the collision are:
- \( V_1 = -0.3 \, \text{m/s} \) (Ball A)
- \( V_2 = 0.5 \, \text{m/s} \) (Ball B)