Two identiacal balls A and B having velocities of 0.5 m/s and `0.3" m"//"s"` respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be

Options

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

Answer

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

### Step 1: Understand the Problem
We have two identical balls A and B with initial velocities:
- Velocity of ball A, \( u_A = 0.5 \, \text{m/s} \)
- Velocity of ball B, \( u_B = 0.3 \, \text{m/s} \)

Since the collision is elastic, both momentum and kinetic energy are conserved.

### Step 2: Use the Conservation of Momentum
The conservation of momentum states that the total momentum before the collision equals the total momentum after the collision.

\[ m u_A + m u_B = m v_A + m v_B \]

Since the masses are identical (let's denote them as \( m \)), we can cancel \( m \) from the equation:

\[ u_A + u_B = v_A + v_B \]

### Step 3: Use the Conservation of Kinetic Energy
The conservation of kinetic energy states that the total kinetic energy before the collision equals the total kinetic energy after the collision.

\[ \frac{1}{2} m u_A^2 + \frac{1}{2} m u_B^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 \]

Again, we can cancel \( \frac{1}{2} m \):

\[ u_A^2 + u_B^2 = v_A^2 + v_B^2 \]

### Step 4: Substitute the Known Values
Now we can substitute the known values into the equations:

1. From the momentum equation:
\[ 0.5 + 0.3 = v_A + v_B \implies 0.8 = v_A + v_B \quad \text{(Equation 1)} \]

2. From the kinetic energy equation:
\[ (0.5)^2 + (0.3)^2 = v_A^2 + v_B^2 \implies 0.25 + 0.09 = v_A^2 + v_B^2 \implies 0.34 = v_A^2 + v_B^2 \quad \text{(Equation 2)} \]

### Step 5: Solve the Equations
Now we have two equations:
1. \( v_A + v_B = 0.8 \)
2. \( v_A^2 + v_B^2 = 0.34 \)

From Equation 1, we can express \( v_B \) in terms of \( v_A \):
\[ v_B = 0.8 - v_A \]

Substituting this into Equation 2:
\[ v_A^2 + (0.8 - v_A)^2 = 0.34 \]

Expanding the equation:
\[ v_A^2 + (0.64 - 1.6v_A + v_A^2) = 0.34 \]
\[ 2v_A^2 - 1.6v_A + 0.64 - 0.34 = 0 \]
\[ 2v_A^2 - 1.6v_A + 0.3 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( v_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -1.6, c = 0.3 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-1.6)^2 - 4 \cdot 2 \cdot 0.3 = 2.56 - 2.4 = 0.16 \]

Now substituting into the quadratic formula:
\[ v_A = \frac{1.6 \pm \sqrt{0.16}}{4} = \frac{1.6 \pm 0.4}{4} \]

Calculating the two possible values:
1. \( v_A = \frac{2.0}{4} = 0.5 \, \text{m/s} \)
2. \( v_A = \frac{1.2}{4} = 0.3 \, \text{m/s} \)

### Step 7: Find \( v_B \)
Using \( v_B = 0.8 - v_A \):
1. If \( v_A = 0.5 \, \text{m/s} \), then \( v_B = 0.3 \, \text{m/s} \).
2. If \( v_A = 0.3 \, \text{m/s} \), then \( v_B = 0.5 \, \text{m/s} \).

### Conclusion
After the collision:
- The velocity of ball A, \( v_A = 0.3 \, \text{m/s} \)
- The velocity of ball B, \( v_B = 0.5 \, \text{m/s} \)

Thus, the final velocities after the collision are:
- Velocity of B after collision: \( 0.5 \, \text{m/s} \)
- Velocity of A after collision: \( 0.3 \, \text{m/s} \)

### Final Answer
The velocities of B and A after the collision respectively will be \( 0.5 \, \text{m/s} \) and \( 0.3 \, \text{m/s} \).

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