Two perfectly elastic particles `A` and `B` of equal masses travelling along a line joining them with velocities `15 m//s` and `10m//s` respectively collide. Their velocities after the elastic collision will be (in m/s) respectively

Options

A. `0 and 25`
B. `5 and 20`
C. `10 and 15`
D. `20 and 5`

Explanation

To solve the problem of two perfectly elastic particles `A` and `B` colliding, we will use the principles of conservation of momentum and conservation of kinetic energy. Here’s the step-by-step solution:

### Step 1: Understand the Initial Conditions
- Let the mass of both particles `A` and `B` be `m`.
- The initial velocity of particle `A` (U1) = 15 m/s.
- The initial velocity of particle `B` (U2) = 10 m/s.

### Step 2: Apply Conservation of Momentum
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Therefore, we can write:
\[ m \cdot U1 + m \cdot U2 = m \cdot V1 + m \cdot V2 \]
Since the masses are equal, we can cancel `m` from the equation:
\[ U1 + U2 = V1 + V2 \]
Substituting the values:
\[ 15 + 10 = V1 + V2 \]
\[ 25 = V1 + V2 \] (Equation 1)

### Step 3: Apply 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:
\[ \frac{1}{2} m U1^2 + \frac{1}{2} m U2^2 = \frac{1}{2} m V1^2 + \frac{1}{2} m V2^2 \]
Again, canceling `m` and `1/2` gives us:
\[ U1^2 + U2^2 = V1^2 + V2^2 \]
Substituting the values:
\[ 15^2 + 10^2 = V1^2 + V2^2 \]
\[ 225 + 100 = V1^2 + V2^2 \]
\[ 325 = V1^2 + V2^2 \] (Equation 2)

### Step 4: Solve the Equations
Now we have two equations:
1. \( V1 + V2 = 25 \)
2. \( V1^2 + V2^2 = 325 \)

From Equation 1, we can express \( V2 \) in terms of \( V1 \):
\[ V2 = 25 - V1 \]

Substituting \( V2 \) into Equation 2:
\[ V1^2 + (25 - V1)^2 = 325 \]
Expanding the equation:
\[ V1^2 + (625 - 50V1 + V1^2) = 325 \]
Combining like terms:
\[ 2V1^2 - 50V1 + 625 - 325 = 0 \]
\[ 2V1^2 - 50V1 + 300 = 0 \]
Dividing the entire equation by 2:
\[ V1^2 - 25V1 + 150 = 0 \]

### Step 5: Use the Quadratic Formula
To find \( V1 \), we will use the quadratic formula:
\[ V1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -25, c = 150 \):
\[ V1 = \frac{25 \pm \sqrt{(-25)^2 - 4 \cdot 1 \cdot 150}}{2 \cdot 1} \]
\[ V1 = \frac{25 \pm \sqrt{625 - 600}}{2} \]
\[ V1 = \frac{25 \pm \sqrt{25}}{2} \]
\[ V1 = \frac{25 \pm 5}{2} \]

Calculating the two possible values:
1. \( V1 = \frac{30}{2} = 15 \) m/s
2. \( V1 = \frac{20}{2} = 10 \) m/s

### Step 6: Find \( V2 \)
Using \( V2 = 25 - V1 \):
1. If \( V1 = 15 \) m/s, then \( V2 = 25 - 15 = 10 \) m/s.
2. If \( V1 = 10 \) m/s, then \( V2 = 25 - 10 = 15 \) m/s.

### Conclusion
Thus, after the collision, the velocities of particles `A` and `B` will be:
- Velocity of particle `A` (V1) = 10 m/s
- Velocity of particle `B` (V2) = 15 m/s

### Final Answer
The final velocities after the elastic collision are:
- Particle A: 10 m/s
- Particle B: 15 m/s