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. `10 ms^(-1), 10ms^(-1)`
B. `15ms^(-1), 15ms^(-1)`
C. `10 ms^(-1), 15ms^(-1)`
D. `15 ms^(-1),10ms^(-1)`

Explanation

To solve the problem of two perfectly elastic particles A and B of equal masses colliding, we can follow these steps:

### Step 1: Define the Variables
Let the mass of both particles be \( m \).
- The initial velocity of particle A, \( u_A = 15 \, \text{m/s} \) (moving in the positive direction).
- The initial velocity of particle B, \( u_B = -10 \, \text{m/s} \) (moving in the negative direction).

### Step 2: Apply Conservation of Momentum
In an elastic collision, momentum is conserved. The equation for conservation of momentum can be written as:
\[ m u_A + m u_B = m v_A + m v_B \]
Where \( v_A \) and \( v_B \) are the final velocities of particles A and B respectively. Since the masses are equal, we can simplify the equation:
\[ u_A + u_B = v_A + v_B \]
Substituting the values:
\[ 15 + (-10) = v_A + v_B \]
\[ 5 = v_A + v_B \quad \text{(Equation 1)} \]

### Step 3: Apply Conservation of Kinetic Energy
In an elastic collision, kinetic energy is also conserved. The equation for conservation of kinetic energy can be written as:
\[ \frac{1}{2} m u_A^2 + \frac{1}{2} m u_B^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_B^2 \]
Again, since the masses are equal, we can simplify:
\[ u_A^2 + u_B^2 = v_A^2 + v_B^2 \]
Substituting the values:
\[ 15^2 + (-10)^2 = v_A^2 + v_B^2 \]
\[ 225 + 100 = v_A^2 + v_B^2 \]
\[ 325 = v_A^2 + v_B^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations:
1. \( v_A + v_B = 5 \)
2. \( v_A^2 + v_B^2 = 325 \)

From Equation 1, we can express \( v_B \) in terms of \( v_A \):
\[ v_B = 5 - v_A \]

Substituting this into Equation 2:
\[ v_A^2 + (5 - v_A)^2 = 325 \]
Expanding the equation:
\[ v_A^2 + (25 - 10v_A + v_A^2) = 325 \]
\[ 2v_A^2 - 10v_A + 25 = 325 \]
\[ 2v_A^2 - 10v_A - 300 = 0 \]
Dividing the entire equation by 2:
\[ v_A^2 - 5v_A - 150 = 0 \]

### Step 5: Use the Quadratic Formula
Using the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -5, c = -150 \):
\[ v_A = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-150)}}{2 \cdot 1} \]
\[ v_A = \frac{5 \pm \sqrt{25 + 600}}{2} \]
\[ v_A = \frac{5 \pm \sqrt{625}}{2} \]
\[ v_A = \frac{5 \pm 25}{2} \]
Calculating the two possible values:
1. \( v_A = \frac{30}{2} = 15 \, \text{m/s} \)
2. \( v_A = \frac{-20}{2} = -10 \, \text{m/s} \)

### Step 6: Find \( v_B \)
Using \( v_B = 5 - v_A \):
1. If \( v_A = 15 \):
\[ v_B = 5 - 15 = -10 \, \text{m/s} \]
2. If \( v_A = -10 \):
\[ v_B = 5 - (-10) = 15 \, \text{m/s} \]

### Conclusion
The final velocities after the collision are:
- \( v_A = -10 \, \text{m/s} \) (particle A moves in the opposite direction)
- \( v_B = 15 \, \text{m/s} \) (particle B moves in the original direction of A)

Thus, the final velocities of particles A and B after the collision are \( -10 \, \text{m/s} \) and \( 15 \, \text{m/s} \) respectively.