Two elastic bodies P and Q having equal mases are moving along the same line with velocities of 16 m/s and 10 m/s respectively. Their velocities after the elastic collision will be in m/s :

Options

A. 0 and 25
B. 5 and 20
C. 10 and 16
D. 20 and 5

Answer

To solve the problem of finding the velocities of two elastic bodies P and Q after an elastic collision, we can follow these steps:

### Step 1: Understand the Conservation of Momentum
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.

Given:
- Mass of body P (m) = Mass of body Q (m) (equal masses)
- Initial velocity of body P (u1) = 16 m/s
- Initial velocity of body Q (u2) = 10 m/s

The equation for conservation of momentum can be written as:
\[ m \cdot u_1 + m \cdot u_2 = m \cdot v_1 + m \cdot v_2 \]

Since the masses are equal, we can simplify this to:
\[ u_1 + u_2 = v_1 + v_2 \]

### Step 2: Substitute the Given Values
Substituting the initial velocities into the equation:
\[ 16 + 10 = v_1 + v_2 \]
\[ 26 = v_1 + v_2 \] (Equation 1)

### Step 3: Understand 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.

The equation for kinetic energy conservation is:
\[ \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 \]

Again, since the masses are equal, we can simplify this to:
\[ u_1^2 + u_2^2 = v_1^2 + v_2^2 \]

### Step 4: Substitute the Given Values for Kinetic Energy
Substituting the initial velocities into the kinetic energy equation:
\[ 16^2 + 10^2 = v_1^2 + v_2^2 \]
\[ 256 + 100 = v_1^2 + v_2^2 \]
\[ 356 = v_1^2 + v_2^2 \] (Equation 2)

### Step 5: Solve the Equations Simultaneously
Now we have two equations:
1. \( v_1 + v_2 = 26 \) (Equation 1)
2. \( v_1^2 + v_2^2 = 356 \) (Equation 2)

From Equation 1, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = 26 - v_1 \]

Substituting this into Equation 2:
\[ v_1^2 + (26 - v_1)^2 = 356 \]
Expanding the equation:
\[ v_1^2 + (676 - 52v_1 + v_1^2) = 356 \]
Combining like terms:
\[ 2v_1^2 - 52v_1 + 676 - 356 = 0 \]
\[ 2v_1^2 - 52v_1 + 320 = 0 \]

### Step 6: Simplify and Solve the Quadratic Equation
Dividing the entire equation by 2:
\[ v_1^2 - 26v_1 + 160 = 0 \]

Using the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 1, b = -26, c = 160 \):
\[ v_1 = \frac{26 \pm \sqrt{(-26)^2 - 4 \cdot 1 \cdot 160}}{2 \cdot 1} \]
\[ v_1 = \frac{26 \pm \sqrt{676 - 640}}{2} \]
\[ v_1 = \frac{26 \pm \sqrt{36}}{2} \]
\[ v_1 = \frac{26 \pm 6}{2} \]

Calculating the two possible values:
1. \( v_1 = \frac{32}{2} = 16 \) m/s
2. \( v_1 = \frac{20}{2} = 10 \) m/s

### Step 7: Find \( v_2 \)
Using \( v_2 = 26 - v_1 \):
1. If \( v_1 = 16 \), then \( v_2 = 26 - 16 = 10 \) m/s
2. If \( v_1 = 10 \), then \( v_2 = 26 - 10 = 16 \) m/s

### Conclusion
After the elastic collision, the velocities of bodies P and Q will be:
- Body P: 10 m/s
- Body Q: 16 m/s

### Final Answer
The velocities after the elastic collision will be:
- Body P: 10 m/s
- Body Q: 16 m/s