A body of mass m moving with velocity `u hati` collides elastically with a body at rest of mass 3m and then the final velocity of body mass m is v `hatj`. Find the relation between v and u

Options

A. u = 2v
B. u = v/2
C. u = `root_2` v
D. u = v/`root_2`

Answer

To solve the problem, we will use the principles of conservation of momentum and conservation of kinetic energy, as the collision is elastic.

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities:**
- The initial velocity of the mass \( m \) is \( \mathbf{u} = u \hat{i} \).
- The initial velocity of the mass \( 3m \) is \( \mathbf{u_2} = 0 \hat{i} \).
- The final velocity of mass \( m \) after the collision is given as \( \mathbf{v_1} = v \hat{j} \).
- Let the final velocity of mass \( 3m \) be \( \mathbf{v_2} = v_{x} \hat{i} + v_{y} \hat{j} \).

2. **Apply Conservation of Momentum:**
- The total initial momentum before the collision is:
\[ \text{Initial Momentum} = m \cdot u \hat{i} + 3m \cdot 0 = mu \hat{i} \]
- The total final momentum after the collision is:
\[ \text{Final Momentum} = m \cdot v \hat{j} + 3m \cdot (v_{x} \hat{i} + v_{y} \hat{j}) = mv \hat{j} + 3m v_{x} \hat{i} + 3m v_{y} \hat{j} \]
- Setting initial momentum equal to final momentum:
\[ mu \hat{i} = 3m v_{x} \hat{i} + (mv + 3m v_{y}) \hat{j} \]

3. **Equate Components:**
- For the \( \hat{i} \) components:
\[ mu = 3m v_{x} \implies u = 3v_{x} \quad \text{(1)} \]
- For the \( \hat{j} \) components:
\[ 0 = mv + 3m v_{y} \implies 0 = v + 3v_{y} \implies v_{y} = -\frac{v}{3} \quad \text{(2)} \]

4. **Determine Final Velocity of Mass \( 3m \):**
- Substitute \( v_{y} \) into the expression for \( \mathbf{v_2} \):
\[ \mathbf{v_2} = v_{x} \hat{i} - \frac{v}{3} \hat{j} \]

5. **Apply Conservation of Kinetic Energy:**
- The initial kinetic energy is:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]
- The final kinetic energy is:
\[ KE_{\text{final}} = \frac{1}{2} m v^2 + \frac{1}{2} (3m) \left( v_{x}^2 + v_{y}^2 \right) \]
- Setting initial kinetic energy equal to final kinetic energy:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v^2 + \frac{3}{2} m \left( v_{x}^2 + v_{y}^2 \right) \]
- Cancel \( \frac{1}{2} m \) from both sides:
\[ u^2 = v^2 + 3(v_{x}^2 + v_{y}^2) \]

6. **Substitute \( v_{x} \) and \( v_{y} \):**
- From equation (1), substitute \( v_{x} = \frac{u}{3} \).
- From equation (2), substitute \( v_{y} = -\frac{v}{3} \).
- Substitute these into the kinetic energy equation:
\[ u^2 = v^2 + 3\left(\left(\frac{u}{3}\right)^2 + \left(-\frac{v}{3}\right)^2\right) \]
- Simplifying gives:
\[ u^2 = v^2 + 3\left(\frac{u^2}{9} + \frac{v^2}{9}\right) \]
\[ u^2 = v^2 + \frac{u^2}{3} + \frac{v^2}{3} \]
- Multiply through by 3 to eliminate the fractions:
\[ 3u^2 = 3v^2 + u^2 + v^2 \]
\[ 3u^2 - u^2 = 4v^2 \implies 2u^2 = 4v^2 \implies u^2 = 2v^2 \]
- Taking the square root gives:
\[ u = \sqrt{2} v \]

### Final Relation:
\[ u = \sqrt{2} v \]