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 need to analyze the elastic collision between two bodies: one of mass \( m \) moving with velocity \( \mathbf{u} \hat{i} \) and another of mass \( 3m \) that is initially at rest. After the collision, the first body has a final velocity of \( \mathbf{v} \hat{j} \). We will find the relationship between \( v \) and \( u \).

### Step-by-Step Solution:

1. **Identify Initial and Final Conditions:**
- Before the collision:
- Mass \( m \) has velocity \( \mathbf{u} \hat{i} \).
- Mass \( 3m \) is at rest, so its velocity is \( 0 \).
- After the collision:
- Mass \( m \) has velocity \( \mathbf{v} \hat{j} \).
- Let the final velocity of mass \( 3m \) be \( \mathbf{V} \) with components \( V_x \) and \( V_y \).

2. **Apply Conservation of Momentum in the x-direction:**
\[ \text{Initial momentum in x-direction} = \text{Final momentum in x-direction} \]
\[ m \cdot u + 3m \cdot 0 = m \cdot 0 + 3m \cdot V_x \]
Simplifying gives:
\[ mu = 3m V_x \implies V_x = \frac{u}{3} \]

3. **Apply Conservation of Momentum in the y-direction:**
\[ \text{Initial momentum in y-direction} = \text{Final momentum in y-direction} \]
\[ m \cdot 0 + 3m \cdot 0 = m \cdot v + 3m \cdot V_y \]
Simplifying gives:
\[ 0 = mv + 3m V_y \implies V_y = -\frac{v}{3} \]

4. **Apply Conservation of Kinetic Energy:**
Since the collision is elastic, kinetic energy is conserved:
\[ \text{Initial kinetic energy} = \text{Final kinetic energy} \]
\[ \frac{1}{2} m u^2 + 0 = \frac{1}{2} m v^2 + \frac{1}{2} (3m) (V_x^2 + V_y^2) \]
Substituting \( V_x \) and \( V_y \):
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v^2 + \frac{3}{2} m \left( \left(\frac{u}{3}\right)^2 + \left(-\frac{v}{3}\right)^2 \right) \]
Simplifying gives:
\[ u^2 = v^2 + 3 \left( \frac{u^2}{9} + \frac{v^2}{9} \right) \]
\[ u^2 = v^2 + \frac{u^2}{3} + \frac{v^2}{3} \]

5. **Rearranging the Equation:**
Multiply through by 3 to eliminate the fractions:
\[ 3u^2 = 3v^2 + u^2 + v^2 \]
\[ 3u^2 - u^2 = 4v^2 \implies 2u^2 = 4v^2 \implies u^2 = 2v^2 \]

6. **Taking the Square Root:**
\[ u = \sqrt{2} v \]

### Final Relation:
Thus, the relation between \( v \) and \( u \) is:
\[ u = \sqrt{2} v \]