A body of mass `1kg` collides heat on elastically with a stationary body of mass `3kg`.After collision,the smaller body reverses its direction of motion and moves with a speed of `2m/s`.The initial speed of the smaller body before collision is_____ ` ms^(-1)`.

Answer

To solve the problem, we will use the principles of conservation of momentum and the fact that the collision is elastic.

### Step-by-Step Solution:

1. **Identify the masses and velocities:**
- Mass of the smaller body (m1) = 1 kg
- Mass of the stationary body (m2) = 3 kg
- Initial velocity of the smaller body (u1) = ? (to be determined)
- Initial velocity of the stationary body (u2) = 0 m/s
- Final velocity of the smaller body (v1) = -2 m/s (it reverses direction)
- Final velocity of the larger body (v2) = ? (to be determined)

2. **Apply the conservation of momentum:**
The total momentum before the collision is equal to the total momentum after the collision.
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the known values:
\[ 1 \cdot u_1 + 3 \cdot 0 = 1 \cdot (-2) + 3 \cdot v_2 \]
This simplifies to:
\[ u_1 = -2 + 3v_2 \quad \text{(Equation 1)} \]

3. **Use the condition of elastic collision:**
In an elastic collision, the relative velocity of separation is equal to the relative velocity of approach.
\[ v_2 - v_1 = u_1 - u_2 \]
Substituting the known values:
\[ v_2 - (-2) = u_1 - 0 \]
This simplifies to:
\[ v_2 + 2 = u_1 \quad \text{(Equation 2)} \]

4. **Substitute Equation 2 into Equation 1:**
From Equation 2, we have:
\[ u_1 = v_2 + 2 \]
Substitute this into Equation 1:
\[ v_2 + 2 = -2 + 3v_2 \]
Rearranging gives:
\[ 2 + 2 = 3v_2 - v_2 \]
\[ 4 = 2v_2 \]
\[ v_2 = 2 \text{ m/s} \]

5. **Find u1 using v2:**
Substitute \(v_2\) back into Equation 2:
\[ u_1 = 2 + 2 = 4 \text{ m/s} \]

### Final Answer:
The initial speed of the smaller body before the collision is **4 m/s**.