A `6 kg` mass travelling at `2.5 ms^(-1)` collides head on with a stationary `4 kg` mass. After the collision the `6kg` mass travels in its original direction with a speed of `1 ms^(-1)`. The final velocity of `4 kg` mass is

Options

A. `1 ms^(-1)`
B. `2.25 ms^(-1)`
C. `2 ms^(-1)`
D. `0 ms^(-1)`

Explanation

To solve the problem, we will use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the first object (m1) = 6 kg
- Initial velocity of the first object (u1) = 2.5 m/s
- Mass of the second object (m2) = 4 kg
- Initial velocity of the second object (u2) = 0 m/s (stationary)
- Final velocity of the first object (v1) = 1 m/s
- Final velocity of the second object (v2) = ? (this is what we need to find)

2. **Write the Momentum Conservation Equation:**
The total momentum before the collision is equal to the total momentum after the collision:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

3. **Substitute the Known Values:**
Substitute the values into the momentum conservation equation:
\[ (6 \, \text{kg} \times 2.5 \, \text{m/s}) + (4 \, \text{kg} \times 0 \, \text{m/s}) = (6 \, \text{kg} \times 1 \, \text{m/s}) + (4 \, \text{kg} \times v_2) \]

4. **Calculate Initial Momentum:**
Calculate the left-hand side (initial momentum):
\[ 6 \times 2.5 = 15 \, \text{kg m/s} \]
So, the equation becomes:
\[ 15 = 6 \times 1 + 4 v_2 \]

5. **Calculate Final Momentum:**
Calculate the right-hand side:
\[ 6 \times 1 = 6 \, \text{kg m/s} \]
Now, the equation is:
\[ 15 = 6 + 4 v_2 \]

6. **Isolate v2:**
Rearranging the equation to solve for \( v_2 \):
\[ 15 - 6 = 4 v_2 \]
\[ 9 = 4 v_2 \]
\[ v_2 = \frac{9}{4} = 2.25 \, \text{m/s} \]

7. **Final Answer:**
The final velocity of the 4 kg mass (v2) is **2.25 m/s**.