The collision of two balls of equal mass takes place at the origin of coordinates. Before collision, the components of velocities are (`v_(x) = 50cm^(-1)` , `v_(y) = 0`) and `(v_(x) = -40cm^(-1)` and `v_(y) = 30cms^(-1)`. The first ball comes to rest after collision. The velocity (components `v_(x)` and `v_(y)` respectively) of the second ball are

Options

A. `10 and 30 cms^(-)`
B. `30 and 10cms^(-1)`
C. `5 and 15cms^(-1)`
D. `15 and 5cms^(-1)`

Explanation

To solve the problem, we will use the principle of conservation of momentum. The momentum before the collision must equal the momentum after the collision since there are no external forces acting on the system.

### Step-by-Step Solution:

1. **Identify the initial velocities of both balls:**
- For the first ball (mass \( m_1 \)):
- \( v_{1x} = 50 \, \text{cm/s} \)
- \( v_{1y} = 0 \, \text{cm/s} \)
- For the second ball (mass \( m_2 \)):
- \( v_{2x} = -40 \, \text{cm/s} \)
- \( v_{2y} = 30 \, \text{cm/s} \)

2. **Express the initial momentum of both balls:**
- The initial momentum of the first ball:
\[ \vec{P_1} = m_1 \vec{v_1} = m \cdot (50 \hat{i} + 0 \hat{j}) = 50m \hat{i} \]
- The initial momentum of the second ball:
\[ \vec{P_2} = m_2 \vec{v_2} = m \cdot (-40 \hat{i} + 30 \hat{j}) = -40m \hat{i} + 30m \hat{j} \]

3. **Calculate the total initial momentum:**
\[ \vec{P_{\text{initial}}} = \vec{P_1} + \vec{P_2} = (50m - 40m) \hat{i} + 30m \hat{j} = 10m \hat{i} + 30m \hat{j} \]

4. **Determine the final momentum after the collision:**
- After the collision, the first ball comes to rest, so its final momentum is:
\[ \vec{P_1'} = m_1 \vec{v_1'} = m \cdot (0 \hat{i} + 0 \hat{j}) = 0 \]
- Let the final velocity of the second ball be \( \vec{v_2'} = v_{2x}' \hat{i} + v_{2y}' \hat{j} \). Thus, the final momentum of the second ball is:
\[ \vec{P_2'} = m_2 \vec{v_2'} = m \cdot (v_{2x}' \hat{i} + v_{2y}' \hat{j}) = mv_{2x}' \hat{i} + mv_{2y}' \hat{j} \]

5. **Apply the conservation of momentum:**
\[ \vec{P_{\text{initial}}} = \vec{P_{\text{final}}} \]
\[ 10m \hat{i} + 30m \hat{j} = 0 + mv_{2x}' \hat{i} + mv_{2y}' \hat{j} \]

6. **Equate the components:**
- For the \( \hat{i} \) component:
\[ 10m = mv_{2x}' \implies v_{2x}' = 10 \, \text{cm/s} \]
- For the \( \hat{j} \) component:
\[ 30m = mv_{2y}' \implies v_{2y}' = 30 \, \text{cm/s} \]

7. **Final answer:**
The components of the velocity of the second ball after the collision are:
\[ v_{2x}' = 10 \, \text{cm/s}, \quad v_{2y}' = 30 \, \text{cm/s} \]