A ball , moving with a speed of `10 sqrt(3) m//s` , strikes an identical stationary ball such that after the collision , the direction of each ball makes an angle of `30^(@)`with the original line of motion. The speeds of two balla after the collision are , respectively.

Options

A. `3 m s^(-1), 3 m s^(-1)`
B. `3sqrt3 ms^(-1), 3sqrt3 ms^(-1)`
C. `3sqrt3 m s^(-1), 3 m s^(-1)`
D. `3 m s^(-1), 3sqrt3 m s^(-1)`

Answer

To solve the problem step by step, we will use the principles of conservation of momentum and the geometry of the collision.

### Step 1: Understand the scenario
We have two identical balls. One ball is moving with a speed of \(10\sqrt{3} \, \text{m/s}\) and strikes a stationary ball. After the collision, both balls move at an angle of \(30^\circ\) with respect to the original line of motion.

### Step 2: Define the variables
Let:
- \(u_1 = 10\sqrt{3} \, \text{m/s}\) (initial speed of the moving ball)
- \(u_2 = 0 \, \text{m/s}\) (initial speed of the stationary ball)
- \(v_1\) = speed of the first ball after the collision
- \(v_2\) = speed of the second ball after the collision

### Step 3: Apply conservation of momentum
We will apply the conservation of momentum in both the x-direction and y-direction.

#### X-direction:
The initial momentum in the x-direction is:
\[ m u_1 + m u_2 = m (10\sqrt{3}) + m(0) = m(10\sqrt{3}) \]
After the collision, the momentum in the x-direction is:
\[ m v_1 \cos(30^\circ) + m v_2 \cos(30^\circ) \]
Setting these equal gives us:
\[ m(10\sqrt{3}) = m v_1 \cos(30^\circ) + m v_2 \cos(30^\circ) \]
Cancelling \(m\) from both sides:
\[ 10\sqrt{3} = v_1 \cos(30^\circ) + v_2 \cos(30^\circ) \]

#### Y-direction:
The initial momentum in the y-direction is zero, since only the first ball is moving initially:
\[ 0 = m v_1 \sin(30^\circ) - m v_2 \sin(30^\circ) \]
This simplifies to:
\[ 0 = v_1 \sin(30^\circ) - v_2 \sin(30^\circ) \]
Thus, we find:
\[ v_1 = v_2 \]

### Step 4: Substitute \(v_2\) in the x-direction equation
Since \(v_1 = v_2\), we can substitute \(v_2\) with \(v_1\) in the x-direction equation:
\[ 10\sqrt{3} = v_1 \cos(30^\circ) + v_1 \cos(30^\circ) \]
This simplifies to:
\[ 10\sqrt{3} = 2 v_1 \cos(30^\circ) \]
Using \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):
\[ 10\sqrt{3} = 2 v_1 \left(\frac{\sqrt{3}}{2}\right) \]
\[ 10\sqrt{3} = v_1 \sqrt{3} \]

### Step 5: Solve for \(v_1\)
Dividing both sides by \(\sqrt{3}\):
\[ v_1 = 10 \, \text{m/s} \]
Since \(v_1 = v_2\), we also have:
\[ v_2 = 10 \, \text{m/s} \]

### Final Answer:
The speeds of the two balls after the collision are:
- \(v_1 = 10 \, \text{m/s}\)
- \(v_2 = 10 \, \text{m/s}\)