A billiard ball moving at a speed of 6.6 ms–1 strikes an identical stationary ball a glancingblow. After the collision, one ball is found to be moving at a speed of 3.3, ms–1 in a direction making an angle of 60º with the original line of motion. The velocity of the other ball is

Options

A. `(3)^(1//2)m//s` at `30^(@)` to the original direction.
B. 1m/s at `60^(@)` to the original direction.
C. `(3)^(1//2)m//s` at `60^(@)` to the original direction.
D. 1 m/s at `30^(@)` to the original direction.

Answer

To solve the problem, we will apply the principle of conservation of momentum. Let's break down the steps:

### Step-by-Step Solution:

**Step 1: Understand the initial and final conditions.**
- Initially, we have one billiard ball moving with a speed of \( v = 6.6 \, \text{m/s} \) and an identical stationary ball (speed = 0).
- After the collision, one ball moves with a speed of \( v' = 3.3 \, \text{m/s} \) at an angle of \( 60^\circ \) to the original line of motion.

**Step 2: Set up the momentum conservation equation.**
- Since the masses of the balls are identical, we can ignore the mass in our calculations. The momentum before and after the collision must be equal:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
This can be expressed as:
\[ m \cdot v = m \cdot v' + m \cdot v_0 \]
where \( v_0 \) is the velocity of the second ball after the collision.

**Step 3: Break down the vectors.**
- We can break the velocities into components. Let’s denote the initial velocity vector as \( \vec{v} \) and the final velocity of the first ball as \( \vec{v}' \).
- The initial momentum vector is:
\[ \vec{p}_{\text{initial}} = m \cdot \vec{v} \]
The final momentum vector is:
\[ \vec{p}_{\text{final}} = m \cdot \vec{v}' + m \cdot \vec{v}_0 \]

**Step 4: Use vector subtraction to find \( v_0 \).**
- Rearranging the momentum conservation equation gives:
\[ \vec{v}_0 = \vec{v} - \vec{v}' \]

**Step 5: Calculate the magnitude of \( v_0 \).**
- We will use the formula for the magnitude of the difference of two vectors:
\[ |\vec{v}_0| = \sqrt{|\vec{v}|^2 + |\vec{v}'|^2 - 2 |\vec{v}| |\vec{v}'| \cos(\theta)} \]
where \( \theta = 60^\circ \).

**Step 6: Substitute the values.**
- Substitute \( |\vec{v}| = 6.6 \, \text{m/s} \), \( |\vec{v}'| = 3.3 \, \text{m/s} \), and \( \cos(60^\circ) = 0.5 \):
\[ |\vec{v}_0| = \sqrt{(6.6)^2 + (3.3)^2 - 2 \cdot (6.6) \cdot (3.3) \cdot 0.5} \]

**Step 7: Calculate each term.**
- Calculate \( (6.6)^2 = 43.56 \)
- Calculate \( (3.3)^2 = 10.89 \)
- Calculate \( 2 \cdot (6.6) \cdot (3.3) \cdot 0.5 = 21.78 \)

**Step 8: Combine the results.**
- Now plug these values back into the equation:
\[ |\vec{v}_0| = \sqrt{43.56 + 10.89 - 21.78} \]
\[ |\vec{v}_0| = \sqrt{32.67} \]
\[ |\vec{v}_0| \approx 5.7 \, \text{m/s} \]

### Final Answer:
The velocity of the other ball after the collision is approximately \( 5.7 \, \text{m/s} \).

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