A particle is projected with velocity 'u' makes an angle `theta` w.r.t. horizontal. Now it breaks in two identical parts at highest point of trajectory. If one part is retrace its path, then velocity of other part is -

Options

A. 3u cos `theta`
B. 2u cos `theta`
C. u cos `theta`
D. u

Explanation

To solve the problem step by step, we will analyze the motion of the particle and apply the principles of conservation of momentum.

### Step 1: Understand the initial conditions
A particle is projected with an initial velocity \( u \) at an angle \( \theta \) with respect to the horizontal. At the highest point of its trajectory, the vertical component of the velocity becomes zero, and only the horizontal component remains.

**Hint:** Remember that at the highest point, the vertical velocity component is zero.

### Step 2: Resolve the initial velocity
The initial velocity \( u \) can be resolved into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

At the highest point, the vertical component \( u_y \) is zero.

**Hint:** Use trigonometric functions to resolve the velocity into horizontal and vertical components.

### Step 3: Analyze the situation at the highest point
At the highest point, the particle breaks into two identical parts. Each part has a mass of \( \frac{m}{2} \) if the original mass is \( m \).

**Hint:** Identify the mass of each part after the break.

### Step 4: Determine the velocities of the parts
One part retraces its path, which means it moves back with the same horizontal velocity but in the opposite direction. Therefore, its velocity is:
\[ v_1 = -u \cos \theta \]

Let the velocity of the other part be \( v_2 \).

**Hint:** The direction of the velocity is important; pay attention to the signs.

### Step 5: Apply the conservation of momentum
According to the law of conservation of momentum, the total momentum before the break must equal the total momentum after the break. The initial momentum in the x-direction is:
\[ P_{\text{initial}} = m \cdot u \cos \theta \]

The final momentum after the break is:
\[ P_{\text{final}} = \frac{m}{2} v_1 + \frac{m}{2} v_2 \]
Substituting \( v_1 \):
\[ P_{\text{final}} = \frac{m}{2} (-u \cos \theta) + \frac{m}{2} v_2 \]

Setting the initial momentum equal to the final momentum:
\[ m u \cos \theta = \frac{m}{2} (-u \cos \theta) + \frac{m}{2} v_2 \]

**Hint:** Remember to factor out the mass \( m \) since it is common in all terms.

### Step 6: Simplify the equation
Cancel \( m \) from both sides:
\[ u \cos \theta = \frac{1}{2} (-u \cos \theta) + \frac{1}{2} v_2 \]

Multiply through by 2 to eliminate the fraction:
\[ 2u \cos \theta = -u \cos \theta + v_2 \]

**Hint:** Isolate \( v_2 \) on one side of the equation.

### Step 7: Solve for \( v_2 \)
Rearranging gives:
\[ v_2 = 2u \cos \theta + u \cos \theta \]
\[ v_2 = 3u \cos \theta \]

Thus, the velocity of the other part is:
\[ v_2 = 3u \cos \theta \]

**Hint:** Ensure you have accounted for all terms correctly when isolating \( v_2 \).

### Final Answer
The velocity of the other part after the break is:
\[ v_2 = 3u \cos \theta \]