A positively charged particle having charge `q_1 = 1 C` and mass `m_1 =40 gm` is revolving along a circle of radius `R=40 cm` with velocity `v_1 = 5 ms^-1` in a uniform magnetic field with center of circle at origin `O` of a three-dimensional system. At `t=0`, the particle was at (0, 0.4m, 0) and velocity was directed along positive x direction. Another particle having charge `q_2 = 1 C` and mass `m_2 = 10g` moving uniformly parallel to positive z-direction with velocity `v_2 = 40//pi ms^-1` collides with revolving particle at `t=0` and gets stuck to it. Neglecting gravitational force and coulomb force, calculate x-, y- and z-coordinates of the combined particle at `t= pi//40 s`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
- The first particle has:
- Charge \( q_1 = 1 \, C \)
- Mass \( m_1 = 40 \, g = 0.04 \, kg \)
- Velocity \( v_1 = 5 \, m/s \)
- Radius of circular motion \( R = 40 \, cm = 0.4 \, m \)
- Initial position at \( (0, 0.4, 0) \)

- The second particle has:
- Charge \( q_2 = 1 \, C \)
- Mass \( m_2 = 10 \, g = 0.01 \, kg \)
- Velocity \( v_2 = \frac{40}{\pi} \, m/s \) in the positive z-direction.

### Step 2: Calculate the velocity of the combined particle after collision
Using the conservation of momentum:
\[ (m_1 + m_2) \cdot v = m_1 \cdot v_1 + m_2 \cdot v_2 \]

Where:
- \( v \) is the velocity of the combined particle after the collision.
- The initial velocity of the first particle is along the x-axis, so \( v_1 = 5 \hat{i} \).
- The second particle's velocity is along the z-axis, so \( v_2 = \frac{40}{\pi} \hat{k} \).

Substituting the values:
\[ (0.04 + 0.01) v = 0.04 \cdot 5 \hat{i} + 0.01 \cdot \frac{40}{\pi} \hat{k} \]
\[ 0.05 v = 0.2 \hat{i} + \frac{0.4}{\pi} \hat{k} \]

Now, solving for \( v \):
\[ v = \frac{0.2 \hat{i} + \frac{0.4}{\pi} \hat{k}}{0.05} = 4 \hat{i} + \frac{8}{\pi} \hat{k} \, m/s \]

### Step 3: Determine the radius of the helical path
The radius of the helix is given by:
\[ R_{helix} = \frac{(m_1 + m_2) v_x}{(q_1 + q_2) v_x} \]
Where \( v_x = 4 \, m/s \).

Substituting the values:
\[ R_{helix} = \frac{(0.04 + 0.01) \cdot 4}{(1 + 1) \cdot 4} = \frac{0.05 \cdot 4}{2 \cdot 4} = \frac{0.2}{8} = 0.025 \, m \]

### Step 4: Calculate the coordinates at \( t = \frac{\pi}{40} \, s \)
1. **X-coordinate**: The x-coordinate will be equal to the radius of the helix:
\[ x = R_{helix} = 0.025 \, m \]

2. **Y-coordinate**: The y-coordinate remains constant since the motion is circular in the x-y plane:
\[ y = R_{helix} = 0.025 \, m \]

3. **Z-coordinate**: The z-coordinate changes uniformly with time:
\[ z = v_z \cdot t = \left(\frac{8}{\pi}\right) \cdot \left(\frac{\pi}{40}\right) = \frac{8}{40} = 0.2 \, m \]

### Final Coordinates
Thus, the coordinates of the combined particle at \( t = \frac{\pi}{40} \, s \) are:
\[ (x, y, z) = (0.025, 0.025, 0.2) \]