A particle of positive charge q and mass m enters with velocity `Vhati` at the origin in a magnetic field `B(-hatk)` which is present in the whole space. The charge makes a perfectely inelastic collision with an identical particle (having same charge) ar rest but free to move at its maximum positive y-coordinate. After collision, the combined charge will move on trajectory where `r=(mV)/(qB)`.

Options

A. `y = (mv)/(qB) (- hat(i))`
B. `(x + r)^(2)+(y - r//2)^(2) = r^(2)//4`
C. `(x-r)^(2) +(y-r)^(2) =r^(2)`
D. `(x-r)^(2)+(y+r//2)^(2) =r^(2)//4`

Explanation

To solve the problem step-by-step, we will analyze the motion of the charged particles in the magnetic field and the effects of the inelastic collision.

### Step 1: Understand the Initial Motion of the Charged Particle
A particle with charge \( q \) and mass \( m \) enters the magnetic field with velocity \( \vec{V} = V \hat{i} \) at the origin. The magnetic field is given by \( \vec{B} = -B \hat{k} \).

**Hint:** The force on a charged particle in a magnetic field is given by the Lorentz force \( \vec{F} = q (\vec{V} \times \vec{B}) \).

### Step 2: Calculate the Force Acting on the Particle
Using the cross product to find the direction of the force:
\[ \vec{F} = q (\vec{V} \times \vec{B}) = q (V \hat{i} \times (-B \hat{k})) \]
Calculating the cross product:
\[ \hat{i} \times \hat{k} = \hat{j} \quad \text{(using the right-hand rule)} \]
Thus,
\[ \vec{F} = -qVB \hat{j} \]
This indicates that the force is acting in the negative \( y \)-direction.

**Hint:** The force acts perpendicular to the velocity, causing circular motion.

### Step 3: Determine the Radius of the Circular Path
The radius \( R \) of the circular path of the charged particle in a magnetic field is given by:
\[ R = \frac{mv}{qB} \]
where \( v \) is the speed of the particle.

**Hint:** This formula arises from equating the centripetal force to the magnetic force.

### Step 4: Analyze the Collision
The particle collides inelastically with another identical particle (charge \( q \), mass \( m \)) that is at rest. After the collision, the two particles will move together with a combined mass of \( 2m \) and charge \( 2q \).

**Hint:** In an inelastic collision, momentum is conserved.

### Step 5: Apply Conservation of Momentum
Before the collision, the momentum of the moving particle is:
\[ p_{\text{initial}} = mv \]
After the collision, the combined momentum is:
\[ p_{\text{final}} = (2m)v' \]
Setting the initial momentum equal to the final momentum:
\[ mv = (2m)v' \implies v' = \frac{v}{2} \]

**Hint:** The new velocity after the collision is half of the original velocity due to the doubling of mass.

### Step 6: Calculate the New Radius After Collision
The new radius \( R' \) of the circular path after the collision can be calculated using the new speed:
\[ R' = \frac{(2m)(v/2)}{(2q)B} = \frac{mv}{qB} = R \]
However, since the momentum is now shared, the effective radius becomes:
\[ R' = \frac{R}{2} \]

**Hint:** The radius decreases because the effective charge moving in the field has doubled while the speed has halved.

### Step 7: Determine the Center of the New Circular Path
The center of the new circular path will be at coordinates:
\[ \text{Center} = (-R', \frac{R'}{2}) = \left(-\frac{R}{2}, \frac{R}{4}\right) \]

**Hint:** The center of the circle is determined by the radius and the direction of the motion.

### Step 8: Write the Equation of the Circle
The general equation of a circle with center \( (a, b) \) and radius \( r \) is:
\[ (x - a)^2 + (y - b)^2 = r^2 \]
Substituting the values:
\[ \left(x + \frac{R}{2}\right)^2 + \left(y - \frac{R}{4}\right)^2 = \left(\frac{R}{2}\right)^2 \]

**Hint:** This equation describes the trajectory of the combined charge after the collision.

### Final Result
The trajectory of the combined charge after the collision can be represented by the equation:
\[ \left(x + \frac{R}{2}\right)^2 + \left(y - \frac{R}{4}\right)^2 = \frac{R^2}{4} \]