(C) Since, gravitational force is present , so conservation principle is only applicable to a system consists of air molecules + earth + meteorite .

        p = mv

`:.`          m= p/v`

Hence , m - v  graph will be rectangular hyperbola

The momentum and kinetic energy depend upon reference frame . In the frame of rain ,the man is in rest. So the momentum and kinetic energy of the man in the  frame of train are zero.

But in the frame of ground, velocity of the man is non -zero , So in the frame of ground , momenum and kinetic energy are not zero .

`(E_(k_1))/(E_(k_2))`

=`((p_1^2)/(2m_1))/((p_2^2)/(2m_2))`

=` m_2/m_1`

=`(4m)/(m)`                                          (`because  p_1 = p_2`)

= 4 :  1

If we consider a system consists of gun plus mass  m. Then no external force is present on the system. So, the acceleration of centre of mass of system will be zero . Since initially the system is in rest . So centre of mass will always remain at rest.

Since , no external force is present on the system . So the centre of mass if system will not be changed.

Hence , (B) is correct .

(C) The car is stopped by the contact force exerted due to road ,

Since , external force on system is zero , so no change takes place in the centre of mass .

`x_(cm) = (m_1x_1 + m_2 x_2)/(m_1 + m_2)`

or ` Delta x_(cm) = (m_1 Delta x_1 + m_2 Delta x_2)/(m_1 + m_2)`

Here       `Delta x_(cm) = 0`

`:.`          `m_1 Delta x_1 + m_2 Delta x_2 = 0`

or            ` M Delta x_1 + M/3 Delta x_2 = 0`

`:.`         `    Delta x_1 = - (M)/(3M) Delta x_2 =  - (Delta x_2)/(3)`........................(i)

But             `x_(rel) = Delta x_1 - Delta x_2`

or            `L= Delta x_1 - Delta x_2`

or             ` L = - (Delta x_2)/(3) - Delta x_2 =  - 4/3 Delta x_2`           [from eq.(i)

`because   Delta x_2 = - 3/4 L`

`:.`            `Delta x_1 = - (Delta x_2)/(3)   = - (-3L)/(4 xx 3) = - L/4`

Negative sign indicates that both moves in opposite directions.