7) A train, originally of mass $M$, is traveling on a frictionless straight horizontal track with constant speed $v$. Snow starts to fall vertically and sticks to the train at a rate of $\rho$, where $\rho$ has units of kilograms per second. The train's engine keeps the train moving at constant speed $v$ as snow accumulates on the train.

What is the rate at which the kinetic energy of the train and snow increases?

A) $0$

B) $Mgv$

C) $\frac{1}{2}Mv^2$

D) $\frac{1}{2}pv^2$

E) $\rho v^2$
1. **Identify the initial kinetic energy of the train:**
   The initial kinetic energy \( K \) of the train with mass \( M \) and velocity \( v \) is given by:
   \[
   K = \frac{1}{2} M v^2
   \]

2. **Determine the rate of change of mass:**
   Snow is falling and sticking to the train at a rate of \( \rho \) kilograms per second. Therefore, the rate of change of the mass of the train is:
   \[
   \frac{dm}{dt} = \rho
   \]

3. **Express the kinetic energy as a function of time:**
   As snow accumulates, the mass of the train increases. Let \( m(t) \) be the mass of the train at time \( t \). Initially, \( m(0) = M \). After time \( t \), the mass of the train is:
   \[
   m(t) = M + \rho t
   \]

4. **Calculate the kinetic energy at time \( t \):**
   The kinetic energy \( K(t) \) of the train at time \( t \) is:
   \[
   K(t) = \frac{1}{2} m(t) v^2 = \frac{1}{2} (M + \rho t) v^2
   \]

5. **Find the rate of change of kinetic energy:**
   To find the rate at which the kinetic energy is increasing, we differentiate \( K(t) \) with respect to time \( t \):
   \[
   \frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2} (M + \rho t) v^2 \right)
   \]
   Since \( M \) and \( v \) are constants, the differentiation yields:
   \[
   \frac{dK}{dt} = \frac{1}{2} v^2 \frac{d}{dt} (M + \rho t) = \frac{1}{2} v^2 \cdot \rho
   \]

6. **Simplify the expression:**
   \[
   \frac{dK}{dt} = \frac{1}{2} \rho v^2
   \]

Conclusion:
The rate at which the kinetic energy of the train and snow increases is:
\[
\boxed{\frac{1}{2} \rho v^2}
\]