Does mass affect velocity when travelling through frictionless medium? I found the following question on an standardized test, and was debating with some friends what the answer would be:
A car of mass M is travelling with a constant velocity through a plane in which friction is non-existent. An object of mass m (m = M/3) that is falling perpendicularly to the car lands inside of it. How will the velocity of the car be affected?
This illustration can help explain the problem.

My initial thought was that the velocity would be the same, given that friction is non-existent and that the momentum of the falling object is perpendicular to that of the car. However, some friends suggested that, since the mass of the car increases, the velocity should decrease.
 A: Your friends are correct.
If there is no force in the left-right direction, then linear momentum will be conserved in that direction. Because the new composite object has more mass than the original object, it will have a lower speed to the right.
What about energy?
Kinetic energy is not conserved in this case, because the collision is inelastic. The kinetic energy lost in the collision (and the downward momentum lost, for that matter) is absorbed by whatever wall is preventing the composite object from continuing to move downward.
A: Let us first simplify a bit, and get rid of the vertical movement of $m$. It is much simpler to assume that $m$ is standing with zero velocity in the path of $M$, which will collide it at $t=0$. This will be the same problem along the horizontal axis, and avoids to superimpose it with another problem in the vertical direction.
The conserved quantities are both the kinetic energy and the momentum. If we call $v$ and $V$ the velocities of $m$ and $M$ respectively, and $V_0$ the initial velocity of $M$, we have that 
$$
mv + MV = MV_0
$$
$$
mv^2 + MV^2 = MV_0^2
$$
I'll leave it for you to solve yourself, but you'll find that the velocities $V$ and $v$ are different: the masses do not travel together.
If you want to model a case where they'll travel together, you'll need to provide with a mechanism dissipating energy in the collision. In that case, the kinetic energy will not be conserved, and in the absence of external forces, you'll have $V=v=\frac{M}{M+m} V_0$. The system will have a lower kinetic energy than previously.
Thanks to @NeuroFuzzy for pointing my previous wrong answer (deleted).
A: The relation $v=\sqrt{Fd/m}$ suggests that $v$ depends on not only mass but also the external force acting on a body multiplied by distance travelled by the body divided by its mass.
