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.

Problem Illustration

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.


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.

  • $\begingroup$ But the conservation of kinetic energy of the horizontal direction is what slows the car. $\endgroup$ – user6972 Apr 1 '14 at 23:14
  • $\begingroup$ @user6972 and Colin: if the system behaves as described in the question, kinetic energy is not conserved and there is also an inelastic collision along the horizontal direction. See my answer below. $\endgroup$ – Joce Apr 2 '14 at 12:08

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).

  • $\begingroup$ I understand the question to basically be if M suddenly weighed M+m, because m is inside M. $\endgroup$ – user6972 Apr 2 '14 at 17:00
  • 1
    $\begingroup$ If $m$ is "magically" added, then you have to ask the magician whether he does that keeping the energy, the momentum, or why not the velocity a constant. The way $m$ is connected to $M$ does matter, e.g. if they are connected by a piece of loose rubber band that tightens at $t=0$, you'll have a system that oscillates around an average velocity. Or in the case drawn in the question, you'd have $m$ banging back and forth in the $M$ container (for ever if there is no dissipation in the banging), and the velocity of $M$ would jump from one value to the other after each bang. $\endgroup$ – Joce Apr 2 '14 at 20:20

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.