To my knowledge, momentum is conserved when net external force equals zero.

In the case of a block $m_1$ attached to a string hitting another block $m_2$, there must be a net force on $m_1$, which would be its centripetal force, pointing inward. Therefore comparing just before $m_1$ hits $m_2$ and just after $m_1$ hits $m_2$, momentum cannot be conserved since net external force exists.

However the answer to this homework problem in my book states that momentum is still conserved.

How is this so?enter image description here


At the moment when $m_1$ collides with $m_2$, net force along X-axis is $0$ so, momentum is conserved in that direction. But momentum is not conserved along Y-axis.


You are right to be concerned. A better demonstration of momentum conservation would have been if the blocks had been sitting on perfectly frictionless ice, or even better: in space. The reason they chose this example because it shows how the famous Newton's cradle works. If you've ever seen a Newton's cradle in real life you know that after a while the balls slow down and come to rest. Especially the smaller ones. So obviously momentum is not totally conserved. But enough of it is conserved that it's still a nice experiment.

The reason the momentum is conserved is that the duration of the collision is very short. In this question I found and estimate that says the collision between two steel balls would last less than a millisecond. In a collision the force is very high but the timespan very short. The contributions of other forces can be approximated by $\Delta \vec p\approx\vec F\Delta t$. So as $\Delta t$ gets small the other contribution vanish.

So to conclude, don't worry that momentum isn't entirely conserved. The actual effect is small and you can still make predictions using momentum conservation. For example if you lift 3 out 5 balls in a Newton's cradle you can predict that 3 will bounce off which is still a nice result (in my opionion).

As a final note I should add that this experiment relies not only on momentum conservation but also on energy conservation. In a closed system momentum is always conserved but energy can still be dissipated as heat/sound etc. If a collision loses all of its energy (imagine throwing two blobs of clay into eachother in space) the two parts will stick to eachother. In a Newton's cradle you can recognise this when the balls start swinging together because the collisions aren't perfectly elastic (=energy conserving).

  • $\begingroup$ So obviously momentum is not totally conserved. No but that has little to do little to do with a ballistic pendulum (I also don't believe The reason they chose this example because it shows how the famous Newton's cradle works) For Newton's cradle there are obvious losses due to friction and even sound... $\endgroup$
    – Gert
    Apr 9 at 15:54
  • $\begingroup$ The I way I interpreted this answer was: because we are looking at times just before and just after the collision of m1 and m2, the time gap is too small to bring about a significant impulse that would cause a change in momentum. Am I correct? $\endgroup$
    – mikeeei
    Apr 9 at 16:05
  • $\begingroup$ But this makes another problem... if the collision takes such little time that the impulse is negligible, the impulse on block m2 would also be negligible, and therefore this would mean that m2 would not move. I need help... $\endgroup$
    – mikeeei
    Apr 9 at 16:28
  • $\begingroup$ The force along along the vertical axis has nothing to do with conservation of momentum along the x axis. However, if the pendulum moves during collision, the tension will have a horizontal component. Here comes into play the short duration of collision. The displacement during collision is small enough to be negleced, if you have a short collision. $\endgroup$
    – nasu
    Apr 9 at 16:49
  • 1
    $\begingroup$ Conservation of momentum has nothing to do with the duration of the forces between the two parts of the system. It simply follows from Newton's 3rd law that the internal forces are equal an opposite. $\endgroup$
    – alephzero
    Apr 9 at 16:50

In this case, conservation of momentum is applied along the individual axes as the given motion may be broken down into mutuallly independent motions along the mutually perpendicular axes. In this case, conservation of momentum is applied only along the x-axis as the net external force acting on the system of blocks along the x-axis is zero. However, since there is a net force acting on the system (the centripetal force on $m_1$) along the y-axis, we cannot apply the conservation of linear momentum along the y-axis.

So yes, we can apply the conservation of linear momentum in this problem but only along the x-axis.

Hope it helps.


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.