How is momentum conserved in the case of a ballistic pendulum collision?

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?

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.

• it seems to me that if we assume an instantaneous collision, then the velocity of $m_1$ is horizontal, both immediately before and immediately after the collision, whereas the velocity of $m_2$ is zero before the collision and horizontal immediately after. Since all non-zero velocities are horizontal immediately before/after the collision, I would say that momentum is conserved in both axes.
– Javi
May 12 at 15:10
• @Javi String provides a tension T such that $T-mg \cos{\theta}$ results in a net centripetal force $mv^2/r$, where $v$ is velocity of $m_1$ at that instant and $\theta$ is angle between string and negative y-axis assuming origin is point of rotation. So net force along y-axis is always present, and hence momentum of the system along y-axis is not conserved. May 12 at 16:31
• I understand your point. But then, if you claim that $p_y$ is not conserved during the instantaneous collision, then you are implying that $p_y$ is not zero just before and after the collision, as it looks in the picture. Since $p_y = p_{y1} + p_{y2}$, would you say that $p_{y1}$ or $p_{y2}$ (or perhaps both) are nonzero immediately before or after the collision?
– Javi
May 12 at 19:02
• @Javi Assuming collision occurs at lowest point, $p_{y2}$ is zero, but $p_{y1}$ is non-zero due to centripetal force, and it would continue its swing (if we were to immediately remove $m_2$ after collision so as to not obstruct it) . If collision is completely inelastic they would stick and both would swing. If collision is perfectly elastic, $v_1 = 0$ after collision and no centripetal force acts. The point is as long as $m_1$ has a velocity at any point during collision which is non-zero, an external force acts on it along y-axis and momentum is not conserved. May 13 at 1:44
• Although I can see that $\sum F_y^{ext} \neq 0$ so that $p_y$ is not conserved, it really seems from the picture that $\vec{v}_1$ is tangential to the circumference, and this holds for the instants before and after the collision. I believe that stating that $p_{y1} \neq 0$ after the collision is against the hypothesis that the collision is instantaneous.
– Javi
May 13 at 12:23

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).

• 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...
– Gert
Apr 9, 2021 at 15:54
• 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? Apr 9, 2021 at 16:05
• 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... Apr 9, 2021 at 16:28
• 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.
– nasu
Apr 9, 2021 at 16:49
• 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. Apr 9, 2021 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.