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I'm a physics tutor for 11th and 12th grade students. I taught my students that momentum is conserved only when the net external force on the system is zero. But in the books we are following, momentum was conserved in the following cases in spite of presence of an external force:

  1. A body (mass m1) is dropped from a tower (height h) and another body (mass m2) is projected vertically upward with initial velocity u. Find the velocities just after collision. (Gravitational force is the external force)

  2. Two bodies projected towards each other on a rough table surface, with all required parameters given asked to find the velocities after collision. (frictional force is the external force)

  3. Body explodes in mid air. Momentum is conserved before and after explosion. (Again gravitational force is the external force)

I explained to my students in all these cases though an external force is present it has negligible effect. But I'm not satisfied with my answer.

EDIT: Thank you for the comments and answer. It is clear that contribution of external force to change in momentum is negligible. Can someone do the calculation of taking change in momentum into consideration and showing that it is negligible?

claws
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  • That force is a property of the system itself, not external. For example, in the case of free fall, you cannot take the freely falling body only as your system. The system is the body in a gravitational field. No other force acts on the system, so that no change in total momentum of the system occurs. – UKH Feb 08 '17 at 12:54
  • @UKH You are right that you can always make the source of the external force part of your system and then correctly assert that momentum is conserved. This does however lead to relatively complex calculations. It is much easier to realize that on the timescale of the collission/explosion, the momentum change due to the external force is negligable. – Crimson Feb 08 '17 at 13:45

3 Answers3

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Consider your example 1 with two billiard balls, mass $0.16 \,\rm kg$, colliding with a red ball falling down with a speed of $5\, \rm m\,s^{-1}$ and colliding with a stationary white ball.
Applying conservation of linear momentum (assuming there are no external force and the collision is elastic) results in the red ball momentarily stopping and the white ball moving downwards at $5\, \rm m\,s^{-1}$.
The impulse (change of momentum) on each ball is $0.16 \times 5 = 0.8 \, \rm N\,s$.

During the collision the gravitational force would have had a effect on the balls but to know how much one must know the collision time.

Peter Bohacek has produced many Direct Measurement Videos and the relevant one for this answer is Billiard Ball Collision three consecutive frames from which are shown below.

enter image description here

This shows that the collision time is less than $0.001 \, \rm s$.

Going back to the falling red and white billiard balls, in a time of $0.001 \, \rm s$ the impulse due to gravity on one of the balls is $\text{~}\, 0.16 \times 10 \times 0.001 = 0.016 \,\rm Ns $ which is very much smaller than the impulse on the balls due to the collision, $ 0.8 \, \rm N\,s$.
So the assumption of a very short collision time resulting in very little effect on the outcome of the collision is a good one.

More time could have been spent analysing the video to get a more accurate upper bound for the collision time which might be the basis of a nice assignment?

Farcher
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  • How do you get 0.16×5=0.75 Ns? 0,16kgx5m/s=0.8kgm/s if you multiply by g to get force units, it results in 0.8*9.8=7.84 Ns. – jpcgandre Apr 18 '22 at 20:01
  • @jpcgandre Thank you for pointing out my error in the evaluation of the change in momentum which I have now corrected. – Farcher Apr 18 '22 at 22:57
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You are basically telling them the right thing. In all three situation if we assume that gravitation is just an external force, momentum is not conserved. The reason is that we don't take the momentum transfer between the object in the system to the source of the external force (in this case the earth) into account.

If two bodies in our system interact for a very short time we can still assume momentum conservation for this interaction. The reason is that the momentum transfer between the objects in our system is much larger than the momentum transfer to the objects outside our system during this brief time period.

Jannick
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    To complement this answer. Note that the error you make, by assuming conservation of momentum, can be estimated by calculating the change in momentum due to the external force during the time of the explosion/collission. When we assume the event to happen in a very short time span, this error becomes negligible. – Crimson Feb 08 '17 at 13:50
  • If your collision problems involve only rigid bodies or point particles, the duration of the collision in your mathematical model is zero, so any finite external force has no effect. In the collision model, the impulse is finite, but that corresponds to "an infinite force acting for zero time" (and yeah, I do know how to turn that arm-waving use of "$\infty \times 0 = \text{something finite}$" into valid mathematics, but that is way beyond 11th or 12th grade!) – alephzero Feb 08 '17 at 14:21
  • @Jannick I if an acrobat jumping off a trampoline catches a monkey on a pole of height h, why would I have to include earth into my system. Actually, even in this question the vertical momentum is assumed to be conserved but the error difference considering gravity and not considering it is not negligible at all. – suiz Apr 04 '18 at 12:52
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Momentum can be conserved in presence of non impulsive forces like gravitational forces and frictional force. In some cases frictional force is impulsive only when normal is impulsive. In presence of non impulsive forces momentum can be conserved since they do not produce finite change in momentum. So it remains constant . So in collision between projectile momentum in both direction is conserved.but external forces can cause change in momentum.

JATIN KAMATHI
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