# Can momentum not be conserved in an inelastic collision?

Consider an inelastic collision in which a ball B of light mass moves towards a ball of clay A and ultimately sticks to it. Suppose that the whole system (A+B) comes to rest. Every book on Newtonian mechanics says that inelastic collision conserves momentum. But it is clear that the momentum is not conserved in this collision. What am I missing?

• You say Suppose that the whole system (A+B) comes to rest but what do you mean by this? Do you mean we're in the centre of mass frame so the two balls come in from opposite sides and come to rest after colliding? If so the total momentum is zero in the COM frame both before and after the collision. – John Rennie Nov 25 '18 at 6:29
• They can't just come to rest after the collision unless some external force is acting e.g. if the two balls are on a table top and there is friction with the table. – John Rennie Nov 25 '18 at 6:33
• @JohnRennie Probably you're right that the final system cannot come to rest. Momentum conservation is an experimentally verified fact for Newtonian mechanics. It cannot be challenged in Newtonian mechanics. – mithusengupta123 Nov 25 '18 at 7:00
• But when you say the fused balls come to rest you're implicitly assuming that the momentum is not conserved. What I'm getting at is you offer no justification for this assumption so as it stands your question is meaningless. – John Rennie Nov 25 '18 at 7:02
• The question isn't valid or useful. To paraphrase, you are asking whether momentum conservation is required in a scenario where you hypothesize that momentum is not conserved. – Rob Jeffries Nov 25 '18 at 10:00

They won't come to rest, unless the momentum of the two masses fused is spent doing some work. Or the rsultatant speed is very small.

Let's say a small meteor hits the earth after losing most of its mass burning in the atmosphere, and becomes one with it.

Its momentum is shared between the sum of earth and the meteor. But the earth is much more massive for the change to be measurable.

Even the part of its momentum spent in the atmosphere to accelerate the air around it and heat it up will affect the momentum of the atmosphere and eventually the earth.

• Let the momentum of the lighter mass is distributed as random kinetic energies of the clay particles. Who stops that? – mithusengupta123 Nov 25 '18 at 7:02

'I can't still rule out the possibility of distributing the momentum of the lighter particle into random internal motion of the particles of the clay'...This is a deeper and more interesting topic. The key here is that when you consider the ball to have only 3 degrees of freedom (velocity of the ball is a vector in 3D space) you are considering the ball as a point object. On the other hand, when you are 'zooming-in' and considering the ball as a system of N particles, each one with its own speed, you are considering a system with 3xN degrees of freedom. The relationship between these two points of view is that the velocity in the first case (macroscopic) is an average of the velocities of the particles from the second case (microspcopic)...and so we enter the field of statistical mechanics. The concept of inelastic collision (loss of kinetic energy) is 'true' only in the macroscopic case, while if we consider the microscopic point of view, the kinetic energy is never lost. It is only transformed from pure mechanical (average with no fluctuation) to thermal (fluctuating with zero average).
So a reformulation of your question would be: given that at microscopic level both momentum and energy are conserved, and given that momentum and kinetic energy are strictly related mathematically (mV vs 1/2mV^2) why at macroscopic level we can consider a non-conservative kinetic energy but we still have conservation of momentum? The simple version of the answer would be that in the macroscopic point of view we do not have the mechanical analogue of temperature but I think there is a deeper answer that has to do with the difference between the average of a vector random variable (momentum) and the average of a scalar quadratic random variable (as kinetic energy) but I cannot give a more rigorous answer right now...

The principal of COM is only true if there is no external net force acting on the system. You say that heavier ball A is stationary, and ball B is the lighter one. In an inelastic collision, we would see that both balls would lump together. However, the presence of external frictional forces prevents the movement of the final lump. What exactly results in the movement of the resultant 2 objects would be the force of collision. If there is friction, then we would certainly then observe that COM is not applicable here.

• You can suspend the clay ball in air by a string so that there is no external friction, and now imagine the same i.e. A and B sticking together and coming to rest. – mithusengupta123 Feb 3 at 13:01

If it is at rest after the collision, it implies that body A has positive momentum and body B negative with the exact proportion $$m_A\, v_A + m_B\, v_B = 0$$

Using conservation of momentum, the expression for the final common velocity $$\vec{v}_f$$ is: $$\vec{v}_f=\frac{m_1\vec{v}_{1i}+m_2\vec{v}_{2i}}{m_1+m_2}$$ where $$\vec{v}_{1i},\vec{v}_{2i}$$ are the initial velocities of the lighter ball of mass $$m_1$$ and heavy clay ball of mass $$m_2$$ respectively. Note that, since $$\vec{v}_{2i}=0$$, $$\vec{v}_f\neq 0$$ i.e., as JohnRennie points out the final system cannot come to rest.

New addition What actually can happen is the following. The lighter ball can will impart its momentum to the particles making up the the clay ball in such a way so as to deform it either by distorting it in shape and by adding mass to it. Although to the naked eye, the heavier ball may not move perceptibly, it centre of mass indeed shifts with velocity $$v_f$$ after they stick. However, since it is a composite object, not a point mass, various internal forces will come into play to quickly make $$v_f=0$$. The workdone by the internal forces then transform into heat.

The total momentum of a system is always conserved . There may be losses of energy due to plastic deformation but no loss of momentum.

• The loss is of kinetic energy. If there is plastic deformation there is temperature change, and/or stress energy increase (imagine a spring compressing but never decompressing.) Total energy is conserved. – Bill N Mar 6 at 16:09
• Yes losses of kinetic energy. – Max Destiny Mar 6 at 16:11