This is a pretty basic conceptual question about the conservation of linear momentum.

Consider an isolated system of 2 fixed-mass particles of masses $m_1$ and $m_2$ moving toward each other with velocities $v_1(t)$ and $v_2(t)$ respectively.

Now conservation of momentum says that at any point during the particles' motion the quantity $$m_1v_1(t) + m_2v_2(t) =constant$$

With non-zero velocities and non-zero masses this constant will be non-zero.

Let us say the particles collide at time $t_0$. At the point of collision, both particles have velocity zero. which would mean that the constant above will be zero. Contradiction.

I realize I might be going wrong in my reasoning at the point of collision.

In fact, I feel defining velocity at that point would not even make sense, since if one considers the displacement functions $x_i(t)$ $i=1,2$ of the particles, then $t_0$ would represent a point of non-differentiability of $x_i(t)$ for $i=1,2$.

So assuming there are no collisions, by following the text-book derivation I can see why

$$m_1v_1(t) + m_2v_2(t) = C_1$$ before the collision and $$m_1v_1(t) + m_2v_2(t) = C_2$$ after the collision

would hold true, but not why $C_1=C_2$

Can someone help me in clearing this up?

  • 1
    $\begingroup$ "With non-zero velocities and non-zero masses this constant will be non-zero." Non-sequitur. It will in fact be zero for equal masses and opposite velocities, which "happens" to be also the only case in which "At the point of collision, both particles have velocity zero" is possible. So even with zero-duration collisions, there would be no contradiction here. $\endgroup$ Commented Mar 30, 2013 at 21:30

2 Answers 2


Instead of treating the collision as happening instantaneously, i.e. taking zero time, assume it takes some unknown but non-zero time and during that time there is some force between the colliding bodies, $F(t)$ that is a function of time.

The change in the momentum of object 1 is just the impulse $J$ given by:

$$ J_1 = \int \space F_{12} dt $$

But Newton's third law tells us that the force on the second object is equal and opposite to the force on the first body so the change of momentum for the second body is the impulse:

$$ J_2 = \int \space -F_{12} dt = -J_1 $$

And therefore the total momentum change during the collision is zero.

Note that we've made no assumptions about how long the collision takes or the forces acting during it. You can let the collision time tend to zero, but if you do this the force will tend to infinity and you're left with the unphysical problem of integrating an infinite force for zero time.

  • $\begingroup$ So treating collision as a zero-time event was the main flaw. So even during the collision one can assume that the velocity function continues to be defined, only that it changes very fast (large accelaration). Is that correct? $\endgroup$ Commented Mar 30, 2013 at 18:50
  • $\begingroup$ Yes. In any real collision the collision takes a finite time while the bodies deform. Even for elementary particles, e.g. in the LHC, the collision isn't instantaneous because the particles interact via a force (strong, weak, etc) over a non-zero distance. $\endgroup$ Commented Mar 31, 2013 at 6:24

During the collision, particle 1 will exert a force on particle 2. Let's call this force $F_{12}(t)$. Now, at the same time, particle 2 will exert a force on particle to, $F_{21}(t)$. According to Newton's third law, $F_{21}(t)=-F_{12}(t)$.

Suppose the collision happens from $t=0$ until $t=\tau$, then $F_{12}(t)=0$ outside this interval.

From Newton's second law of motion, we obtain for the first particle: $$m_1 \frac{dv_1}{dt}=F_{21}=-F_{12}$$

Or, in other words

$$m_1v_1(\tau)-m_1v_1(0)=-\int_0^\tau F_{12}dt$$

And, for particle 2

$$m_2v_2(\tau)-m_2v_2(0)=\int_0^\tau F_{12}dt$$

The term on the right hand side is also referred to as impulse.

Now, check out what happens if we add the last two equations:

$$m_1v_1(\tau)-m_1v_1(0)+m_2v_2(\tau)-m_2v_2(0)=-\int_0^\tau F_{12}dt+\int_0^\tau F_{12}dt$$

Or, in other words


And, thus



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