# Confusion regarding a case of head-on elastic collision of two bodies of equal mass

If two bodies collide elastically, it's common to hear that the velocities, speeds and momenta are interchanged. I thought this was obvious. But if we consider two bodies moving with equal mass and speed, moving towards each other and colliding, they would both come to rest, to conserve momentum. In this case, neither of the three aforementioned things are 'interchanged'. Where am I going wrong?

• If it's an elastic collision, the two bodies don't come to rest ... they bounce off each other. A perfectly inelastic collision is required for the two bodies to come to rest. Commented Oct 29, 2021 at 15:45

One of the things which you must remember in the case of elastic collisions is that with momentum, energy is also conserved. In all e<1 collision, some energy is wasted as sound, heat, etc Now, consider that the two objects have equal mass, and speed, and are moving towards each other. Here, there are two viable paths to conserve momentum, since initial momentum is zero.

1. Final velocity of both objects is 0, so net momentum will be 0, which is the one you mentioned.
2. The velocities are interchanged. Assume the objects are traveling along x-axis Initially, the body B-1 has speed S along the positive x-axis, and the body B-2 has speed S along negative x-axis. After colliding, the velocities are interchanged, while the momentum remains zero. Let the mass of each body be M, speed=S However, since energy is scalar and has to be conserved(Elastic Collision),

                                **Initial Energy=Final Energy**.


Also, let it be clear that we are only dealing with kinetic energy here. Since initial energy= MS^2, Final Energy should also be equal to MS^2, which is only possible in the second case. So, the velocities are interchanged.

## Prerequisites

Momentum conservation states that:

$$m_{1} \mathbf v_{1i} + m_{2} \mathbf v_{2i}=m_{1} \mathbf v_{1f} + m_{2} \mathbf v_{2f}$$

where $$m_1$$ and $$m_2$$ are the masses of the objects, $$\mathbf v_{1i}$$ and $$\mathbf v_{2i}$$ are the initial velocities and $$\mathbf v_{1f}$$ and $$\mathbf v_{2f}$$ are the initial velocities. In your specific cas the momentum equation reduces to

\begin{align} m\mathbf v + m (-\mathbf v) & = m\mathbf v _1 + m\mathbf v _2\\ 0&=m\mathbf v _1 +m \mathbf v _2\\ 0&=\mathbf v _1 + \mathbf v _2 \tag{1} \end{align}

where $$m$$ is the mass of the objects, $$\mathbf v$$ is the initial velocity of both the objects and $$\mathbf v_1$$ and $$\mathbf v_2$$ are the final velocities.

Now there are two variables of interest ($$\mathbf v_1$$ and $$\mathbf v_2$$) but there is only one equation. So as you can see, momentum conservation cannot be used alone to predict the final velocities of two colliding objects. You need to apply some other constraint/equation which will then help you in uniquely determining the final velocities.

In this case, since you are talking about elastic collisions, we can apply two equivalent constraints, one being energy conservation other being $$e=1$$ (where $$e$$ is the coefficient of restitution).

## Energy conservation

Note that from $$(1)$$, $$|\mathbf v_1| = |\mathbf v_2 |$$. From now on, I will denote the magnitudes of velocities as $$v_1$$, $$v_2$$ and $$v$$. Thus applying energy conservation, we get

\begin{align} \frac 1 2 m v^2+ \frac 1 2 m v^2 &= \frac 1 2 m v_1 ^2 + \frac 1 2 m v_2 ^2\\ v^2 &= v_1^2 =v_2^2\\ |\mathbf v|&=|\mathbf v_1| = |\mathbf v_2 | \end{align}

It follows trivially that the initial velocities have reversed their directions and thus the velocities have been exchanged.

## Coefficient of restitution

The coefficient of restitution is defined as the ratio of relative velocity of separation and relative velocity of approach. Thus

\begin{align} e&=\frac{|\mathbf v_1 -\mathbf v_2|}{|\mathbf v -(-\mathbf v)|}\\ 1&=\frac{|\mathbf v_1 -\mathbf v_2|}{2|\mathbf v|}\\ 2|\mathbf v|&=|\mathbf v_1 -\mathbf v_2| \end{align}

Since this is a one dimensional collision, we can convert the magnitude of difference in velocities to the difference in magnitudes of velocities (with proper sign convention):

$$2v=v_1-v_2$$

Solving this and equation $$(1)$$ gives us the final velocities. Again you would notice that the initial velocities have reversed and exchanged.

## Fallacy in your argument

The case which you are thinking about, happens when the collision is perfectly inelastic, which implies that the coefficient of restitution is zero.

Elastic collision means Collison without loss of kinectic energy . In this collision Collison , e=1 means relative velocity of approach=. relative velocity of separation
If you know this now you will understand easily

If we take formulae for boys conservation of momentum why?? As we know both Collison follow this.let both mass m)

(mu1+mu2=mv1+mv2) If we take equation (u1-u2=v2-v1 ) V1= (m +m)/m-m)u1+ 2mu2/m+m Now m-m= o so this variable tend to zero (V1= u2)

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Commented Oct 29, 2021 at 11:08