What would be the final momentum if two bodies of mass m each and moving with velocities in each in opposite direction collide?

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Table of Contents Show

Energy conservation
Coefficient of restitution
Fallacy in your argument

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.

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CONCEPT:

Conservation of Momentum: According to this principle, in an isolated system, the vector sum of linear momenta of all bodies of a system is conserved and is not affected due to their mutual action and reaction.
- If two bodies of different masses collide with each other and there is no external force acted on it then,

The initial momentum of bodies (P1) = Final momentum of bodies (P2)