I'm confused over the Wiki equation for an elastic collision. Does anyone know how equations (1) and (2) are formed to result in (3)? I think I'm overlooking some simple algebra.

Consider particles 1 and 2 with masses $m_1$, $m_2$, and velocities $u_1$, $u_2$ before collision, $v_1$, $v_2$ after collision. The conservation of the total momentum before and after the collision is expressed by:

$$m_{1}u_{1}+m_{2}u_{2} \ =\ m_{1}v_{1} + m_{2}v_{2} \tag 1$$

Likewise, the conservation of the total kinetic energy is expressed by:

$$\tfrac12 m_1u_1^2+\tfrac12 m_2u_2^2 \ =\ \tfrac12 m_1v_1^2 +\tfrac12 m_2v_2^2 \tag 2$$

These equations may be solved directly to find $v_1,v_2$ when $u_1,u_2$ are known.

$$ \begin{array}{ccc} v_1 &=& \dfrac{m_1-m_2}{m_1+m_2} u_1 + \dfrac{2m_2}{m_1+m_2} u_2 \\[.5em] v_2 &=& \dfrac{2m_1}{m_1+m_2} u_1 + \dfrac{m_2-m_1}{m_1+m_2} u_2 \end{array} \tag 3 $$


2 Answers 2


Note that the following procedure is valid only for the one-dimensional elastic collision. If elastic collision happens in two dimensions, then conservation of momentum (your Eq. (1)) gives 2 equations while conservation of energy (your Eq. (2)) gives only 1 equation, in which case you have 3 equations and 4 unknowns (x and y components for two velocities). This means that in the two-dimensional elastic collision you need to know at least one magnitude or one angle for one of the two velocities after the collision in order to be able to solve for other unknowns.

One-dimensional elastic collision

From your Eq. (1) it follows

$$m_1 (u_1 - v_1) = -m_2 (u_2 - v_2) \tag {1a}$$

From your Eq. (2) it follows

$$\frac{1}{2} m_1 (u_1^2 - v_1^2) = -\frac{1}{2} m_2 (u_2^2 - v_2^2)$$

$$m_1 (u_1 - v_1) (u_1 + v_1) = -m_2 (u_2 - v_2) (u_2 + v_2) \tag{2a}$$

From my Eqs. (1a) and (2a) it follows

$$u_1 + v_1 = u_2 + v_2$$

$$\boxed{u_1 - u_2 = -(v_1 - v_2)} \tag {3a}$$

This equation tells you that relative velocity before and after elastic collision has the same magnitude and opposite direction, which is important property of elastic collisions.

You can now combine my Eqs. (1a) and (3a) to reach your Eq. (3)

$$v_2 = u_1 - u_2 + v_1 \qquad \text{and} \qquad v_2 = u_2 + \frac{m_1}{m_2} (u_1 - v_1) \tag {4a}$$

$$\text{give} \qquad v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2 m_2}{m_1 + m_2} u_2$$

Procedure is similar to get the expression for velocity $v_2$.

  • $\begingroup$ Is (3a) correct? I thought it would be $u_1 + v_1 = u_2 + v_2$ after dividing (1a) by (2a). $\endgroup$
    – Nick
    Mar 30, 2022 at 14:55
  • $\begingroup$ @Nick Yes, you get exactly that. But then you can move $u_2$ on the left side, and $v_1$ on the right side, and you get my Eq. (3a). $\endgroup$ Mar 30, 2022 at 14:56
  • $\begingroup$ Re: combining (1a) and (3a) to reach (3). Are you multiplying (3a) by a constant and adding each side to (1a)? I'm having trouble understanding how combination results in $v_1, v_2$ $\endgroup$
    – Nick
    Mar 30, 2022 at 15:32
  • $\begingroup$ No, I am not multiplying Eq. (3a) by a constant. In Eq. (3a) you leave $v_2$ on one side and put everything else on the other, and you do the same with Eq. (1a). That is how you get my Eq. (4a). From the two equations in my Eq. (4a) you can write $u_1 - u_2 + v_1 = u_2 + \frac{m_1}{m_2} (u_1 - v_1)$ and from this you can easily solve for $v_1$. Once you know expression for $v_1$, use $v_2 = u_1 - u_2 + v_1$ to solve for $v_2$. $\endgroup$ Mar 30, 2022 at 15:34
  • $\begingroup$ From the two equations in (4a) placed in an equality, I get $v_1 = 2 u_2 + \frac {m_1}{m_2} (u_1 - v_1)$, but do not see how to arrange this further to solve for $v_1$. $\endgroup$
    – Nick
    Mar 30, 2022 at 17:07

Usually you work with the differences $𝑚_1(𝑢_1-𝑣_1)=m_2(𝑣_2-u_2)$ and $m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)$ and divide the second by the first to get $u_1+v_1=u_2+v_2$ than put this in the first equation to get the result, you should be able to do the rest.


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