Derivation of the Elastic Collision Formula

In the scenario of an one-dimensional elastic collision between two objects, 11 and 22, their final velocities, v1v_1 and v2v_2 can be found with the following formula knowing their individual masses, m1m_1 and m2m_2, and their initial velocities, u1u_1 and u2u_2.


In an elastic collision, momentum is conserved:

pf=pim1v1+m2v2=m1u1+m2u2m1v1m1u1=m2u2m2v2m1(v1u1)=m2(u2v2)\begin{aligned} \sum p_f&=\sum p_i\\ m_1v_1+m_2v_2&=m_1u_1+m_2u_2\\ m_1v_1-m_1u_1&=m_2u_2-m_2v_2\\ \textcolor{#3D99F6}{m_1(v_1-u_1)}&=\textcolor{#3D99F6}{m_2(u_2-v_2)} \end{aligned}

In an elastic collision, kinetic energy is conserved:

Kf=Ki12m1v12+12m2v22=12m1u12+12m2u22m1v12+m2v22=m1u12+m2u22m1v12m1u12=m2u22m2v22m1(v12u12)=m2(u22v22)m1(v1u1)(v1+u1)=m2(u2v2)(u2+v2)m2(u2v2)(v1+u1)=m2(u2v2)(u2+v2)v1+u1=v2+u2v2=u1+v1u2\begin{aligned} \sum K_f&=\sum K_i\\ \frac12m_1v_1^2+\frac12m_2v_2^2&=\frac12m_1u_1^2+\frac12m_2u_2^2\\ m_1v_1^2+m_2v_2^2&=m_1u_1^2+m_2u_2^2\\ m_1v_1^2-m_1u_1^2&=m_2u_2^2-m_2v_2^2\\ m_1(v_1^2-u_1^2)&=m_2(u_2^2-v_2^2)\\ \textcolor{#3D99F6}{m_1(v_1-u_1)}(v_1+u_1)&=m_2(u_2-v_2)(u_2+v_2)\\ \textcolor{#3D99F6}{m_2(u_2-v_2)}(v_1+u_1)&=m_2(u_2-v_2)(u_2+v_2)\\ v_1+u_1&=v_2+u_2\\ \textcolor{#20A900}{v_2}&=\textcolor{#20A900}{u_1+v_1-u_2} \end{aligned}

pf=pim1v1+m2v2=m1u1+m2u2m1v1+m2(u1+v1u2)=m1u1+m2u2m1v1+m2u1+m2v1m2u2=m1u1+m2u2m1v1+m2v1=m1u1+2m2u2m2u1(m1+m2)v1=(m1m2)u1+2m2u2v1=(m1m2m1+m2)u1+(2m2m1+m2)u2\begin{aligned} \sum p_f&=\sum p_i\\ m_1v_1+m_2\textcolor{#20A900}{v_2}&=m_1u_1+m_2u_2\\ m_1v_1+m_2(\textcolor{#20A900}{u_1+v_1-u_2})&=m_1u_1+m_2u_2\\ m_1v_1+m_2u_1+m_2v_1-m_2u_2&=m_1u_1+m_2u_2\\ m_1v_1+m_2v_1&=m_1u_1+2m_2u_2-m_2u_1\\ (m_1+m_2)v_1&=(m_1-m_2)u_1+2m_2u_2\\ \textcolor{#D61F06}{v_1}&=\textcolor{#D61F06}{\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2} \end{aligned}

v2=u1+v1u2v2=u1+(m1m2m1+m2)u1+(2m2m1+m2)u2u2v2=(1+m1m2m1+m2)u1+(2m2m1+m21)u2v2=(m1+m2m1+m2+m1m2m1+m2)u1+(2m2m1+m2m1+m2m1+m2)u2v2=(2m1m1+m2)u1+(m2m1m1+m2)u2\begin{aligned} v_2&=u_1+\textcolor{#D61F06}{v_1}-u_2\\ v_2&=u_1+\textcolor{#D61F06}{\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2}-u_2\\ v_2&=\left(1+\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}-1\right)u_2\\ v_2&=\left(\frac{m_1+m_2}{m_1+m_2}+\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}-\frac{m_1+m_2}{m_1+m_2}\right)u_2\\ v_2&=\left(\frac{2m_1}{m_1+m_2}\right)u_1+\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2 \end{aligned}

v1=(m1m2m1+m2)u1+(2m2m1+m2)u2v2=(2m1m1+m2)u1+(m2m1m1+m2)u2\therefore \boxed{ \begin{aligned} v_1&=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2\\ v_2&=\left(\frac{2m_1}{m_1+m_2}\right)u_1+\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2 \end{aligned} }

or, equivalently:

[v1v2]=1m1+m2[m1m22m22m1m2m1][u1u2]\boxed{ \def\arraystretch{1.5} \begin{bmatrix} v_1\\v_2 \end{bmatrix}= \frac1{m_1+m_2}\begin{bmatrix} m_1-m_2&2m_2\\ 2m_1&m_2-m_1 \end{bmatrix} \begin{bmatrix} u_1\\u_2 \end{bmatrix} }

Note by Gandoff Tan
1 year, 7 months ago

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