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.

Derivation:

In an elastic collision, momentum is conserved:

∑pf=∑pim1v1+m2v2=m1u1+m2u2m1v1−m1u1=m2u2−m2v2m1(v1−u1)=m2(u2−v2)\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+m2u22m1v12−m1u12=m2u22−m2v22m1(v12−u12)=m2(u22−v22)m1(v1−u1)(v1+u1)=m2(u2−v2)(u2+v2)m2(u2−v2)(v1+u1)=m2(u2−v2)(u2+v2)v1+u1=v2+u2v2=u1+v1−u2\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+v1−u2)=m1u1+m2u2m1v1+m2u1+m2v1−m2u2=m1u1+m2u2m1v1+m2v1=m1u1+2m2u2−m2u1(m1+m2)v1=(m1−m2)u1+2m2u2v1=(m1−m2m1+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+v1−u2v2=u1+(m1−m2m1+m2)u1+(2m2m1+m2)u2−u2v2=(1+m1−m2m1+m2)u1+(2m2m1+m2−1)u2v2=(m1+m2m1+m2+m1−m2m1+m2)u1+(2m2m1+m2−m1+m2m1+m2)u2v2=(2m1m1+m2)u1+(m2−m1m1+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=(m1−m2m1+m2)u1+(2m2m1+m2)u2v2=(2m1m1+m2)u1+(m2−m1m1+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[m1−m22m22m1m2−m1][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} }

#Mechanics

Note by Gandoff Tan
1 year, 5 months ago

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