Question

show that the total energy is conserved in elastic collision?

Answer 1

Answer:

The conservation of the total momentum before and after the collision is expressed by:

{\displaystyle \,\!m_{1}u_{1}+m_{2}u_{2}\ =\ m_{1}v_{1}+m_{2}v_{2}.} {\displaystyle \,\!m_{1}u_{1}+m_{2}u_{2}\ =\ m_{1}v_{1}+m_{2}v_{2}.}

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

{\displaystyle {\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}\ =\ {\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}.} {\displaystyle {\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}\ =\ {\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}.}

These equations may be solved directly to find {\displaystyle v_{1},v_{2}} v_{1},v_{2} when {\displaystyle u_{1},u_{2}} {\displaystyle u_{1},u_{2}} are known:

{\displaystyle {\begin{array}{ccc}v_{1}&=&\displaystyle {\frac {m_{1}-m_{2}}{m_{1}+m_{2}}}u_{1}+{\frac {2m_{2}}{m_{1}+m_{2}}}u_{2}\\[.5em]v_{2}&=&\displaystyle {\frac {2m_{1}}{m_{1}+m_{2}}}u_{1}+{\frac {m_{2}-m_{1}}{m_{1}+m_{2}}}u_{2}\end{array}}} {\displaystyle {\begin{array}{ccc}v_{1}&=&\displaystyle {\frac {m_{1}-m_{2}}{m_{1}+m_{2}}}u_{1}+{\frac {2m_{2}}{m_{1}+m_{2}}}u_{2}\\[.5em]v_{2}&=&\displaystyle {\frac {2m_{1}}{m_{1}+m_{2}}}u_{1}+{\frac {m_{2}-m_{1}}{m_{1}+m_{2}}}u_{2}\end{array}}}

If both masses are the same, we have a trivial solution:

{\displaystyle \ v_{1}=u_{2}} {\displaystyle \ v_{1}=u_{2}}

{\displaystyle \ v_{2}=u_{1}} {\displaystyle \ v_{2}=u_{1}}.

This simply corresponds to the bodies exchanging their initial velocities to each other.