Two masses m_{a} and m_{b} moving with veloci | vedclass

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(C) In an elastic collision,both linear momentum and kinetic energy are conserved.Let the initial velocities be $v_{a}$ and $-v_{b}$ (since they move

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(C) In an elastic collision,both linear momentum and kinetic energy are conserved.Let the initial velocities be $v_{a}$ and $-v_{b}$ (since they move in opposite directions).After the collision,the velocities are $-v_{b}$ and $v_{a}$ respectively.According to the law of conservation of linear momentum:$m_{a}v_{a} - m_{b}v_{b} = m_{a}(-v_{b}) + m_{b}v_{a}$Rearranging the terms:$m_{a}v_{a} + m_{a}v_{b} = m_{b}v_{a} + m_{b}v_{b}$$m_{a}(v_{a} + v_{b}) = m_{b}(v_{a} + v_{b})$Since $(v_{a} + v_{b}) 
eq 0$,we can divide both sides by $(v_{a} + v_{b})$:$m_{a} = m_{b}$Therefore,the ratio $\frac{m_{a}}{m_{b}} = 1$.

Two masses $m_{a}$ and $m_{b}$ moving with velocities $v_{a}$ and $v_{b}$ in opposite directions collide elastically. After the collision,$m_{a}$ and $m_{b}$ move with velocities $v_{b}$ and $v_{a}$ respectively. The ratio $m_{a} : m_{b}$ is:

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