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Relative Velocity

Derive an exp...

Question

Derive an expression for the velocity of the two masses $m_{1}$ and $m_{2}$ with speeds $u_{1}$ and $u_{2}$ undergoing elastic collision in one direction.

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Solution

Let two bodies of masses $m_{1}$ and $m_{2}$ moving with velocities $u_{1}$ and $u_{2}$ along the same straight line.

And consider the two bodies collide and after collision $v_{1}$ and $v_{2}$ be the velocities of two masses.

Before collision

Momentum of mass $m_{1} = m_{1} u_{1}$
Momentum of mass $m_{2} = m_{2} u_{2}$

Total momentum before collision
$p_{1} = m_{1} u_{1} + m_{2} u_{2}$

Kinetic energy of mass $m_{1} = \frac{1}{2} m_{1} {u_{1}}^{2}$
Kinetic energy of mass $m_{2} = \frac{1}{2} m_{2} {u_{2}}^{2}$

Thus,

Total kinetic energy before collision is
$K . E = \frac{1}{2} m_{1} {u_{1}}^{2} + \frac{1}{2} m_{2} {u_{2}}^{2}$

After collision

Momentum of mass $m_{1} = m_{1} v_{1}$
Momentum of mass $m_{2} = m_{2} v_{2}$

Total momentum before collision
$P_{f} = m_{1} v_{1} + m_{2} v_{2}$

Kinetic energy of mass $m_{1} = \frac{1}{2} m_{1} {v_{1}}^{2}$
Kinetic energy of mass $m_{2} = \frac{1}{2} m_{2} {v_{2}}^{2}$

Total kinetic energy after collision
$K_{f} = \frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2}$

So, according to the law of conservation of momentum

$\begin{matrix} m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \Rightarrow m_{1} (u_{1} - v_{1}) = m_{2} (v_{2} - u_{2}) \end{matrix}$ $. . . . . . . . . . . . (1)$

And according to the law of conservation of kinetic energy

$\begin{matrix} \frac{1}{2} m_{1} {u_{1}}^{2} + \frac{1}{2} m_{2} {u_{2}}^{2} = \frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2} \Rightarrow m_{1} ({u_{1}}^{2} - {v_{1}}^{2}) = m_{2} ({v_{2}}^{2} - {v_{1}}^{2}) \end{matrix}$ $. . . . . . . . . . . . . (2)$

Now dividing the equation

$\begin{matrix} (u_{1} + v_{1}) = (v_{2} + u_{2}) \Rightarrow u_{1} - u_{2} = v_{2} - v_{1} \end{matrix}$

Therefore, this is, relative velocity of approach is equal to relative velocity of separation.

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