Consider a body at rest in the L-frame, which explodes into two fragments of masses $m_{1}$ and $m_{2}$ .
Calculate energies of the fragments of the body

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Solution

Since the body is initially at rest, its total momentum is zero. As a result of the explosion, the two fragments will separate in opposite directions with momenta $p_{1}$ and $p_{2}$ such that $p_{1}^{'} + p_{2}^{'} = 0$ ; magnitude of momentum will be same, i.e., $p_{1}^{'} = p_{2}^{'}$ .
On substituting,
$E_{k}^{'} = \frac{p_{1}^{' 2}}{2 m_{1}} + \frac{p_{2}^{' 2}}{2 m_{2}} = \frac{1}{2} (\frac{1}{m_{1}} + \frac{1}{m_{2}}) p_{1}^{'}$
$E_{k} = 0$
In Eq. (iv) of kinematics of reaction, we get
$\frac{1}{2} (\frac{1}{m_{1}} + \frac{1}{m_{2}}) p_{1}^{' 2} = Q$
On rearranging this equation, we get
$p_{1}^{'} = {(\frac{2 m_{1} m_{2}}{m_{1} + m_{2}} Q)}^{1 / 2} = (2 μ_{12} Q)^{1 / 2}$
where $μ_{12}$ is the reduced mass of the system. The kinetic energies of the fragments are $\frac{p_{1}^{2}}{2 m_{1}}$ and $\frac{p_{2}^{2}}{2 m_{2}}$ with $p_{1}^{'} = p_{2}^{'}$ . Hence,
$E_{k, 1}^{'} = \frac{p_{1}^{2}}{2 m_{1}} = \frac{m_{1} Q}{m_{1} + m_{2}}$ and $E_{k, 2}^{'} = \frac{p_{2}^{2}}{2 m_{2}} = \frac{m_{2} Q}{m_{1} + m_{2}}$

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Consider a body at rest in the L-frame, which explodes into two fragments of masses m1 and m2.Calculate energies of the fragments of the body

Consider a body at rest in the L-frame, which explodes into two fragments of masses $m_{1}$ and $m_{2}$ .
Calculate energies of the fragments of the body