A moving particle of mass 'm' makes a head-on collision with a particle of mass '2m' initially at rest. If the collision is perfectly elastic, the percentage loss of energy of the colliding particle is

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Solution

The correct option is C
Kinetic energy of projectile before collision $K_{i} = \frac{1}{2} m_{1} u_{1}^{2}$
Kinetic energy of projectile after collision $K_{f} = \frac{1}{2} m_{1} v_{1}^{2}$
Kinetic energy transferred from projectile to target DK = decrease in kinetic energy in projectile
$Δ K = \frac{1}{2} m_{1} u_{1}^{2} - \frac{1}{2} m_{1} v_{1}^{2} = \frac{1}{2} m_{1} (u_{1}^{2} - v_{1}^{2})$ Fractional decrease in kinetic energy
$\frac{Δ K}{K} = \frac{\frac{1}{2} m_{1} (u_{1}^{2} - v_{1}^{2})}{\frac{1}{2} m_{1} u_{1}^{2}} = 1 - {(\frac{v_{1}}{u_{1}})}^{2}$
We can substitute the value of $v_{1}$ from the equation
$v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1} + \frac{2 m_{2} u_{2}}{m_{1} + m_{2}}$
If the target is at rest i.e. $u_{2} = 0$ then $v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1}$
From Equation (i) $\frac{Δ K}{K} = 1 - {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} o r \frac{Δ K}{K} = \frac{4 m_{1} m_{2}}{(m_{1} + m_{2})^{2}}$
Percentage loss of energy= $\frac{4 m_{1} m_{2}}{(m_{1} + m_{2})^{2}} \times 100$
= $\frac{4 m \times 2 m}{(2 m + m)^{2}} \times 100 = \frac{800}{9} = 88.9 %$
Hence the correct choice is (c).

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