Question

During the head on collision of two masses of $1$ $kg$ and $2$ $kg$ , the maximum energy of deformation is $\frac{100}{3}$ $J$ . If before collision the masses are moving in the opposite direction, then their velocity of approach before the collision is

• A. $10$ $m/s$
• B. $5$ $m/s$
• C. $20$ $m/s$
• D. $10\sqrt{2}$ $m/s$

Answer 1

Answer: (A)

By Reduced mass concept
$\Delta E=\frac{1}{2}\mu v_{r}^{2}$
$\Delta E=$ Maximum energy of deformation
$\mu =$ Reduced mass
$v_{r}=$ Velocity of approach
also, $\mu =\frac{m_{1} m_{2}}{m_{1} + m_{2}}$
on substituting values from question
$\frac{100}{3}=\frac{1}{2}\times \frac{1 \times 2}{1 + 2}\times v_{r}^{2}$
$\Rightarrow v_{r}=\sqrt{100}$
$v_{r}=10m/s$