Question

During the head on collision of two masses 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. 111 m/sec
• B. 5 m/sec
• C. 10 m/sec
• D. $10\sqrt{2}$ m/sec

Answer 1

The correct option is C 10 m/sec

$\frac{1}{2}\mu {V_{ret}}^2=energystored=\frac{100}{3}$

$\frac{1}{2}\times \frac{2}{3}\times {v_{ret}}^2=\frac{100}{3}$

$v_{ret}=10m/s$

Momentum conserve$,$

$1.v_1-2V_2=0$

$v_1=2V_2$

$V_1-\left(V_2\right)=10$

$V_1+V_2=10$

velocity of approach$=10m/s$

Hence,

option $\left(C\right)$ is correct answer.