Question

A thin uniform bar lies on a friction less horizontal surface and is free to move in any direction on the surface. Its mass is 0.16 kg and length is $\sqrt{3}m$. Two particles, each of mass is 0.08 kg are moving on the same surface and towards the bar in a direction perpendicular to the bar, one with velocity 10 ms${}^{-1}$ and the other with 6 ms${}^{-1}$ as shown in the Figure. The first particle strikes the bar at point A and the other at point B. Point A and B are at a distance of 0.5 m from the center of the bar. The particle strikes the bar at the same instant of time and sticks to the bar after collision. Then, the loss of kinetic energy of the system in the above collision process is :

• A. 0.16 J
• B. 2.56 J
• C. 2.72 J
• D. 5.44 J

Answer 1

Answer: (D)

5.44 J

$K.E_i=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$

$=\frac{1}{2}\times 0.08(100+36)$

$=0.04\left(136\right)$

$K.E_i=5.44J$

$K.E_f=0$

$\therefore$ loss $=K{.E}_f-K.E_i=-5.44J$

-ve sign indicates loss.