Q. A block of mass 0.50 kg is moving with a speed of 2.00$ms^{- 1}$on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body .The energy loss during the collision is
A) 0.16 J
B) 1.00 J
C) 0.67 J
D) 0.34 J
Explanation
(c) Law of conservation of momentum $0.5 \times 2 + 1 \times 0 = 1.5 \times v$
(assumed that 2$^{nd}$ body is at rest)
Or $v = \frac{2}{3}$
The energy loss during the collision
$\mathrm{\Delta}K = K_{f} - K_{i}$
$= \frac{\frac{3}{2} \times \left( \frac{2}{3} \right)^{2}}{2} - \left( \frac{1}{2} \right) \times \frac{2^{2}}{2}$
$= - \frac{2}{3}J = - 0.67J$
So, energy lost is 0.67 J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.