Q. If a skater of weight $3\ kg$ has initial speed $32\ m/s$ and second one of weight $4\ kg$ has $5\ m/s$. After collision, they have speed (couple) $5\ m/s$. Then the loss in K.E. is
A) $48\ J$
B) $96\ J$
C) Zero
D) None of these
Explanation
(d) Loss in K.E. = (initial K.E. – Final K.E.)of system
$\frac{1}{2}\ m_{1}u_{1}^{2} + \frac{1}{2}\ m_{2}u_{2}^{2} - \frac{1}{2}\left( m_{1} + m_{2} \right)V^{2}$
$= \frac{1}{2}\ 3 \times (32)^{2} + \frac{1}{2} \times 4 \times (5)^{2} - \frac{1}{2} \times (3 + 4) \times (5)^{2}$
$= 1498.5\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.