Q. Two trolleys of mass $m$ and $3m$ are connected by a spring. They were compressed and released once, they move off in opposite direction and comes to rest after covering distances $S_{1}$ and $S_{2}$ respectively. Assuming the coefficient of friction to be uniform, the ratio of distances $S_{1}:S_{2}$ is
A) $1\ :9$
B) $1\ :3$
C) $3\ :1$
D) $9\ :1$
Explanation
(d) When trolley are released then they possess same linear momentum but in opposite direction. Kinetic energy acquired by any trolley will dissipate against friction
$\therefore\mu\ mg\ s = \frac{P^{2}}{2m} \Rightarrow s \propto 1/m^{2}$ [As $P\ $and$\ u$ are constants]
$\Rightarrow \frac{s_{1}}{s_{2}} = \left( \frac{m_{2}}{m_{1}} \right)^{2} = \left( \frac{3}{1} \right)^{2} = \frac{9}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.