Question

Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a massless spring and placed at a smooth surface. The spring initially stretched and released. then

• A. The momentum of each particle remains constant separately
• B. The magnitude of momentum of both bodies are same to each other
• C. The mechanical energy of system remains constant
• D. Both (b) and (c) are correct.

Answer 1

Answer: (D)

Since, no external force is present on the system so, conservation principle of momentum is applicable
$\therefore \overset{ \rightarrow }{p}_{i}=\overset{ \rightarrow }{p}_{f}=\overset{ \rightarrow }{p}_{1}+\overset{ \rightarrow }{p}_{2}$
$\therefore \, \left(\overset{ \rightarrow }{p}\right)_{1}=-\left(\overset{ \rightarrow }{p}\right)_{2} \, \, \left(\right.\left(\because \overset{ \rightarrow }{p}\right)_{i}=0\left.\right)$
$\therefore \, \, \left|\overset{ \rightarrow }{p}_{1}\right|=\left|- \overset{ \rightarrow }{p}_{2}\right|$
$\therefore \, \, \overset{ \rightarrow }{p}_{1}=\overset{ \rightarrow }{p}_{2}$
From this point of view, it is clear that momenta of both particles are equal in magnitude but opposite in direction
Also, friction is absent. So total mechanical energy of system remains conserved