Question

When two blocks A and B coupled by a spring on a frictionless table are stretched and then released, then
A.Kinetic energy of body at any instant after releasing is inversely proportional to their masses
B.Kinetic energy of body at any instant may or may not be inversely proportional to their masses
C.\[\dfrac{{K.E.\;of\;B}}{{K.E.\;of\;A}} = \dfrac{{mass\;of\;B}}{{mass\;of\;A}}\]
D.Both (B) and (C) are correct.

Answer 1

Hint: When two bodies are coupled each block will perform SHM and force on each block will be of same magnitude and spring is the same since the table is frictionless, energy throughout will be conserved and you can definitely apply momentum conservation as there is no external force.

Complete step by step answer: let's solve this in centre of mass frame
let at any instant its velocity of block A is \[{v_A}\] and of block B is \[{v_B}\].
now in centre of mass frame total momentum will be zero so,
\[{m_A}{v_A} + {m_B}{v_B} = 0\]
\[ \Rightarrow {m_A}{v_A} = - {m_B}{v_B}\]
So magnitude of momentum of both the blocks is same
Let's call this common momentum p then kinetic energies
\[K{E_A} = \dfrac{{{p^2}}}{{2{m_A}}}\]and
\[K{E_B} = \dfrac{{{p^2}}}{{2{m_B}}}\]taking ratio of both the kinetic energies we have,
\[\dfrac{{K{E_A}}}{{K{E_B}}} = \dfrac{{\dfrac{{{p^2}}}{{2{m_A}}}}}{{\dfrac{{{p^2}}}{{2{m_B}}}}} = \dfrac{{{m_B}}}{{{m_A}}}\]
As we can see ratio of kinetic energies of the blocks is inverse ratio of the masses of the blocks
Hence we can conclude that at any instant kinetic energy is inversely proportional to mass of the blocks when two blocks are coupled by a spring in a frictionless table.

Therefore, Option-A is correct.

Note: Here if you solve this question in the frame of centre of mass it is little easy but you can definitely solve it in ground frame that way it will be little tough because they you will also have to use energy conservation to find the relationship between kinetic energies as there will some elastic potential stored in the spring at almost all instant.