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In a two block system in figure match the following.
image `{:("Column1","Column2"),("A Velocity of center of mass","P Keep on changing all the time"),("B Momentum of center of mass","Q First decreases then become zero"),("C Momentum of 1kg block","R Zero"),("D Kinetic energy of 2kg block","S Constant"):}`

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Correct Answer - A `rightarrow`r,s, B`rightarrow`r s, C`rightarrow`q, D`rightarrow`q
`v_(CM)` = `(m_(1)v-(1) + m_(2)v-(2))/(m_(1)+m-(2)` = `((1)(10) + (2)(-5))/(3)` = 0`
Similarly, `p_(CM)` =0`
Net force on the system is zero, hence `V_(CM)` and `p_(CM)` will remain constant.
Velocity of 1 kg and 2kg blocks keep on changing initially and finally both of them stop as `V_(CM)` was zero.

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