Two blocks `M_(1)` and `M_(2)` having equal mass are free to move on a horizontal frictionless surface. `M_(2)` is attached to a massless spring as shown in figure. Initially `M_(2)` is at rest and `M_(1)` is moving toward `M_(2)` with speed `upsilon` and collides head-on with `M_(2)`.

Options

A. While spring is fully compressed all the K.E of `M_(1)` is stored as P.E of spring
B. While spring is fully compressed the system momentum is not conserved , though final momentum is equal to initial momentum .
C. If spring is masless, the final state of the `M_(1)` is state or rest .
D. If the surface on which blocks are moving has friction , then collision cannot be elastic .

Answer

C, D

Explanation

Let `v_(1) and v_(2)` be the final velocities of blocks `M_(1) and M_(2)`
Using law of conservation of momentum ,
`mv = mv_(1) +mv_(2)`
`rArr " " v = v_(1)+v_(2)`
or `v^(2) = v_(1)^(2) +v_(2)^(2) +2v_(1)v_(2)" " …(i)`
Using conservation of energy , we get
`1/2 mv^(2) =1/2 mv_(1)^(2) +1/2mv_(2)^(2) `
` :. v^(2) = v_(1)^(2) +v_(2)^(2)....(ii)`
Using (i) and (ii) , we can write
` 2v_(1) v_(2) =0 `
After collision , bodies of equal masses interchange their speeds .
` :. " " v_(1) = 0 `
The final state of mass `M_(1) ` is at rest .
When the surface on which blooks are moving is not frictionless , some of the kinetic energy will be lost and hence collision cannot be elastic .