Two blocks M _1 and M _2 having equal mass are free to move on a … - Vidyadip

MCQ

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. Initially $M _2$ is at rest and $M _1$ is moving toward $M _2$ with speed $v$ and collides head$-$on with $M _2$.

A
While spring is fully compressed all the $KE$ of $M_1$ is stored as $PE$ of spring.
B
If the surface on which blocks are moving has friction, then collision cannot be elastic.
C
If spring is massless, the final state of the $M_1$ is state of rest.
✓
Both $B$ and $C$

✓

Answer

Correct option: D.

Both $B$ and $C$

If there is not specified we always consider the collision elastic. When two bodies of equal masse collides elastically, their velocities are interchanged in these types of collision.
Kinetic energy and linear momentum remain conserved. According to the above diagram when $m _1$ comes in contact with the spring, $m_1$ is retarded by the spring force and $m_2$ is accelerated by the spring force.
$a.$ The spring will continue $*$to compress until the two blocks acquire common velocity. So some of kinetic energy of block $M_x$ store into $P.E$ and some part of it stores into $K.E$ of block $M_2$. So option $(a)$ is incorrect.
$b.$ As surfaces are frictionalless momentum of the system will be conserved. So option $(b)$ is also incorrect.
$c.$ The two bodies of equal mass exchange their velocities in a head on elastic collision between them. So, if spring is massless, the final state of the $M _1$ is state of rest.
$d.$ Since there is a loss of $K.E$ when the blocks collides on the rough surface. Hence, the collision is inelastic.

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