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  • 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

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 M2 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) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.
(c) If spring is massless, the final state of the M_{1} is state of rest.
(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.

Answers (2)

The answer is the option (c)

a) the kinetic energy of M_{1} is not fully transferred to the spring as its potential energy, and hence, the option a is incorrect

b) law of conservation of mass is valid here since the surface is frictionless; hence option b is incorrect.

c) if we consider the case, where the spring is totally massless, then all the kinetic energy of M_{1} gets transferred to M_{2}. As a result, m1 comes to rest, and M_{2} acquires a velocity v and starts moving. Hence, option c is correct.

d) even when the force of friction is not involved, a collision can be inelastic. Hence, we reject option d.

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infoexpert24

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The answer is the option (c)

Let M_{1} come into contact with the spring of M_{2}, such that the velocity of M_{2} increases and that of m1 decreases. As, the velocities adjust to being at par, both the star of the block moving with the same velocity.

a) the kinetic energy of M_{1} is not fully transferred to the spring as its potential energy, and hence, the option a is incorrect

b) law of conservation of mass is valid here since the surface is frictionless; hence option b is incorrect.

c) if we consider the case, where the spring is totally massless, then all the kinetic energy of M_{1} gets transferred to M_{2}. As a result, m1 comes to rest, and M_{2} acquires a velocity v and starts moving. Hence, option c is correct.

d) even when the force of friction is not involved, a collision can be inelastic. Hence, we reject option d.

Posted by

infoexpert24

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