Two blocks M1 and M2 having equal mass are free to move on a horizontal frictionless surface. M2 is attached to a massless spring as shown in Fig. 6.10. Initially M2 is at rest and M1 is moving toward M2 with speed v and collides head-on with M2.
(a) While spring is fully compressed all the KE of M1 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 M1 is state of rest.
(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.
This is a multiple choice answer as classified in NCERT Exemplar
(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force.
a) So spring will continue compress until the blocks moves with the same velocity.
b) As the surfaces are frictionless momentum of the system will conserved.
c) If the spring is massless whole energy of M1 will be imparted and M1 will be rest .
d) collision is elastic even if friction is not involved.
<p><span data-teams="true">This is a multiple choice answer as classified in NCERT Exemplar</span></p><div> (c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force. </div><div> </div><div>a) So spring will continue compress until the blocks moves with the same velocity. </div><div> </div><div>b) As the surfaces are frictionless momentum of the system will conserved. </div><div> </div><div>c) If the spring is massless whole energy of M1 will be imparted and M1 will be rest . </div><div> </div><div>d) collision is elastic even if friction is not involved.</div><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745405189phphMra6C_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745405189phphMra6C.jpeg" alt="" width="325" height="100"></picture></div></div>
This is a multiple choice answer as classified in NCERT Exemplar
(b) conserving energy between “O” ans ”A”
Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1=………….2
From eqn 1 and 2
So v1 = v’/=v2(v’/2)
v1>v’/2
hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’
(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path
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This is a multiple choice answer as classified in NCERT Exemplar
(b) conserving energy between “O” ans ”A”
Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1=………….2
From eqn 1 and 2
So v1 = v’/=v2(v’/2)
v1>v’/2
hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’
(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path, followed will change and the bullet reaches at point B instead of A
(f) as the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . therefore their internal energy increases.
This is a multiple choice answer as classified in NCERT Exemplar
(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.
Hence total work done =-mgL + mgL=0
As the point of application of the contact forces does not move hence work done by reaction forces will be zero.
This is a multiple choice answer as classified in NCERT Exemplar
(b) First velocity of the iron sphere increases and after sometimes becomes constant called terminal velocity. Hence according first KE increases and then becomes constant which is best represented by b.
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