two blocks of masses m and 2m respectively attached at opposite end of a spring of force constant k, placed on a smooth horizontal surface. Spring is initially at its natural length l. A is given a velocity 2vo and B given vo as shown. separation between 2m and centre mass of the system?
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Answer:
ANSWER
Consider as shown in figure momentum in X-direction conserve.
mv1+2mv2=mu
where v1 and v2 be the speed at any instant of block A and B.
v1+2v2=u→(1)
at time of maximum Kinetic energy of B spring is at neutral position let us consider block A is rest at that time.
So, v1=0 , v224
Kinetic energy =2×21(m(24))2=4mu2
But initially kinetic energy =4mu2=2mu2
if means block A never goes kinetic energy zero everywhere.So A is related to 1.
(B) Velocity of Block B goes zero when spring is at mean position and total kinetic energy with A situation given like in fig (1).
The system repeats as fig 2,3,4 and 5 velocities will be zero of B at the situation given in fig(5) B related to 2.
(C) The kinetic energy of the two block system is never zero as we can see in fig(1) to fig(5) a liner momentum always conserve and kinetic energy is minimum of a system when potential energy of the system is maximum, therefore kinetic energy at maximum compression or mix elongation.So
C→ 1 and 3
D→ 2 and 4 (As fig 4 and 5)