Two blocks A and B of masses m and 2m, respectively, are connected by a massless spring of constant K. This system lies over a smooth horizontal surface. At t=0, block A has velocity u towards right as shown, while the speed of block B is zero, and the length of the spring is equal to its natural length at that instant. In each situation of column I, certain statements are given and corresponding results are given in column II. Match the statements in column I with the corresponding result in column II.
Answers
Answer:
Consider as shown in figure momentum in X-direction conserve.
mv
1
+2mv
2
=mu
where v
1
and v
2
be the speed at any instant of block A and B.
v
1
+2v
2
=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, v
1
=0 , v
2
2
4
Kinetic energy =2×
2
1
(m(
2
4
))
2
=
4
mu
2
But initially kinetic energy =
4
mu
2
=
2
mu
2
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)
solution