Two blocks of M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s.
(a) What is the speed of the block of mass M?
(b) Find the original elastic energy in the spring if M = 0.30 kg.
Solution
Given,
The initial velocity of both the bodies V
initial
=0
Initial momentum is zero.
Final Velocity of 3M block V
B
=2m/s
Mass of block A=M
Mass of block B=3M
Apply conservation of momentum
Final momentum = initial momentum
3MV
B
+ MV
A
= 0
V
A
= -3V
B
=-−3×2=−6m/s
Given,
Mass M=0.30kg
Energy in spring = final kinetic Energy of both Masses
K.E=
2
1
MV
A
2
+
2
1
(3M)V
B
2
K.E=
2
1
0.3×(−6)
2
+
2
1
(3×0.3)2
2
=7.2 J
(a) Speed of block mass M is 6m/s
(b) Total Kinetic Energy is 7.2J