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