Question

Block A has a mass 1.00kg, and block B has a mass 3.00 kg. the blocks are forced together, compressing a spring S between them; then the system is released from rest on a level, frictionless surface. The spring which egligible mass, is not fastened to either block and drops to the surface after it has expanded Block B acquires a speed of 1.20m/s.
a)- what is the final speed of block A ?
b)- how much potential energy was stored in the compressed spring ?

Answer 1

The law of conservation of linear momentum
0 = m1•v1 – m2•v2,
v2 = m1•v1/m2 = 1•1.2/3 = 0.4.
The law of conservation of energy
PE = KE1 +KE2 = m1•v1^2/2 + m2•v2^2/2 =
= 1•1.2^2/2 + 3• (0.4)^2/2 =0.96 J.

Answer 2

I got it already

Thanks so much

Answer 3

(I know this is a dead post, but) you mixed up your mass and velocity variables. the equation should look like this: Ma*Va = Mb*Vb.
Va =(Mb*Vb)/Ma