Physics question.?

Question

A block of mass m is pushed against a spring that has a spring constant k. The spring is compressed by a distance d, and then the block is released. We observe that the block is launched by the spring along a horizontal frictionless surface with a final speed v.

If a second block, this one having a mass of 4m, is pushed against the spring and released, we observe that it gains a final speed of 2v. By what distance was the spring compressed in this second case?

d
2d
4d
16d
25d

Answer 1

Lets do some state equations
at first just as Bekki has pointed out

potential energy of spring = kinetic energy of the block
.5 kd^2=.5 mv^2

so
d=sqrt(m v^2/k)
or for case #2
d2=sqrt( 4m (2v)^2/k)
from the first case k = m v^2/d1^2 substituting we have
d2=sqrt( 4m (2v)^2/(m v^2/d1^2) )
d2=sqrt ( 4 x 4 d1^2) = 4d1

the answer is 4d

Answer 2

Conservation of energy:

initial spring energy = 1/2 kx^2
=
final kinetic energy = 1/2 mv^2

k is constant. x is proportional to v and to square root of m. So figure out what happens to x if m goes to 4m and v goes to 2v.