Ph1a: Solution to Quiz 4
Alejandro Jenkins, Fall 2004
Problem 1 (6 points)  Collision on a Spring
A linear spring has a free length
D
. When a mass
m
is hung on one end, the spring has an
equilibrium length
D
+
`
. While it is hanging motionless with an attached mass
m
, a second
mass
m
is dropped from a height
`
onto the ﬁrst one. The masses collide inelastically and
stick together.
(a) (1 point)
What is the new equilibrium length of the spring?
Solution:
The equilibrium length of the spring is that for which the spring’s
restoring force exactly cancels the weight of the attached mass. If a linear
spring with elastic constant
k
is stretched by an amount
`
(measured from its
free length
D
), then the restoring force is
k`
. If a mass
m
was attached, then,
at equilibrium,
k`
=
mg
.
If the weight is doubled, then the spring’s restoring force at its new equilibrium
length must be twice what it was before: 2
mg
=
k
2
`
. The spring will now be
stretched by 2
`
with respect to its free length. The total length of the spring
at its new equilibrium is
D
+ 2
`
.
(b) (1 point)
What is the period of the resulting motion?
Solution:
The spring’s stiﬀness, characterized by the elastic constant
k
, has not
changed. The mass attached is now 2
m
. Therefore the period is
T
= 2
π
s
2
m
k
.
But
k
is not speciﬁed in the problem, so we must express it in terms of other
quantities given. Since the spring stretched by an amount
`
under the weight
of
m
, we have
k`
=
mg
, or
k
=
mg/`
. Therefore,
T
= 2
π
s
2
`
g
.
(c) (2 points)
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 Fall '07
 Goodstein
 Angular Momentum, Mass, equilibrium length, new equilibrium length, Alejandro Jenkins

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