Notes on the two remaining problems from 09/27
Problem 1. Problem 2-12 from Marion&Thornton.
Solution. Let the z axis point up. Then the force of gravity is F~g = m~g = −mg~e3 . The
retarding force is
!
~
v
2
F~r = −mkv
= −mkv ~v
v
√
√
with ~v = ż~e3 , but note that v = |~v | = ~v 2 = ż 2 = |ż| so that the retarding force is
F~r = −mk|ż| ż ~e3
Now plug the forces into Newton’s law
m~a = F~g + F~r
which has only one non-trivial component: along the z-axis:
mz̈ = −mg − mk|ż| ż
and cancelling the m factor,
z̈ = −g − k|ż| ż
(1)
In order to proceed, we must remove the absolute value sign. To this end, we divide the motion
into two stages:
1. “going up”, then ż > 0 and |ż| = ż
2. “coming down”, then ż < 0 and |ż| = −ż
Notice that the differential equation looks different in these two cases. Now solve each case one
at a time:
Going up. Substitute |ż| = ż in eq. (1)
z̈ = −g − k ż 2
and turn it into an equation for ż(z) instead of ż(t):
z̈ = ż
dż
= −g − k ż 2
dz
This is an equation with separable variables which can be easily integrated:
Z
żdż
= −k
g/k + ż 2
Z
dz + C
g
1
ln
+ ż 2 = −kz + C
2
k
The constant can be fixed from the initial condition ż(z = 0) = v0 , so that we get
1
ln
2
g
k
g
k
+ ż 2
+ v02
1
!
= −kz
How high will the particle get? At the highest point it will stop, i.e. ż(hmax ) = 0, which allows
us to find hmax :
!
g
+ v02
1
k
hmax =
ln
g
2k
k
Coming down. Now we consider the second stage, where a particle falls down from a height
hmax and we want to find the velocity at the bottom. Substituting |ż| = −ż in eq. (1) we obtain
the equation of motion as
z̈ = −g + k ż 2
We can solve this equation in exactly the same way:
1
g
− ln
− ż 2 = −kz + C
2
k
Now the constant C is determined from the initial condition ż(hmax ) = 0
1 g 1
C = − ln + ln
2 k 2
g
k
+ v02
g
k
!
and substitute into the solution above:
1 g 1
g
1
− ż 2 = −kz − ln + ln
− ln
2
k
2 k 2
g
k
+ v02
g
k
!
and solve for ż(z):
g
ż 2 = −
k
2
g
k
g
k
+ v02
e2kz
Notice that the terminal velocity is given by
vt2 = ż 2 (z = −∞) =
g
k
so that the solution above can be rewritten as
2
ż =
vt2
vt4
e2kz
− 2
2
vt + v0
Then the velocity upon return to the ground (z = 0) is given by
ż 2 = vt2 −
vt4
vt2 v02
=
vt2 + v02
vt2 + v02
Problem 2. Problem 2-26 from Marion&Thornton. Given: L = 2 m, m = 2 kg, v0 =
4 m/s, k = 6 N/m, µk = 0.2.
(a) This part is very easy. There is no energy loss due to friction, so all of the initial kinetic
energy
1
E = mv02
2
2
is converted into potential energy due to the compression of the spring:
1
U = k(∆x)2
2
Energy conservation then implies
1 2 1
mv = k(∆x)2
2 0 2
and solving for ∆x we get
∆x =
s
mv02
k
(b) Friction implies energy losses which can be calculated as the work done by the frictional
force over a distance L + ∆x (note that as the spring compresses, the friction continues to work
∆E =
Z
~ =−
F~f riction · dx
Z
Ff riction dx = −µk N
Z
0
L+∆x
dx = −µk mg(L + ∆x)
and this is equal to the change in the total energy of the particle:
1
1
f inal − initial = k(∆x)2 − mv02 = −µk mg(L + ∆x)
2
2
and solve for ∆x.
3