Environmental Engineering Reference
In-Depth Information
(g) The motion is a combination of simple harmonic motion and linear
motion of the centre of mass at constant speed. It is possible to show
from the above solutions that there are times when one of the cars is
instantaneously at rest in the lab frame. To an observer in the lab one
car moves forward, stops then appears to pull the other car after it.
(h) The kinetic energy due to the simple harmonic motion is
2
kA
2
,i.e.the
potential energy in the spring at maximum extension. This is added to
the energy due to motion of the centre of mass:
1
m)v
0
/
4 to obtain
the mechanical energy of the system. Substitution for
A
and
k
using the
results from (f) proves that the total mechanical energy is
mv
0
/
2asit
must be since this is a conservative system.
3.4 The momentum of the rope at any instant is
λyv.
Differentiating with respect
to time gives the rate of change of the momentum of the rope to be
λv
2
.
By
Newton's Second Law, this must be equal to the sum of the gravitational
and applied forces acting on the rope. The gravitational force is
gλy
from
which we deduce that the applied force is
λv
2
1
2
(m
+
gλy.
We can integrate this
to obtain the work done in raising the rope:
λv
2
y
+
gλy
2
/
2
.
The product
of the applied force and the speed gives the power used to raise the rope:
λv
3
+
gλyv.
If we compute the rate of change of mechanical energy, directly
from the potential energy (
gλy
2
/
2) and the kinetic energy (
λyv
2
/
2) we obtain
λv
3
/
2
+
gλyv.
This is not equal to the power. The discrepancy comes from
not including the energy that must be lost through friction in the coil of rope,
in order that the coil does not rotate.
3.5 Calculate the reduced mass directly to obtain
(
10
/
3
)
kg
.
The centre of mass
velocity is
(
11
/
3
)
j
ms
−
1
.
The momentum of particle with mass
m
1
, relative
to the centre of mass, is
+
10
3
j
)
kg ms
−
1
.
µ(
v
1
−
v
2
)
=
(
10
i
−
The momentum of the other particle relative to the centre of mass has equal
magnitude but opposite direction to the first, i.e.
−
µ
v
r
.
3.6 Use
F
=−
d
U/
d
r
to find the point at which
F
=
0
.
This gives the solution
r
r
0
, which when substituted into the expression for the potential gives the
depth to be
=
ε.
The range of
r
is obtained upon realising that the
total energy is equal to
K
max
−
−
U(r
0
)
=
ε
and that
r
must be such that the kinetic
energy is always positive, i.e.
U(r) <
K
max
.
3.7 The proton has initial velocity
v
and recoils at a speed of 2
v/
3. Conservation
of momentum gives
−
ε
+
5
3
m
p
v
=
m
u
v
u
,
where
m
p
the mass of the proton,
m
u
is the mass of the unknown particle
and
v
u
is its final velocity. Conservation of kinetic energy gives
5
9
m
p
v
2
m
u
v
u
,
=
from which we get
m
u
=
5
m
p
.
