PHY4222/Spring 08: CLASSICAL MECHANICS II
HOMEWORK ASSIGNMENT #6: SOLUTIONS
due by 12:50 p.m. Wed 02/27
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
Every (algebraic) final result must be supplemented by a check of units. Without such a
check, no more than 75% of the credit will be given even for an otherwise correct solution.
1. Thornton & Marion, Problem 9-10. [33 points]
When the shell explodes, the center of mass continues to move in the horizontal direction. In the CM system,
the sum of the x components of the momenta of the fragments is equal to zero:
p1c + p2c = 0.
Enrgy conservation in the CM system reads
E=
p21c
p2
p2
+ 2c = 1c ,
2m1
2m2
2µ
where
µ≡
m1 m2
.
m1 + m2
Finding p1c
p1c =
p2c
p
2µE
p
= − 2µE.
The x− components of the velocities of the CM system are
p
v1c =
2µE/m1
p
v2c = − 2µE/m2
The velocities in the LAB system are
p
2µE/m1 + vc
p
= v2c + vc = − 2µE/m2 + vc
v1 = v1c + vc =
v2
where vc = v0 cos θ is the x component of the CM velocity. The fragments will fall at horizontal distances from
the point of explosion
x1 = v 1 t
x2 = v2 t,
where t is the time of fall from the maximum elevation of the shell t = v0 sin θ/g. The difference
p
p
∆x = x1 − x2 = (v1 − v2 ) t =
2µE/m1 + 2µE/m2 v0 sin θ/g.
2. Thornton & Marion, Problem 9-54. [34 points]
Solution:
Velocity of the rocket
v = u ln
m0
m
2
Momentum
m0
p = mv = mu ln
m
m0 d
d m
dp
m ln
= −u
= u
m ln
dm
dm
m
dm
m0
m
= −u ln
+1 =0→
m0
m/m0 = e−1 ≈ . 367 88...
3. A single-stage rockect is taking off from Earth.
a) Find the height of the rocket at burnout. [20 points]
Solution
Equation of motion in the gravitational field for a constant burnout rate (v (0) = 0)
uα
dv
=
−g
dt
m
dv
dv dm
dv
=
= −α
dt
dm dt
dm
dv
g
u
=
−
dm
αZ m
Z m
dm
m0
g
g m
0
dm − u
= u ln
− (m0 − m)
v =
α m0
m
m
α
m0
Z m0
Z t
Z m(t)
dt0
0
v
(m
)
=
α
dm0 v (m0 )
y =
dt0 v (t0 ) =
dm0
0
dm
m
0
m0
Z m0
h
i
m0
g
−1
0
0
= α
dm u ln 0 − (m0 − m )
m
α
m
Z m0
Z 1
Z 1
m0
0
−1
dm ln 0 = m0
dz ln z = −m0
dz ln z = −m0 z (ln z − 1) |1m/m0
m
m
m/m0
m/m0
m0
m
+ m0 − m
− 1 = −m ln
= m0 + m ln
m0
m
Z m0
Z m0 −m
1
2
dm0 (m0 − m0 ) =
dss = (m0 − m)
2
m
0
uh
m0 i
g
2
y (m) =
m0 − m − m ln
− 2 (m0 − m)
α
m
2α
(1)
The last equation gives the elevation of the rocket as a function of its mass. If the mass at burnout is mb , then
the elevation at burnout is
u
m0
g
2
yb = y (mb ) =
m0 − mb − mb ln
− 2 (m0 − mb )
α
mb
2α
b) How much further in height will the rocket go after burnout? [13 points]
Solution
According to Eq.(1), the velocity at burnout is
vb = u ln
m0
g
− (m0 − mb )
mb
α
The additional elevation after burnout is found from the energy conservation
mb g∆y = mb vb2 /2
∆y =
vb2 /2g
2
m0
g
= u ln
− (m0 − mb ) /2g
mb
α
3
4. Thornton & Marion, Problem 9-54. [34 points]
5. Bonus: Thornton & Marion, Problem 9-56. [35 points]
Solution:
The rate of condensation is proportional to the cross-sectional area
dm
= kA = πkr2
dt
4
πρr3
3
dm
dr
= 4πρr2
dt
dt
dr
k
k
=
→ r = r0 + t
dt
4ρ
4ρ
m =
Equation of motion
d
(mv)
dt
dv dm
m
+
v
dt
dt
dv
πkr2
v
+ 4
3
dt
3 πρr
dv 3 k 1
+
v
dt
4 ρ r0+ k t
= mg
= mg
= g
= g
4ρ
Setting τ = 4ρr0 /k , the equation becomes
v
dv
+3
= g.
dt
t+τ
For late time, i.e., for t τ, the equation is simplified to
dv
v
+ 3 = g,
dt
t
which can be solved by an obvious substitution
v = ct
c + 3c = g → c = g/4
For a general case, use a substitution
v (t) = v0 (t) C (t) ,
where v0 is a partial solution of the homogeneous equation (with g = 0)
dv0
3
+
v0 = 0 →
dt
t+τ
−3
v0 = (t + τ ) .
dv
dv0
dC
=
C + v0
.
dt
dt
dt
Substituting this back into the equation for v, we obtain
v0
dC
=g
dt
4
which need to be solved with the initial condition C (0) = 0 (I assume that the droplet has zero velocity when
it enters the cloud).
dC
g
3
=
= g (t + τ )
dt
v0
i
gh
4
(t + τ ) − τ 4
C =
4
Finally,
−3
v = (t + τ )
"
#
i g
gh
τ4
4
4
(t + τ ) − τ =
t+τ −
3 .
4
4
(t + τ )
For late times (t τ ), the result reduces to
v=
g
t
4