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Problem 4 Solutions Part A
Assume a rocket of mass M + ∆m is moving with a speed v. During some time,
∆t, a mass ∆m is ejected out the back with a velocity −vr (1000 m/s) with
respect to the rocket (v − vr relative to an outside observer). As a result of
this ejected mass, the rocket (now only of mass, M ) is moving slightly faster,
v + ∆v.
Because this is a sort of inelastic collision (energy is lost in the burning of
the fuel), we can investigate the problem using conservation of momentum. The
initial momentum (deﬁning the direction of motion of the rocket to be positive)
of the system is,
pi = (M + ∆m)v
And the ﬁnal momentum is,
pf = ∆m(v − vr ) + M (v + ∆v)
So expanding these expressions and setting them equal to each other, we have
that
M v + ∆mv = ∆mv − ∆mvr + M v + M ∆v
We can cancel both terms on the right hand side, then and we have:
0 = −∆mvr + M ∆v
or,
∆mvr = M ∆v
Now, because this takes some time, ∆t, to happen, we can divide by ∆t on both
sides of the previous equation, leaving us with,
∆m
∆v
vr = M
∆t
∆t
Now we have some familiar terms here. The ﬁrst, ∆m , is the rate at which the
∆t
gas cloud behind the rocket is gaining mass. This is equal to the rate at which
the rocket is losing mass. So we can write that the current mass of the rocket is
M (t) = M0 − Rt
1 where M0 is the original mass of the rocket (5500 kg), R is the mass loss rate
( ∆m = R = 100 kg/s).
∆t
We also have ∆v , which is the acceleration of the rocket. So we have:
∆t
M a = Rvr
Where we recognize M a as the total force on an object. So the thrust on the
rocket is:
Fthrust = Rvr
Then the acceleration of the rocket is:
Rvr
M
We can replace the mass, M , on the right hand side with M (t), the mass of the
rocket at any time, and get:
Rvr
a=
M (t)
So at time t = 0, we are given that the rocket has a mass of 5500 kg. This gives
an acceleration of
100kg/s × 1000m/s
= 18.2m/s2 .
a=
5500kg
a= At the time of burnout, the rocket has reduced its mass by 80%, so the mass at
burnout is 1000 kg(0.20 × 5500kg = 1100kg). This gives an acceleration of:
100kg/s × 1000m/s
= 90.9m/s2 .
1100kg a= Now, substituting in the full form of M (t), we have:
Rvr
M0 − Rt
Integrating with respect to time gives us:
a= tf tf a dt
t0 =
t0 v(tf ) − v(t0 )
v(tf ) Rvr
dt
M0 − Rt = −vr ln(M0 − Rt)
= −vr ln M (t) t0
tf t0
tf = −vr (ln M (tf ) − ln M (t0 ))
M (tf )
= −vr ln
M (t0 )
M (t0 )
= vr ln
M (tf )
Now, in this case, M tf , is twenty percent of M (t0 ), so their ratio is 5:
vf = vr ln 5 = 1000m/s ∗ ln 5 = 1609m/s
2 Part B
This problem can be done two ways: both involve conservation of momentum
and one of of which uses the center of mass.
Center of Mass Method
We are going to measure everything from position of the spacestation. The
spaceship has an unknown mass distribution, so we can just say that its center
of mass located a distance, d, away from the station and its total mass is M .
We also know that the astronaut is located 1m from the station with a mass, m.
When she pushes oﬀ, she will cause the rocket to move in the opposite direction.
The initial momentum in this reference frame is zero, so the center of mass does
not move. So we can calculate how far the ship can possibly move.
Initially, the center of mass of the astronautrocket system is located at
x = M d + m(1m)
¯
After the astronaut has hit the back of the rocket, she has traveled a distance
of the length of the rocket, minus however far the rocket has moved, s, so her
new position is (1m + l − s). The rocket’s new position is then (d − s). So the
center of mass afterwards is located at
x = M (d − s) + m(1m + l − s)
¯
Because the center of mass cannot move, we know that these positions are equal:
M d + m(1m) = M (d − s) + m(1m + l − s) = M d − M s + m(1m) + ml − ms
Simplifying, this leaves:
0 = −M s + ml − ms
Solving for s, the distance the ship has moved,
s =
=
= ml
M +m
100kg × 11m
100kg + 1000kg
1m So the ship barely makes it.
Momentum Conservation and Kinematics
Another method for this problem actually uses the information about the astronauts speed that was given. We again have to use momentum conservation, but
in a more transparent way (instead of just saying that it preserved the location
of the center of mass). 3 The astronaut pushes oﬀ the forward wall of the spaceship with a speed,
v = 10m/s, causing the spaceship to recoil at a speed vs , related by momentum
conservation:
mv − M vs = 0
or,
vs = − m
100kg
v=
10m/s = −1m/s
M
1000kg So the relative velocity between the spaceship and the astronaut is v − vs .
Because the astronaut travels a distance, l = 11m relative to the spaceship, the
time this takes is:
t =
=
= l
v − vs
11m
10m/s + 1m/s
1s The spaceship is moving at a speed 1m/s towards the space station for 1s, going
1m. Again, we see that the rocket barely gets there. 4 Problem 5 Solution Note: Deﬁne right as the positive xdirection and up as the positive ydirection. 1 Part (a) The ﬁnal kinetic energy of the system cannot be greater than the initial kinetic energy of the system otherwise we’d
be creating energy in the system. Since we have v1 = −v0 we know their magnitudes are the same, v1 = v0 .
≥ KEf
1
1
1
2
2
2
≥
mv1 + mv2 + mv3
2
2
2
1
1
2
2
0 ≥
mv2 + mv3
2
2
Since the right hand side can never be negative, we see that v2 = v3 = 0 and the balls 2 and 3 must remain at
rest. It is clearly not possible for these balls to be completely unaﬀected by the collision. Quantitatively, we note
that momentum must be conserved, and given the above condition mv0 x = −mv0 x. Momentum would not be
ˆ
ˆ
conserved.
=⇒ No, it is not possible for ball 1 to bounce backwards with the same speed.
KEi
1
mv 2
2 0 2 Part (b) Using momentum conservation in the x and y directions we have
px : mv0
v0
py : 0
v2,y =
=
=
= mv2,x + mv3,x
v2,x + v3,x
mv2,y − mv3,y
v3,y The only constraint on the xcomponents of v2 and v3 is that they add up to v0 and they need not be equal. The
ycomponents of v2 and v3 must always be equal. Because of this, the angles can be diﬀerent.
=⇒ Yes, it is possible for the angles to be diﬀerent. 3 Part (c) Again, using momentum conservation and that v1  = v2  = v3  = v
py : 0
0
sin θ2
θ2
px : mv0
v0
v0 =
=
=
=
=
=
= mv2 sin θ2 − mv3 sin θ3
v sin θ2 − v sin θ3
sin θ3
θ3 = θ
−mv1 + mv2 cos θ2 + mv3 cos θ3
−v + v cos θ + v cos θ
(2 cos θ − 1)v The only constraint is that the term (2 cos θ − 1) must be positive since v is positive. We can see there are many
angles for θ that will fulﬁll this condition.
=⇒ Yes, it is possible for them all to have the same speed but diﬀerent directions.
1 Problem 6 LIN
Part A)
Conservation of momentun: m2v = (m1 + m2 )v 2
v2 = !
! m2v
m1 + m2 After collision, Use Conservation of Energy: 1
1
2
(m1 + m2 )v 2 = kd 2
2
2
Use v2 from above and solve for d ! ! m2v
k(m1 + m2 ) d= PartB)
Energy initial = Energy final + Energy Lost 1
1
Energy Lost = m2v 2 " kd 2
2
2 ! Energy Lost
m2
m1
= 1"
=
InitialKE
m1 + m2 m1 + m2 PartC)
_ ! !
!
! "p m2"v
=
"t
"t
m2v
"v =
#v
m2 + m1
"m1m2v
Favg =
#t(m1 + m2 )
Note: Just magnitude accepted as well.
Favg = ...
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This test prep was uploaded on 04/01/2008 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Lanzara
 Physics

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