Solution Key - Drexel University

Homework 3 - Solutions
Lise Wills
Phys 113 Contemporary Physics
Drexel University
20 October 2014
1. You are standing at the top of a 10m cliff, throwing stones over the edge. You throw a 0.2kg stone at
an angle of 30 degrees ( π6 ) above the horizontal with a speed of 10 m
s .
(a) What is the x-component of the stone’s velocity immediately after it leaves your hand?
vx
v cos θ
10 ms cosp π6 q
?
10 23 ms
?
5 3 ms
vx
8.66 ms
(b) What is the y-component of the stone’s velocity immediately after it leaves your hand?
vy
v sinpθq
10 ms sinp π6 q
10 12 ms
vy
5 ms
(c) How long after you throw it does it reach it’s maximum height? At the maximum height, vy
v0 at
v v0
t
a
0 5m
9.8 ms
s
t 0.51s
v
1
0.
(d) How long after you throw the stone does it take to hit the ground?
There are two ways to do this, one of which involves solving a quadratic. This doesn’t. We already
have the time to reach the maximum height, so we just need to find the time to fall from there.
Begin by figuring out how high it falls from hmax . Remember that hmax is on top of the height of
the cliff, 10m.
∆y
v0 t
1 2
at
2
5 ms 0.5s
3.73m
hmax 13.73m
1
m
p
9.8 2 p0.5sq2 q
2
s
Now you need only find out how long it takes to fall from there, and add this to c). Remember
v0 0 m
s at the top!
hmax
t2
v0 t
1 2
at
2
2hmax
a
t
c 2h
c 2 a 13.37
max
9.8
1.65s
ttotal 0.51s 1.65s
ttotal 2.16s
(e) How far from the edge of the cliff does the rock hit?
As above, only analyze the direction in question, x. It’s simpler, as there is no acceleration in this
direction, and we can set the base of the cliff as x0 0.
x x0
v0 t
m
5 3 s p2.16sq
?
x 18.7m
2. Consider the earth (m 6 1024 kg) traveling around the sun (M
of 1.5 1011 m.
2 1030 kg) with an orbital radius
(a) What is the gravitational force between the two?
FG
m
GM
r2
3
24 2 1030 kg
6.67 1011 kgms2 6 10p1.5 kg
1011 mq2
2
3.56 1022 N
FG
(b) What is the centripetal acceleration on the Earth?
Here, FG FC .
mac
FG
ac m
FG
10 N
3.56
6 1024 kg
22
ac
0.006 sm2
(c) What is the circular velocity of the Earth?
2
vrc
?
vc ac r
c m
0.006 s2 1.5 1011 m
ac
vc
3 104 ms
(d) What is the magnitude of the momentum of the Earth?
Here, magnitude means were are simply not worrying about it’s direction.
p mvc
3 104 ms 6 1024 kg
p 1.8 1029
kg m
s
(e) Using the relation we found in class, how fast is the sun orbiting around the center of mass of the
Earth-Moon system?
The momentum of the Sun-Earth system must be conserved, and the relation from class derives
from there.
vC mC
vC mC
v@ m@
6 1024 kg 2.97 2.97 104 m
s
2 1030 kg
v@ m@
v@
3
0.09 ms
(f) What is the wobble of the sun as caused by Jupiter (E)?
We can rewrite the conservation relationship above in a more enlightening way to help us with any
kind of question, by subbing in the definition of orbital velocity, vorbit
v@
b
GM
r .
E mE
vm
@d
mE Gm@
m
@ rE
mE
d
G
m@ rE
1.89 10
27
d
6.67 1011 kgms2
3
1.98 1030 kg 7.78 1011 km
kg
v@
12.4 ms
(g) What would the wobble be if Jupiter were placed in the orbit of Mercury (A)?
This is what makes the above equations so useful - now we just plug in one different number!
mE
c
G
m@ rA
1.89 10
27
kg
d
6.67 1011 kgms2
3
1.98 1030 kg 5.79 1010 km
v@
45.6 ms
(h) Imagine a planet 1.2 AU from the Sun. What is the length of its year?
This question is wonderfully simple in AU and Earth years! You can also do a gut-check here the planet is not very much further away from the Sun than Earth, so it’s period should be close
to Earth’s
r3
T r
p1.2AU q
T2
3
2
T
3
2
1.3 yr
3. 3.P.57 You and a friend each hold a lump of wet clay. Each lump has a mass of 30g. You toss your
lump of clay into the air, where the lumps collide and stick together. Just before the impact, the
m
velocity of one lump was 3, 3, 3 ¡ m
s and the velocity of the other lump was 3, 0, 3 ¡ s .
(a) What was the total momentum before the collision?
Remember to always check your units! 30g 0.03kg.
4
p~total
~v1 m1 ~v2 m2
p0.03kgq 3, 3, 3 ¡ ms p0.03kgq 3, 0, 3 ¡ ms
p0.03kgq 0, 3, 6 ¡ kgsm
p~total
0, 0.09, 0.18 ¡ kgsm
(b) What was the momentum of the stuck-together lump just after the collision?
Momentum is always conserved.
p~total
0, 0.09, 0.18 ¡ kgsm
(c) What is the velocity of the stuck together lump just after the collision?
When two bodies collide and then stick together, we have a perfectly inelastic collision - momentum
is conserved and we now treat the two masses as one.
p~ ~v mtotal
p~
~v mtotal
1
0, 0.09, 18 ¡ kgsm
0.06kg
~v
4. You have a spring with constant k
(a) Angular Frequency?
0, 1.5, 3 ¡ ms
200 mN attached to a 3kg mass. What is the:
ck
ω
dm
ω
N
200 m
3kg
8.16 rad
s
(b) Frequency?
2πf
ω
f
2π
ω
rad
s
8.16
2π rad
1.3 1s
5
1.3Hz
f
(c) Period? There are two relations you can use, since ω
to calculate.
2πf , though one is oh so slighly simpler
f1 2π
ω
T
1
1.3Hz
0.77s
T
(d) How long would you have to make a pendulum such that it swung at exactly the same rate as the
oscillator?
For pendulums, T 2π gl . This is a good relationship to keep in mind.
b
T
2π
d
l
g
T 2g
4π 2
m
2
p 0.77s
q
9.8 2
2π
s
L
L 0.15m
(e) If you hung the mass from the spring, how long would the spring stretch?
kx
ma
ma kx
ma
x
k
F
s
3kg20010
N
m
2
m
x 0.15m
5. 4.P.46 Two blocks of mass m1 and m3 , connected by a rod of mass m2 , are sitting on a low-friction
surface, and you push to the left on the right block (mass m1 ) with a constant force.
x
(a) What is the acceleration, dv
dt , of the blocks?
This is as simple as it seems - the systems are connected and are treated as one body of total mass
M. Just becareful with your signs! F is in the x
ˆ direction.
M ax
M dvdtx
dvx
F
ΣFx
dt
6
M
dvx
dt
m mF
1
2
m3
(b) What is the compression force in the rod (mass m2 ) near its right end? Near it’s left end?
Since we’re treating the system as one, all the parts will have the same acceleration. Becareful
balancing your forces on either side! Now you have to keep track of what mass is where.
For the right end, we balance m1 on one side and m2 and m3 on the other. Here, FR is the reaction
force from the left, or the compression you’re looking for.
m1
dvx
dt
FR
F
FR
F
F
FR
dvx
dt
m1 F
m1 m2 m3
m1
F p m m2m m3m q
1
2
3
Same idea to the left, only this time the compression is just from m3 . Remember F is only acting
on m1 , and only reaction forces are acting on the other blocks.
FL
m3 dvdtx
FL
m
1
m3 F
m2 m3
(c) How would these results change if you pull to the left on the left block (mass m3 ) with the same
force, instead of pushing on the right block?
Pulling turns compression forces into tension forces. Note that the changed the side that’s beeing
effected as well!
FR
m1 dvdtx
FR
m
1
m1 F
m2 m3
As you can see, the roles are merely reveresed now.
m3
dvx
dt
F
F
FL
dvx
dt
F m mm3 F m
1
2
3
m1 m2
FR F p
q
m1 m2 m3
FL
7
m3
6. 4.X.56 Bob is pushing a box across the floor at a constant speed of 1 m
s , applying a horizontal force of
magnitude 20N. Alice is pushing an identical block across the floor at a constant speed of 2 m
s , applying
a horizontal force.
(a) What is the magnitude of the force that Alice is applying to her box?
This is a trick question - notice that you don’t have enough information to calculate things. To
push a box at constant speed, Fapp Ff r . Since Alice and Bob are pushing identical boxes, the
friction on each box is the same - they are both pushing horizontally - and so their applied forces
are identical.
Fapp
20N
(b) With the two boxes starting from rest, explain qualitatively what Alice and Bob did to get their
boxes moving at different speeds.
There are a couple of ways to explain this. One is to use equations to help us understand. Starting
with F m∆v
∆t , we can use the following information: Ff inal in both cases is the same, but ∆V is
different. It is more illuminating to look at it in the following sense:
F ∆t
m
∆v
F
Either, m
const, so if ∆v increases, ∆t must also increase for the relation to remain constant.
So, for Alice’s box to be moving faster, she pushed with the same force for twice the time than Bob.
However, if you hold ∆t constant, than she pushed with twice the force.
7. 4.P.82 It was found that a 20 g mass hanging from a particular spring had an oscillation period of 1.2
seconds.
(a) When two 20 g masses are hung from this spring, what would you predict for the period in seconds?
Explain briefly.
From question 4, we know that P0 2π m
k , so if we double the mass, we merely get that the
?
period increases by 2:
a
c 2m
2π k
?
2P0
?
2 1.2s
P 1.70s
P
(b) When one 20g mass is supported by two of these vertical, parallel springs, what would you predict
for the period in seconds? Explain briefly.
Adding
? another identical spring has the effect of doubling the spring constant, or reducing the period
by 2.
8
2π
P
cm
2k
?P0
2
1.2s
?
2
0.85s
P
(c) Suppose you cut one spring into two equal lengths, and you hang one 20g mass from this half
spring. What would you predict for the period in seconds? Explain briefly.
Each half-spring’s spring constant (k.5 ) must add up to the original. We add them in the following
way:
k1
1
k
.5
1
k.5
k2
.5
2k
k0.5
This is the same spring constant we had in the last question, so the answer is the same.
P
0.85s
(d) Suppose that you take a single (full-length) spring and a single 20g mass to the Moon and watch
the system oscillate vertically there. Will the period you observe on the Moon be longer, shorter,
or the same as the period you measured on Earth? (The gravitational field strength on the Moon
is about one-sixth that on the Earth.) Explain briefly.
Note how there is nothing in our equation that has to do with gravity. The period remains unchanged.
P
1.2s
8. You push a 5kg box across the floor with a force of 40N at an angle of 30 degrees with respect to the
horizontal. The coefficient of kinetic friction is µk 0.2.
(a) Draw a free body diagram for the box. Break the applied force into x- and y- components and
compute the normal force explicitly. Please use g 10 m
s throughout.
FN
Fg Fy
mg N F sinp30q N
50N 21 40N
FN
70N
(b) What is the acceleration on the block?
Remember, acceleration will only be in the x direction, and the box starts from rest so v0
9
0 ms .
ma
ΣFx
a
m
ΣFx
F cosp305kgq µFN
34.6N 14N
5kg
a 4.12
m
s2
(c) Using this constant acceleration, how long do you need to push the block 8m
∆x 8m
v0 t
c
t
d
1 2
at
2
2 8m
a
16m
4.12 sm2
t 1.97s
(d) How fast will it be moving at the end of that time?
v v0
a
v at
4.12 sm2
t
v
10
1.97s
8.12 ms

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