Contemporary Physics I – HW 3 Solutions
HW 3 - Solutions
Due October 19, 2015
1. You are standing at the top of a 10m cliff (as shown), throwing stones over the edge.
You throw a 0.2 kg stone at an angle of 30 degrees (π/6) above the horizontal with a speed of 10m/s.
(a) What is the x-component of the stone’s velocity immediately after it leaves your hand?
vix = vi cos
π
6
= 8.66m/s
= 10m/s(0.866)
(b) What is the y-component of the stone’s velocity immediately after it leaves your hand?
viy = vi sin
π
6
= 10m/s(0.5)
= 5m/s
(c) How long after you throw it does it reach it’s maximum height?
vfy = viy − gt
vf − vi y
0 − 5m/s
→ th = y
=
−g
−9.81m/s2
= 0.51s
(d) How long after you throw the stone does it take to hit the ground?
First calculate the maximum height the ball reaches
1
hmax =yi + viy th − gt2h
2
1
=(10m) + (5m/s)(0.51s) − (9.81m/s2 )(0.51s)2
2
=11.3m
Then how long the ball falls from that height
1
yground =hmax + viy tground − gt2ground
2
1
0 =11.3m + (0m/s)tg − (9.81m/s2 )t2g
2
s
→ tg =
2(11.3m)
9.81m/s2
=1.52s
So the total time is this sum of this time to that height, and the time from that height to the
ground
ttotal =th + tg = 0.51s + 1.51s
= 2.02s
(e) How far from the edge of the cliff does the rock hit?
1
xf =xi + vix tg + ax t2g
2
1
=(0m) + (8.66m/s)(2.02s) + (0m/s2 )(2.02s)2
2
= 17.5m
2. Consider the earth (m = 6 × 1024 kg) traveling around the sun (M = 2 × 1030 kg) with an orbital radius
of 1.5 × 1011 m.
(a) What is the gravitational force between the two?
FG =G
m1 m2
r2
=(6.67 × 10−11
m2 (6 × 1024 kg)(2 × 1030 kg)
)
kg s2
(1.5 × 1011 m)2
FG = 3.56 × 1022 N
(b) What is the centripetal acceleration on the earth? Here we say FC = FG
F =mac
FG
ac =
m
3.56 × 1022 N
=
6 × 1024 kg
= 6 × 10−3 m/s = 5.93mm/s2
(c) What is the circular velocity of the earth?
v2
ac = c
r
√
vc = ac r
p
= (5.93 × 10−3 m/s2 )(1.5 × 1011 m)
vc = 3.0 × 104 m/s
(d) What is the magnitude of the momentum of the earth?
p =mvc
=(6 × 1024 )(3 × 104 m/s)
p = 1.8 × 1029
kgm
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?
Conservation of momentum implies
v m =v⊕ m⊕
v ⊕ m⊕
v =
m
=
(3.0 × 104 m/s)(6 × 1024 )
2 × 1030 kg
v = 0.09m/s
(f) What is the wobble of the sun as caused by Jupiter?
We can rewrite the conservation relationship above in a more enlightening
way to help with this
q
GM
question, by subbing in the definition of orbital velocity, vorbit =
r
v m
v = X X
m
s
mX Gm
=
m
rX
s
G
=mX
m rX
27
s
=(1.89 × 10 kg)
3
m
6.67 × 10−11 kgs
2
(2 × 1030 kg)(7.78 × 1011 m)
v = 12.4m/s
(g) What would the wobble of the sun be if Jupiter were placed in the orbit of Mercury?
s
v =mX
G
m r'
s
3
m
6.67 × 10−11 kgs
2
27
=(1.89 × 10 kg)
(2 × 1030 kg)(5.79 × 1010 m)
v = 45.4m/s
(h) Imagine a planet 1.2 AU from the sun. What is the length of its year?
This question is very simple in AU and earth years! You can also notice that the planet is not much
farter than Earth from the sun so it shouldn’t vary too much from Earth’s
T 2 =r3
3
T =r 2
3
=(1.24AU ) 2
T = 1.3yr
3. 3.P.52) A bullet of mass 0.105 kg traveling horizontally at a speed of 300 m/s embeds itself in a block
of mass 2 kg that is sitting at rest on a nearly frictionless surface. What is the speed of the block after
the bullet embeds itself in the block?
We can use conservation of momentum to solve this
p~f =~
pi
(mb + mB )vf =mb vbi + mB vBi
mb vbi + mB vBi
vf =
(mb + mB )
(0.105kg)(300m/s) + (2kg)(0m/s)
vf =
0.105kg + 2kg
vf = 15m/s
4. 3.P.63) Suppose that all the people of the Earth go to the North Pole and, on a signal, all jump straight
up. Estimate the recoil speed of the Earth. The mass of the Earth is 6 ×1024 kg, and there are about
6 billion people (6 × 109 )
Let’s assume that the average mass of a person is ≈ 70 kg, and the jumping speed is ≈ 2.00 m/s
MP =NP mp = (6 × 109 )(70kg) = 4.2 × 1011 kg
Me Ve =MP vP
Ve =
MP
4.2 × 1011 kg
vP =
(2.00m/s)
Me
6 × 1024 kg
Ve = 1.4 × 10−13 m/s
5. You have a spring with constant k = 200N/m attached to a 3kg mass. What is the:
(a) Angular frequency?
r
k
ω=
m
s
N
200 m
=
3kg
ω = 8.16rad/s
(b) Frequency?
ω =2πf
ω
f=
2π
8.16rad/s
=
2πrad
1
f = 1.3 = 1.3Hz
s
(c) Period?
1
2π
=
f
ω
1
=
1.3Hz
T = 0.77s
T =
(d) How long would you have to make aq
pendulum such that it swung at exactly the same rate as the
oscillator? For a pendulum, T = 2π
l
g,
so we can solve for the pendulum length as
T 2g
4π 2
2
0.77s
=
× 9.8m/s2
2π
l=
l = 0.15m
(e) If you hung the mass from the spring, how much would the spring stretch?
F = − kx = ma
→ ma = − kx
ma
x=
−k
3kg × −9.8m/s2
=
−200N/m
x = 0.15m
6. 4.P.41) 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, but be careful with your signs! The force is pushing to the left, in
the -x̂ direction
F =ma = m
dvx
dt
dvx F
=
dt
m
dvx
F
=−
dt
m1 + m2 + m3
(b) What is the vector force F~net,3 exerted by the rod on the block of mass m3 ? What is the vector
force F~net,1 exerted by the rod on the block of mass m1 ?
All parts of the system will be under the same acceleration. For the right end, balance m1 on one
side, with m2 and m3 on the other side.
Fnet,3
=m3 a = m3 −
F~net,3 = −
F
m1 + m2 + m3
m3 F
m1 + m2 + m3
Fnet,1
F
= − (m2 + m3 )a = −(m2 + m3 ) −
m1 + m2 + m3
Fnet,1 =
(m2 + m3 )F
m1 + m2 + m3
(c) Suppose that instead of pushing on the right block (mass m1 ), you pull to the left on the left block
(mass m3 ) with a constant force of magnitude F . Draw a diagram illustrating this situation. Now
what is the vector force F~net,3 exerted by the rod on mass m3 ?
Pulling the third block will turn the compression forces into tension forces. We can see that this
has the effect of reversing our answer from before.
Fnet,3 =
(m1 + m2 )F
m1 + m2 + m3
7. 4.P.56) It was found that a 20 gram mass hanging from a particular spring had an oscillation period
of 1.2 seconds.
(a) When two 20 gram masses are hung from this spring, what would you predict for the period in
seconds? Explain briefly.
This is essentially doubling the mass on the spring, and we know the period of a spring is
r √
2m √
m
= 2 2π
= 2(1.2s)
k
k
r
P =2π
P = 1.70s
(b) When one 20 gram mass is supported by two of these vertical, parallel springs, what would you
predict for the period in seconds? Explain briefly.
Springs in parallel are added with kt ot = k1 + k2 , thus we are effectively doubling the spring
constant
r
P =2π
m
1
=√
2k
2
r m
1
= √ (1.2s)
2π
k
2
P = 0.85s
(c) Suppose that you cut one spring into two equal lengths and hang one 20 gram mass from this half
spring. What would you predict for the period in seconds? Explain briefly.
To find the new spring constant we consider that the original spring would be two of the new
springs in series
1
1
1
2
=
+
=
kT k0.5
k0.5
k0.5
→ k0.5 =2kT
This is the same problem as before, we’ve doubled the spring constant
P = 0.85s
(d) Suppose that you take a single (full-length) spring and a single 20 gram 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.
There is nothing in our previous equations that depends on gravity, thus the period would remain
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. Please break down the applied force into x- components
and y- components, and compute the normal force explicitly. Please use g = 10m/s2 throughout
the problem.
FN =Fg + Fy
=mg + F sin 30◦
=(5kg)(10m/s2 ) + (40N )(0.5)
FN = 70N
(b) What is the acceleration on the block?
Accleration will only be in the x direction, and the box starts from rest (vi = 0 m
s )
ΣFx =ma
ΣFx
a=
m
F cos(30◦ ) + µk FN
=
m
34.6N − 14N
=
5kg
a = 4.12m/s2
(c) Using this constant acceleration, how long do you need to push to move the block 8m?
∆x =8m
1
=vi t + at2
2
s
t=
2(8m)
4.12m/s2
t = 1.97s
(d) How fast will it be moving at the end of that time? endenumerate
v − vi
a
v =at
t=
=(4.12m/s2 )(1.97s)
v = 8.12m/s