Contemporary Physics I – HW 4
HW 4
Due Wednesday, October 26, 2016
Please answer all questions clearly and concisely. While you need not transcribe the question completely, it
should be clear from your answer alone what you are talking about.
You are strongly encouraged to discuss the homework with your classmates, but you must complete the
written homework by yourself, and of course, the material you submit must be your own.
Remember, show all of your work!
1. 5.P.10) A ball of mass 450 g hangs from a spring whose stiffness is 110 N/m. A string is attached to
the ball and you are pulling the string to the right, so that the ball hangs motionless. In this situation
the spring is stretched, and its length is 15 cm. What would be the relaxed length of the spring, if it
were detached from the ball and laid on a table?
The vertical component of the spring’s restoring force balances with the weight of the ball.
mg = Fs cos (θ)
mg = ks cos (θ)
mg
s=
k cos (θ)
From the Pythagorean Theorem:
(15 cm)2 = l2 + (8 cm)2
p
l = (15 cm)2 − (8 cm)2
l = 12.69 cm
→ cos (θ) =
12.69 cm
= 0.846
15 cm
Plugging in:
(0.45 kg)(9.81 m/s2 )
= 0.047 m
(110 N/m)(0.846)
L=x−s
s=
L = 0.15 m − 0.047 m
L = 0.103 m = 10.3 cm
2. 5.P.35) What is the minimum speed v that a roller coaster car must have in order to make it around
an inside loop and just barely lose contact with the track at the top of the loop? The center of the car
moves along a circular arc of radius R. Include a carefully labeled force diagram. Design a plausible
roller coaster loop, including numerical values for v and R.
To find the minimum speed to barely lose contact we look at the force body diagram on the car at
the top of the loop. If the car barely loses contact then the only forces on the car are the “centrifugal
force” balanced by the force of gravity.
Fc = Fg
v2
= mg
R
p
→ v = gR
m
3. I drop a 0.1 kg ball from a height of 10 m (assuming no air resistance).
(a) How much work does gravity do on the ball on its way to the ground?
From the definition for work,
2
W = F~ · d~ = mgh = (0.1 kg)(9.8 m/s )(10 m)
W = 9.8 J
(b) What is the speed of the ball when it hits the ground?
Since work is equal to the change in kinetic energy,
1
mv 2
2
1
mgh = mv 2
2
q
p
2
v = 2gh = 2(9.8 m/s )(10 m)
W = ∆K =
v = 14 m/s
(c) Using your projectile motion equations (not using energy), how long did the ball take before hitting
the ground?
1
y = y0 + v0 t + at2
2
Starting from rest and ending on the ground, y = 0 and v0 = 0, so
1
0 = y0 − gt2
2 s
r
2y0
2(10 m)
=
t=
2
g
9.8 m/s
t = 1.43 s
(d) At constant acceleration, and only using your result from part (c), what will the speed of the ball
be when it hits the ground? Compare this answer to that found in part (b).
At constant acceleration,
2
v = at = (9.8 m/s )(1.43 s)
v = 14 m/s
4. Consider a 6 kg block moving at 0.5c.
(a) What is the momentum of the block?
Using the formula for relativistic momentum,
p = γmv = q
mv
(6 kg)(0.5c)
=q
2
2
1 − vc
1 − 0.5c
c
p = 1.04 × 109 kg m/s
(b) What is the rest energy of the block?
The rest energy is
E0 = mc2 = (6 kg)c2
E0 = 5.4 × 1017 J
(c) What is the kinetic energy of the block? (Hint: do not use the relation you learned in high school.)
Relativistic kinetic energy is the difference between the total relativistic energy and the rest energy.
K = γmc2 − mc2 = mc2 (γ − 1)




1
1

K = mc2  q
− 1 = (6 kg)c2  q
−1
2
v
0.5c 2
1− c
1− c
K = 8.35 × 101 6 J
(d) If I then do 5 × 1017 J of work on the block (admittedly, quite a lot — its about the same amount
of radiant energy that Earth gets from the Sun every 3 seconds), what will its speed be afterwards?
Again, work is equal to the change in kinetic energy.
W = ∆K = mc2 (γf − 1) − mc2 (γi − 1) = mc2 (γf − γi )
W
= γf − γi
mc2
W
γf =
+ γi
2
mc
r
v 2
1
f
1−
= W
c
mc2 + γi
p
(W + γi mc2 )2 − (mc2 )2
v
=
c
W + γi mc2
q
2 −1/2 (6 kg)c2 )2 − ((6 kg)c2 )2
(5 × 1017 J + (1 − ( 0.5c
c ) )
=
2 −1/2 (6 kg)c2
5 × 1017 J + (1 − ( 0.5c
c ) )
v = 0.877c = 2.63 × 108 m/s
5. A tiny rocket ship with a tiny crew has a mass of 2 kg, and is initially at rest. A constant force of 60N
is applied to the ship.
(a) What is the initial acceleration of the particle?
F = ma
a = F/m
a = 60 N / 2kg
a = 30 m/s2
(b) How far has the particle traveled after 5 seconds?
1
∆x = v0 t + at2
2
1
= (0)(5s) + (30 m/s2 )(5s)2
2
∆x = 375 m
(c) The force continues to be applied for a total of 107 sec. What is the momentum of the particle
after that time?
∆p = F ∆t
= (60 N )(107 s)
∆p = 6 × 108
kg m
s
(d) What is the speed of the particle after that time?
v=q
p
m
1+
6×108 kgm/s
2 kg
=r
1+
=
2
p
mc2
6×108 kgm/s
(2 kg)(3×108 m/s)
2
3 × 108 m/s
√
1 + 12
v = 2.12 × 108 m/s
(e) What is the γ-factor of the particle after that time?
γ=q
1
1−
v 2
c
1
=r
1−
2.12×108 m/s
3×108 m/s
2
γ = 1.41
(f) What is the kinetic energy of the particle after 107 sec?
KE = mc2 (γ − 1)
= (2 kg)(3 × 108 m/s)2 (1.41 − 1)
KE = 7.38 × 1016 J
(g) Assuming the particle started at rest, and reminding yourself of the relationship between work and
kinetic energy, how long will the ship travel by the end of 107 sec? (This requires the use of the
work equation; it is a relativistic particle).
W = ∆KE
= F ∆x
→ ∆x =
W
7.38 × 1016 J
=
F
60 N
∆x = 1.23 × 1015 m/s
(h) How fast will the particle go if you accelerate for twice as long?
p = F (2∆t)
= (60 × 2 × 107 )
= 1.2 × 109
v=q
kg m
s
kg m
s
p
m
1+
2
p
mc2
1.2×109 kgm/s
2 kg
=r
1+
1.2×109 kgm/s
(2 kg)(3×108 m/s)
2
v = 2.7 × 108 m/s
6. 6.P.44) A nucleus whose mass is 3.499612 × 10−25 kg undergoes spontaneous alpha decay. The original
nucleus disappears and there appear two new particles: a He-4 nucleus of mass 6.640678 × 10−27 kg (an
“alpha particle” consisting of two protons and two neutrons) and a new nucleus of mass 3.433132 ×
10−25 kg.
(a) When the alpha particle has moved far away from the new nucleus (so the electric interactions
are negligible), what is the combined kinetic energy of the alpha particle and new nucleus?
→ We use the equation for the energy of a multiparticle system.
Esys = mN c2 = (Kα + mα c2 ) + (Kn + mn c2 ) + Uαn
:0
mN c2 = (mα )c2 + Kα + (mn )c2 + KnU
αn
→ KT = Kα + Kn = (mN )c2 − (mα )c2 − (mn )c2
KT = [(3.499612 × 10−25 ) − (6.640678 × 10−27 ) − (3.433132 × 10−25 )] (3 × 108 m/s)2
KT = 6.5898 × 10−13 J
(b) How many electron volts is this? In contrast to this nuclear reaction, chemical reactions typically
involve only a few eV.
KT = (6.5898 × 10−13 J)
1 eV
1.6 × 10−19 eV
KT = 4.12 × 106 eV = 4.12 M eV