Download PHYSICS 113: Contemporary Physics – Midterm Exam Solution Key

PHYSICS 113: Contemporary Physics – Midterm Exam Solution Key
1. [40 points] Short Answers (8 points each)
(a) The ∆++ particle is a baryon, like the proton and neutron, but it
has a charge of +2. What quarks make up the ∆++ ?
Sol.
The up quark has a charge of +2/3, and since all baryons have 3
quarks, a ∆++ must have (uuu) .
Strictly speaking, u, c, or t all have +2/3, so we’ll accept any combination of 3 of them, but if you’re curious, the actual particle is just
made of ups.
(b) What are the charge, baryon number and lepton number of a positron,
e+ ?
Sol.
Q = +1, B = 0, L = −1
the exact opposite of an electron.
(c) Protons in the Large Hadron Collider (LHC) get up to 0.99999999c.
What is the γ factor in the LHC?
E.C. (2 points) A proton has a mass of 1.67 × 10−27 kg. Based on
this, what is the energy of a proton on the LHC?
Sol.
This is a matter of plug and chug:
γ=p
1
1 − (0.99999999)2
= 7070
Thus, a proton has an energy of:
E
= mc2 γ
=
1.67 × 10−27 kg(3 × 108 m/s)2 · 7070
=
1.06 × 10−6 J
=
6.6 TeV
(d) You are in a 1000 kg space station (including your mass and everything else inside) which is initially at rest. In order to get moving,
you throw a 50kg mass out of the back. How fast do you need to
throw the mass relative to you in order to get the station moving 5
m/s?
Sol.
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The initial momentum is zero. To move 950kg at 5 m/s requires:
pship = 960 kg × 5 m/s = 4, 750 kg m/s
Thus, pmass = −4, 750 kg m/s and so:
vmass =
pmass
= −95 m/s
m
Strictly speaking, the relative speed is the difference between the two,
and thus:
vrel = 100 m/s
(e) A 0.1kg mass is attached to a spring which oscillates with a Period
of π/5 s. What is the spring constant?
Sol.
First calculate the angular velocity:
ω=
and thus:
2π
= 10 s−1
P
k
= ω2
m
so
k = mω 2 = 0.1 kg · (10 s−1 )2 = 10N/m
2. [36 points] A small box, of mass 1kg sits at rest on a rough ground with
coefficient of friction µ = 0.25 (You may assume the same for either static
or kinetic). At some initial time, you apply a force of 10N at an angle of
θ = 36.87◦ with respect to the horizontal, as shown:
Why such a strange angle? You may find it helpful to know that:
sin(36.87◦ ) = 0.6 ; cos(36.87◦ ) = 0.8
(a) (8 points) Draw a free body diagram of the system, solving for all
forces, and be sure to break the forces up into x- and y- components.
Sol.
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(b) (7 points) What is the momentum of the box after 10 seconds?
Sol.
The net force on the block is 4N, so:
pf = F ∆t = 40 kg m/s
(c) (7 points) What is the kinetic energy of the box after 10 seconds?
Sol.
There are multiple ways to do this.
i) From the momentum directly (certainly the easiest):
K=
1600 kg2 m2 /s2
p2
=
= 800 J
2m
2 kg
ii) From the acceleration
a=
F
4N
=
= 4 m/s2
m
1 kg
and thus:
vf = at = 4 m/s2 10 s = 40 m/s
and thus:
K=
1
mv 2 = 800 J
2
and so on.
(d) (7 points) Based on your previous two answers, what distance does
the box travel in the first 10 seconds?
Sol.
The simplest approach is to simply say:
d=
W
K
800 J
=
=
= 200 m
F
F
4N
3
but you could also have used the acceleration approach:
d=
1 2
1
at = 4 m/s2 (10 s)2 = 200 m
2
2
(e) (7 points) You continue to push the box for 109 seconds. Ignoring
all of the complications that might arise for pushing for that long
(the plane wouldn’t be long enough, air resistance would be insanely
important, and so on). What is the momentum after 109 seconds?
Sol.
This is the exact same question as part b. We simply have:
pf = F ∆t = 4N × 109 s = 4 × 109 J
(f) E.C. (4 points) How fast is the box moving at that moment?
Sol.
This part is tougher. The simplest approach is to use the equation
for velocity inversion:
v=q
p/m
1+
p 2
mc
In this case, p/mc = 13.33. We quickly get:
v = 0.9972 = 2.99 × 108 m/s
3. [24 points] A car moves at a speed of 50 m/s (about 110 mph) around a
circular track of radius 200 m, as shown:
(a) What is the net acceleration on the car? Be sure to indicate the
direction.
Sol.
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This is the centripetal acceleration formula:
ac =
v2
(50m/s)2
=
= 12.5 m/s2
r
200m
and the acceleration is inward.
(b) Suppose you were the driver with a mass of 50kg. What is the total
magnitude of the normal forces exerted by the seat and interior of
the car?
Sol.
The force of gravity is downwards, and has a magnitude of 500N.
The normal force upwards is equal and opposite.
The force from the interior of the car is 50N × 12.5m/s2 = 625N
The two forces are at right angles to one another so:
q
2 + F 2 ' 800 N
Ftot = Fup
in
(c) The car experiences air resistance. Draw the free body diagram as
seen from above and at the position shown in the diagram. Be sure
to add vectors indicating i) The direction of air resistance on the car,
ii) The net force on the car, and iii) the direction of force that the
friction in the road must exert on the tires.
This is a sketch. The exact magnitudes aren’t expected (of course),
but the approximate relative directions are.
Sol.
Vectorally:
F~W ind + F~T ires = F~tot
and the wind must oppose the direction of motion.
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