Download Midterm Exam 3

Physics 130
General Physics
Fall 2012
Midterm Exam 3
November 29, 2012
Name:
Instructions
1. This examination is closed book and closed notes. All your belongings
except a pen or pencil and a calculator should be put away and your
bookbag should be placed on the floor.
2. You will find one page of useful formulae on the last page of the exam.
3. Please show all your work in the space provided on each page. If you
need more space, feel free to use the back side of each page.
4. Academic dishonesty (i.e., copying or cheating in any way) will
result in a zero for the exam, and may cause you to fail the
class.
In order to receive maximum credit,
each solution should have:
1.
2.
3.
4.
5.
6.
7.
8.
9.
A labeled picture or diagram, if appropriate.
A list of given variables.
A list of the unknown quantities (i.e., what you are being asked to find).
One or more free-body or force-interaction diagrams, as appropriate,
with labeled 1D or 2D coordinate axes.
Algebraic expression for the net force along each dimension, as appropriate.
Algebraic expression for the conservation of energy or momentum equations, as appropriate.
An algebraic solution of the unknown variables in terms of the
known variables.
A final numerical solution, including units, with a box around it.
An answer to additional questions posed in the problem, if any.
1
Physics 130
General Physics
Fall 2012
1. Fred (mass 70 kg) is running with the football at a speed of 7.0 m{s when he is met
head-on by Brutus (mass 130 kg), who is moving at 5.0 m{s. Brutus grabs Fred in a
tight grip, and they fall to the ground. Which way do they slide, and how far? The
coefficient of kinetic friction between football uniforms and Astroturf is 0.25.
Solution:
This is an inelastic collision in which Fred and Brutus are moving toward each other.
Assume that Brutus is moving in the positive x-direction and Fred is moving in the
negative x-direction. Because momentum is conserved, we have
pf x
mB pvi qB mF pvi qF pmB
pix
mF qpvf q
(1)
(2)
Solving for the final velocity of Brutus and Fred,
vf
mB pvi qB mF pvi qF
pmB mF q
p130 kgqp5.0 m{sq p70 kgqp7.0 m{sq
p130 kg 70 kgq
0.80 m{s
(3)
(4)
(5)
Now that we know that Brutus and Fred moved to the right after the collision, we
can find the distance that they traveled. Friction slows them down, and therefore
the force of friction must act in the negative x-direction. The net force is equal to
the total mass (the mass of both players) times the acceleration.
¸
Fx
µpmB
fk
µn
mF qg
ña
2
pmB mF qa
pmB mF qa
pmB mF qa
µg
p0.25q p9.8 m{s2q
2.45 m{s2
(6)
(7)
(8)
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
Finally, we can use the kinematic equation to find the distance traveled with vf
vf2
ñd
vi2 2ad
vi2
2a
0.80 m{s2 q2
2 p
p2.45 m{s2q
0.13 m
13 cm
3
0:
(12)
(13)
(14)
(15)
(16)
Physics 130
General Physics
Fall 2012
2. A two-stage rocket is traveling at 1100 m{s with respect to the earth when the first
stage runs out of fuel. Explosive bolts release the first stage and push it backward with
a speed of 25 m{s relative to the second stage. The first stage is four times as massive
as the second stage. What is the speed of the second stage after the separation?
Solution:
The rocket is traveling at an initial speed before the explosion. The two parts separate
after the explosion. To find the speed of the second stage, we’ll use the conservation
of momentum equation.
pm1
p4m2
pf
m2 qvi
m2 qvi
5m2 vi
5vi
pi
m1 v1f m2 v2f
4m2 v1f m2 v2f
m2 p4v1f v2f q
4v1f v2f
m2p4v1f
v2f q
(1)
(2)
(3)
(4)
(5)
The fact that the first stage is pushed backward at 35 m{s relative to the second can
be written
v1f
5vi
5vi
25 m{s v2f
4p25 m{s v2f q
100 m{s 5v2f
v2f
(6)
(7)
(8)
Solving for the velocity of the second stage v2f
v2f
5vi
100 m{s
5
4
(9)
Physics 130
General Physics
5 1100 m{s
5
1.1 km{s
5
Fall 2012
100 m{s
(10)
(11)
Physics 130
General Physics
Fall 2012
3. A package of mass m is release from rest at a warehouse loading dock and slides down
a 3.0 m high frictionless chute to a waiting truck. Unfortunately, the truck driver went
on a break without having removed the previous package, of mass 2m, from the bottom
of the chute.
(a) Suppose the packages stick together. What is their common speed after the collision?
(b) For an elastic collision, what quantities are conserved?
(c) Write out the conservation of momentum equation for this problem.
(d) Write out the conservation of energy equation for this problem.
(e) Which quantities are not known in these equations?
(f) Explain how to obtain the formulas for these unknown quantities.
(g) If the collision between the packages is perfectly elastic, to what height does the
package of mass m rebound?
Solution:
(a) From conservation of energy, we can find the speed of the package at the bottom
of the chute.
Ki
0
mgh
v
Ui
Kf Uf
1 2
mv
2
a
(1)
2gh (2)
b
2p9.8m{s2 qp3.0mq 7.668m{s
(3)
If the packages stick together, the collision is inelastic and conservation of momentum is conserved.
pi
m1 v1i m2 v2i
mv1i 0
6
pf
p m1
3mvf
m2 qvf
(4)
(5)
(6)
Physics 130
General Physics
Fall 2012
Cancelling the m’s
v1i
3vf
(7)
Solving for vf , we find
vf
v1i
3
{s 2.56m{s
7.668m
3
(8)
(9)
The packages move together with a speed of 2.56m{s.
(b) In an elastic collision, both energy and momentum are conserved.
(c) In this conservation of momentum equation, we’ll use the velocity calculated
above for the initial velocity before the collision. Also, the package at the
bottom is at rest before the collision. The conservation of momentum equation
gives
pi
m1 v1i m2 v2i
m1 v1i 0
mv1i
v1i
v1f
pf
m1 v1f m2 v2f
m1 v1f m2 v2f
mv1f p2mqv2f
v1f 2v2f
v1i 2v2f
(10)
(11)
(12)
(13)
(14)
(15)
Kf Uf
1
m1 pv1f q2
2
(16)
(d) The conservation of energy equation is
1
m2 pv2i q2 2
Ki
1
m1 pv1i q2
2
Ui
1
m2 pv2f q2
2
(17)
(18)
(e) The quantities that are not known in these equations are v1f and v2f .
(f) We have two unknowns and two equations. The unknowns are v1f and v2f .
To find the final velocity of the second package, we can solve for v1f in the
momentum equation and plug this into the energy equation to eliminate v1f .
This will give us the equation for v2f . Once we have an expression for v2f , we
can plug this into the momentum equation to get an expression for v1f .
From the derivations above, the equation relating the final speed of the first
mass to it’s initial speed is
v1f
2m v 1 p7.668m{sq 2.56m{s
m
1i
m 2m
3
(19)
(20)
7
Physics 130
General Physics
Fall 2012
The package rebounds and goes back up the ramp. To determine how high the
package goes, we can use the conservation of energy.
Ki Ui
1 2
mv
0
2
h
Kf
Uf
(21)
0
mgh
(22)
v2
2g
2.56m{sq
p
0.33 m
2p9.8m{s2 q
8
2
(23)
Physics 130
General Physics
Fall 2012
4. Bob can throw a 700 g rock with a speed of 32 m/s. He moves his hand forward 0.8 m
while doing so.
(a) How much work does Bob do on the rock?
(b) How much force, assumed to be constant, does Bob apply to the rock?
(c) What is Bob’s maximum power output as he throws the rock?
Solution:
(a) Calculate work
W
W
W
W
W
∆K
1
1
mpvf q2 mpvi q2
2
2
1
mpvf q2 0
2
1
0.7kgp32m{sq2
2
358 J
(1)
(2)
(3)
(4)
(5)
(b) Calculate force
F
F ∆r F ∆r
W
358 J
∆r
.8m
448 N
P
P
P
F v Fv
448 Np32m{sq
14, 336 W
W
F
(6)
(7)
(8)
(c) Calculate power
9
(9)
(10)
(11)
Physics 130
General Physics
Fall 2012
5. Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose
gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling resistance as the truck
tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes
upward at 8.0 and the coefficient of rolling friction is 0.42. Use work and energy to find
the length of a ramp that will stop a 16, 000 kg truck that enters the ramp at 30 m/s
( 70 mph).
Solution:
We’ll identify the truck and the loose gravel as the system. We need the gravel inside
the system because friction increases the temperature of the truck and the gravel.
We will also use the model of kinetic friction and the conservation of energy equation.
Kf
0
Uf
Uf
∆Eth
∆Eth
Ki
Ki
Ui Wext
0 0
(1)
(2)
The thermal energy created by friction is
∆Eth
fk p∆xq
pµk FN qp∆xq
pµk mg cos θqp∆xq
pµk mg cos θqp∆xq
(3)
(4)
(5)
(6)
Using geometry, the final gravitationl potential energy of the truck is
Uf
mgyf
mg p∆xq sin θ
(7)
(8)
Finally, the initial kinetic energy is simply
Ki
10
1 2
mv
2 i
(9)
Physics 130
General Physics
Fall 2012
Plugging everything into the conservation of energy equation, we get
mg p∆xq sin θ
µk mg cos θp∆xq
g∆xpsin θ
µk cos θq
ñ ∆x 1 2
mv
2 i
1 2
v
2 i
2g psin θ
(10)
(11)
vi2
(12)
µk cos θq
p30 m{sq2
(13)
2 p9.8 m{s2 q psin 8 0.42 cos 8 q
82.7 m
(14)
0.083 km.
(15)
11
Physics 130
General Physics
Fall 2012
6. In a hydroelectric dam, water falls 28 m and then spins a turbine to generate electricity.
(a) What is ∆U of 1.0 kg of water?
(b) Suppose the dam is 78% efficient at converting the water’s potential energy to
electrical energy. How many kilograms of water must pass through the turbines
each second to generate 60 MW of electricity?
Solution:
(a) The change in the potential energy of 1.0 kg of water falling 28 m is
∆Ug
mgh
p1.0 kgqp9.8 m{s2qp28 mq
274 J.
(1)
(2)
(3)
(b) The dam is required to produce 60 MW 50 106 W of power every second,
or in other words 60 106 J of energy per second. If the dam is 78% efficient at
converting the water’s potential energy into electrical energy, then every kilogram of
water that falls produces 0.78 274 J 214 J of electrical energy. Therefore, the
total mass of water needed every second is
60 106 J 1.0 kg
214 J
2.5 105 kg.
This is a typical value for a small hydroelectric dam.
12
(4)
Physics 130
General Physics
Fall 2012
7. Water flowing out of a 15 mm diameter faucet fills a 1.0 L bottle in 8 s. At what distance
below the faucet has the water stream narrowed to 12 mm in diameter?
Solution:
We treat the water as an ideal fluid obeying Bernoulli’s equation. The pressure at
point 1 is p1 and the pressure at point 1 is p2 . Both p1 and p2 are atmospheric
pressure. The velocity and the area at point 1 are v1 and A1 , and at point 2 they
are v2 and A2 . Let d be the distance of point 2 below point 1.
First, we use the time it takes to fill a 1.0 L bottle to compute the flow rate, Q, the
rate at which water is flowing out of the faucet, remembering that one liter equals
103 m3 :
Q
p
1.0 Lq p103 m3 {Lq
8s
1.25 104 m3{s.
(1)
(2)
Next, we use this result to find the velocity, v1 , with which the water leaves the
faucet through the D1 15 103 m diameter aperture (at point 1):
Q
ñ v1
v1 A1
Q
A1
Q
π pD1 {2q2
1.25 104 m3 {s
π p7.5 103 mq2
0.707 m{s.
13
(3)
(4)
(5)
(6)
(7)
Physics 130
General Physics
Fall 2012
Now, from the continuity equation we have
v2 A2
ñ v2
v1 A1
A1
v1
A2
π pD1 {2q2
v1
π pD2 {2q2
(8)
(9)
(10)
2
D1
D2
v1
0.707 m{s
1.11 m{s.
15 103 m
12 103 m
2
(11)
(12)
(13)
Finally, we turn to Bernoulli’s equation to find the height, d:
p1
ρgy1
ρg py1 y2 q
gd
d
1 2
ρv
2 1
p2
1 2
ρv
2 2
ρgy2
1 2
ρpv2 v12 q
2
1 2
p
v2 v12 q
2
pv22 v12q
2g
p1.11 m{sq2 p0.707 m{sq2
2 9.8 m{s2
0.037 m
3.7 cm.
14
(14)
(15)
(16)
(17)
(18)
(19)
(20)
Physics 130
General Physics
Fall 2012
8. A 120 g ice cube at 12 C is placed in an aluminum cup whose initial temperature is
65 C. The system comes to an equilibrium temperature of 15 C. What is the mass of
the cup?
Solution:
There are two interacting systems: aluminum and ice. The system comes to thermal
equilibrium in four steps: (1) the ice temperature increases from 12 C to 0 C;
(2) the ice becomes water at 0 C; (3) the water temperature increases from 0 C to
15 C; and (4) the cup temperature decreases from 65 C to 15 C.
Since the aluminum and ice form a closed system, we have
Q Q1
Q2
Q3
Q4
0.
(1)
Each term is as follows:
Q1
Q2
Q3
Q4
Mice cice ∆T
p0.120 kgq r2090 J{pkg Kqs p12 Kq
3010 J.
Mice Lf
p0.120 kgq p3.33 105 J{kgq
40, 000 J.
Mice cwater ∆T
p0.120 kgq r4190 J{pkg Kqs p15 Kq
7540 J.
MAl cAl ∆T
MAl r900 J{pkg Kqs p50 Kq
p45, 000 J{kgqMAl.
(2)
(3)
(4)
(5)
Inserting everything into equation (1), we get
50, 550 J p45, 000 J{kgqMAl 0
ñ MAl 1.1 kg.
15
(6)
Physics 130
General Physics
Kinematics and Mechanics
Energy
xi vxit 12 axt2
vxf vxi ax t
2
vxf
vxi2 2axpxf xiq
1 2
yf yi vyi t
ay t
2
vyf vyi ay t
2
vyi2 2ay pyf yiq
vyf
1
θf θi ωi ∆t
α∆t2
2
ωf ωi α∆t
ωf 2 ωi 2 2α∆θ
s rθ
c 2πr
vt ωr
v2
ar ω2r
r
Kf
xf
v
2πr
T
Ugf
∆K
Fall 2012
Ki
∆U
Ugi
∆Emech
∆Eth
Kf
Uf ∆Eth
1
K mv 2
2
U mgy
Ki
Ui
∆Eth
Wext
fk ∆s
~ ~d
W F
~ ~v
P F
P ∆E {∆t
∆Eth
Fluids and Thermal Energy
F
P
A
m
ρ
V
Q vA
v2 A2
v1 A1
p1
Qnet
1 2
ρv
2 1
Q1
ρgy1
Q2
p2
... 0
1 2
ρv
2 2
MLf for freezing/melting
Q Mc∆T
Q
Forces
F~net ΣF~
ma
v2
F~net ΣF~r ma m
r
Fg mg
0 fs µs FN
f k µk F N
F~AonB F~BonA
Momentum
p~f
p~ m~v
m1 m2
pvfx q1 m
pvix q1
m
p~i
1
2
pvfx q2 m 2m1m pvix q1
1
2
∆Esys Wext
Constants
9.8 m{s2
Mearth 5.98 1024 kg
rearthsun 1.50 1011 m
ρwater 1000 kg {m 3
ρair 1 .28 kg {m 3
J
cice 2090
kg K
g
cwater
4190
900
J
kg K
J
kg K
J
Lf 333 , 000
ice to water
kg
cAl
16
ρgy2