Download Seat: PHYS 1500 (Fall 2006) Exam #2, V1 Name: 1. Two objects are

Seat:
PHYS 1500 (Fall 2006) Exam #2, V1
5 pts
1. Two objects are moving horizontally with no external forces on them. The 1 kg object
moves to the right with a speed of 1 m/s. The 2 kg object moves to the left with a
speed of 1 m/s. The two objects collide and stick together. The combined 3 kg object
(a)
(b)
(c)
(d)
5 pts
Name:
moves to the left.
moves to the right.
is stationary.
moves in a way that can not be determined from the given information.
2. Because of the Earth’s rotation about its axis, your weight is slightly different at the
equator than at the North or South Pole. Is your weight more or less at the equator?
Explain.
Your weight is the normal force exerted by the ”floor” to keep you in place.
On the equator, you weigh slightly less than at the pole because the normal
force minus mg equals 0 at the pole. At the equator, the normal force minus
mg equals a small negative number (the centripetal force).
5 pts
3. Travelling with a velocity of 2.0 m/s in the x-direction, a 5.0 kg flugle explodes into
2 pieces. After the explosion, piece 1 has a mass of 3.0 kg and a velocity with xcomponent of 4.0 m/s and y-component of −6.0 m/s. What is the velocity vector of
piece 2 after the explosion? (Ignore all external forces.)
Since there are no external forces, the total momentum is conserved. The
momentum before the explosion equals the momentum after.
Before : px = 10 kg m/s
Before : py = 0 kg m/s
After : px = m1 v1x + m2 v2x = 12 kg m/s + 2 kg v2x
After : py = m1 v1y + m2 v2y = −18 kg m/s + 2 kg v2y
v2x = −1 m/s
Before : px = 18 kg m/s
Before : py = 0 kg m/s
Before : py = 0 kg m/s
(V 1)
After : px = m1 v1x + m2 v2x = 24 kg m/s + 2 kg v2x
After : py = m1 v1y + m2 v2y = −12 kg m/s + 2 kg v2y
v2x = −3 m/s
Before : px = 28 kg m/s
v2y = 9 m/s
v2y = 6 m/s
(V 2)
After : px = m1 v1x + m2 v2x = 40 kg m/s + 2 kg v2x
After : py = m1 v1y + m2 v2y = −20 kg m/s + 2 kg v2y
v2x = −6 m/s
v2y = 10 m/s
(V 3)
10 pts
4. You are dragging a 20 kg box across the floor with a rope tied around the box. The
rope makes an angle of 37◦ with the vertical. The box starts from rest. The tension
in the rope is 40 N. In dragging the box from x = 2 m to x = 6 m, the box’s speed
decreases from 3 m/s to 2 m/s. (a) How much work does the rope do on the box? (b)
What is the force from friction?
(a) The work that the rope does on the box can be found from W =
(F cos θ)∆x. The angle θ is the angle between the force and the direction that the box moves. (b) The force from friction can be found from
Wf ric + Wrope = Wnet = ∆KE; the work done by friction is the −Ff ric ∆x where
Ff ric is the magnitude of the friction force and ∆x is the distance moved
(the − is because the friction force is in the opposite direction that the box
is moving).
T = 40 N θ = 53◦
Wnet
∆x = 4 m
1
= ∆KE = 20 kg
2
÷
Wrope = 40 N cos 53◦ 4 m = 96 J
m
2
s
¸2
·
m
− 3
s
¸2 !
= −50 J
Wfric = Wnet − Wrope = −50 J − 96 J = −146 J = −Ffric 4 m
T = 20 N θ = 37◦
Wnet
∆x = 3 m
1
= ∆KE = 10 kg
2
÷
Wrope = 20 N cos 37◦ 3 m = 48 J
m
1
s
¸2
·
m
− 2
s
¸2 !
= −15 J
Wfric = Wnet − Wrope = −15 J − 48 J = −63 J = −Ffric 3 m
T = 10 N θ = 53◦
Wnet
∆x = 5 m
1
= ∆KE = 40 kg
2
÷
Ffric = 36.5 N (V 1)
Ffric = 21 N
(V 2)
Wrope = 10 N cos 53◦ 5 m = 30 J
m
1
s
¸2
·
m
− 3
s
¸2 !
= −160 J
Wfric = Wnet − Wrope = −160 J − 30 J = −190 J = −Ffric 5 m
Ffric = 38 N (V 2)
5 pts
5. Two balls, identical except for their color, are moving with equal speed on the same
horizontal surface. The red ball is rolling; the blue ball is sliding so that it isn’t rotating
at all. The energy of the red ball is
(a)
(b)
(c)
(d)
5 pts
greater than that of the blue ball.
smaller than that of the blue ball.
is the same as that of the blue ball
relative to the blue ball can’t be determined from the information given.
6. A girl is standing on a raft that is floating on a pond. Both the girl and the raft are
initially stationary. The girl walks from the end of the raft away from the shore to the
end of the raft closest to the shore. Ignore the drag from air and water. Does the raft
move? If yes, what direction does it move? Explain.
When the girl steps toward the shore, the boat exerts a force on her toward
the shore. By Newton’s 3rd law, she exerts a force on the boat away from
the shore. Thus, the boat moves away from the shore.
5 pts
7. A 10.0 kg vrabn is dropped from a height of 1.5 m above the floor. When it hits the
floor, it has a speed of 4.0 m/s. How much work was done by air resistance and other
non-conservative forces?
The work done by all other forces is Wnc = Wnet − Wg where Wg is the work
done by gravity and Wnet is the total work done. The total work done can
be found from Wnet = ∆KE. In this problem, KEi = 0 and Wg = mgh because
the vrabn moves down.
µ
1
m
Wnc = 10 kg 4
2
s
¶2
µ
Wnc
1
m
= 20 kg 3
2
s
Wnc
1
m
= 30 kg 2
2
s
µ
− 10 kg 10
m
1.5 m = 80 J − 150 J = −70 J
s2
¶2
(V 1)
− 20 kg 10
m
3 m = 90 J − 600 J = −510 J
s2
(V 2)
− 30 kg 10
m
3 m = 60 J − 900 J = −840 J
s2
(V 3)
¶2
10 pts
8. A 5 kg bucket hangs from a line that is wrapped around a cylinder that has a mass
of 20 kg, moment of inertia of 10 kg·m2 , and a radius of 2 m. The cylinder rotates on
a frictionless axle. If the bucket and cylinder start from rest, how far has the bucket
descended if its speed is 8 m/s?
Since the only force that is acting in this problem is gravity, then the total
work done Wnet = ∆KE = Wg . Because the bucket is dropping, the work
done by gravity is Wg = mgh. There are two source of KE: translational
kinetic energy of the bucket, KE = (1/2)mv 2 , and the rotational kinetic
energy of the cylinder, KE = (1/2)Iω 2 , with ω = v/r.
8 m/s
rad
ω=
=4
2m
s
Ã
1
rad
= 10 kg m2 4
2
s
KEcyl
Wg = 5 kg 10
1
rad
= 15 kg m2 2
2
s
Wg = 10 kg 10
ω=
= 80 J
!2
1
rad
= 18 kg m2 2
2
s
Wg = 2 kg 10
h=
¶2
= 160 J
240 J
= 4.8 m
50 N
1
= 10 kg
2
= 30 J
!2
m
8
s
Wnet = 160 J + 80 J = 240 J
m
h = 210 J
s2
8 m/s
rad
=2
4m
s
Ã
KEcyl
!2
KEbuc
Ã
KEcyl
KEbuc
m
h = 240 J
s2
6 m/s
rad
ω=
=2
3m
s
µ
1
= 5 kg
2
m
h = 100 J
s2
m
6
s
¶2
= 180 J
Wnet = 180 J + 30 J = 210 J
h=
210 J
= 2.1 m
100 N
1
KEbuc = 2 kg
2
= 36 J
µ
(V 1)
µ
8
m
s
(V 2)
¶2
= 64 J
Wnet = 64 J + 36 J = 100 J
h=
100 J
=5m
20 N
(V 3)
5 pts
9. Circle all of the correct answers. One J (Joule) of energy equals
(a) 1 W/s
(d) 1 W·s
5 pts
(b) 1 N/m
(e) 1 N·m
(c) 1 kg·m2
(f) 1 kg/m2
10. You throw 3 identical lead weights from a height of 2 m. All three leave your hand
at exactly the same height and with the same speed. Weight 1 is thrown at an angle
45◦ above the horizontal, weight 2 is thrown horizontally, and weight 3 is thrown at
an angle 45◦ below the horizontal. Neglecting air resistance, rank the speeds of the
weights just before they reach the ground from slowest to fastest. Explain.
The change in kinetic energy is related to the total work done through
Wnet = ∆KE. In this situation, only gravity is doing work. Therefore,
Wnet = mgh which is the same for all three weights. Because the change
in kinetic energy is the same for all three weights and they all start with
the same kinetic energy, they all finish with the same kinetic energy. This
means they all have the same speed when they hit the ground.
5 pts
11. You have a 8 m long, uniform metal rod of mass 3 kg balanced on its midpoint. You
hang a 8 kg mass 1 m from one end and you pull down on a string at the other end so
that the rod is horizontal and motionless. What is the tension in the string? What is
the force exerted on the rod by the wedge?
Because the rod is motionless, the sum of the torques has to be 0 and the
sum of the forces has to be 0. I will take the axis of rotation to be at the
wedge which means only the mass hanging from the left and the string give
nonzero torque; the weight of the rod and the wedge act at the axis and
give zero torque. The forces are only in the y-direction so there is only one
relation between the forces. The forces are from the mass, the string, the
weight of the rod, and the wedge.
m
m
m
τ : 3 m 8 kg 10 2 + −4 m T = 0
F : −8 kg10 2 + −T + −3 kg10 2 + Fw = 0
s
s
s
T = 60 N
Fw = 170 N
(V 1)
τ : 4 m 15 kg 10
τ : 2 m 9 kg 10
m
m
m
+ −5 m T = 0
F : −15 kg10 2 + −T + −7 kg10 2 + Fw = 0
2
s
s
s
T = 120 N
Fw = 340 N
(V 2)
m
m
m
+
−3
m
T
=
0
F
:
−9
kg10
+
−T
+
−2
kg10
+ Fw = 0
s2
s2
s2
T = 60 N
Fw = 170 N
(V 3)
10 pts 12. This problem takes place on earth. A 3 kg mass is tied to a string with a length of
2 m. The mass is rotating clockwise in a vertical circle. When the string is horizontal
and the mass is to the right of the axis of rotation, the mass has a speed of 8 m/s. At
this point, (a) what is the angular velocity? (b) what is the angular acceleration? (c)
what is the net force acting on the mass? (d) When the mass is at the bottom of the
circle, what is its velocity?
(a) Since the mass is rotating clockwise, the angular velocity is ω = −v/r.
(b) The angular acceleration can be found from α = at /r. Gravity is the
only force acting along the circle when the string is horizontal. Therefore,
the tangential acceleration has a magnitude g. Because the torque is in
the clockwise direction, the angular acceleration should be negative. The
angular acceleration is α = −g/r. (c) The net force acting on the mass
is minus the tension in the string for the x-direction and −mg for the ydirection. The tension in the string is the only force that is making the
mass go in a circle (when the string is horizontal) therefore it must equal
mv 2 /r. (d) Since there is hardly any friction in this situation, the only force
doing work on the mass is gravity. This means the amount of net work
done is mgr for this situation. This can be related to the change in kinetic
energy to give KEf = KEi + mgr. The final velocity is vf2 = vi2 + 2gr for this
problem.
(a) ω = −
8 m/s
rad
= −4
2m
s
(b) α = −
10 m/s2
rad
= −5 2
2m
s
(8 m/s)2
m
= −96 N
Fy = −3 kg 10 2 = −30 N
2m
s
µ
¶2
2
√
m
m
m
(d) vf2 = 8
+ 2 10 2 2 m = 104 2 = 104 m/s
s
s
s
(c) Fx = −3 kg
12 m/s
rad
(a) ω = −
= −3
4m
s
(V 1)
10 m/s2
rad
(b) α = −
= −2.5 2
4m
s
(12 m/s)2
m
= −72 N
Fy = −2 kg 10 2 = −20 N
4m
s
µ
¶2
m
m
m2 √
(d) vf2 = 12
+ 2 10 2 4 m = 224 2 = 224 m/s
s
s
s
(c) Fx = −2 kg
(a) ω = −
rad
20 m/s
= −4
5m
s
(b) α = −
(V 2)
10 m/s2
rad
= −2 2
5m
s
(20 m/s)2
m
= −320 N
Fy = −4 kg 10 2 = −40 N
5m
s
µ
¶2
2
√
m
m
m
(d) vf2 = 20
+ 2 10 2 5 m = 500 2 = 500 m/s
s
s
s
(c) Fx = −4 kg
(V 3)
Equations
Basic Mathematic Formulas
tan θ = ho /ha h2 = h2o + h2a
Acirc = πr2
4π 3
Circum of circ = 2πr
Vsph =
r
Asur of sph = 4πr2
3
cos 37◦ = 4/5 sin 37◦ = 3/5 cos 53◦ = 3/5 sin 53◦ = 4/5
sin θ = ho /h
cos θ = ha /h
Chapter 2
g = 10 m/s2
Avg speed = distance/elapsed time
∆x = xf − xi
v̄ = ∆x/∆t
1
∆x = (v0 + v)t
2
v = v0 + at
ā = (vf − vi )/(tf − ti ) = ∆v/∆t
1
∆x = v0 t + at2
2
v 2 = v02 + 2a∆x
Chapter 3
Ax = A cos θ
Ay = A sin θ
∆~r = ~rf − ~ri
A=
q
~vav = ∆~r/∆t
A2x + A2y
tan θ = Ay /Ax
~aav = (∆~v)/∆t
vx = v0x + ax t
1
∆x = (v0x + vx )t
2
1
∆x = v0x t + ax t2
2
2
vx2 = v0x
+ 2ax ∆x
vy = v0y + ay t
1
∆y = (v0y + vy )t
2
1
∆y = v0y t + ay t2
2
2
vy2 = v0y
+ 2ay ∆y
Chapter 4
X
~ = m~a
F
m1 m2
r2
~ 21
= −F
G = 6.67259 × 10−11
Fg = G
~ 12
F
f s ≤ µs n
N · m2
kg2
w = mg
f k = µk n
Chapter 5
W = (F cos θ)∆x
PE = mgy
1
KE = mv 2
2
Wnet = KEf − KE0
Wnc = KEf − KEi + PEf − PEi
P̄ =
Wg = mg(yi − yf )
W
∆t
Chapter 6
~p = m~v
~I = F∆t
~
~
F∆t
= ∆~p = m~vf − m~vi
m1~v1i + m2~v2i = m1~v1f + m2~v2f
P̄ = F v̄
Chapter 7
θf − θi
∆θ
=
tf − ti
∆t
1
∆θ = (ωi + ω)t
2
ωav =
ω = ωi + αt
vt = rω
at = rα
ωf − ωi
∆ω
=
tf − ti
∆t
1
∆θ = ωi t + αt2
ω 2 = ωi2 + 2α∆θ
2
ac =
αav =
v2
= rω 2
r
T2 =
4π 2 3
r
GMS
Chapter 8
τ = rF sin θ
I≡
X
mr2
Wnc = ∆KEt + ∆KEr + ∆P E
X
τ = Iα
L ≡ Iω
1
KEr = Iω 2
2
X
∆L
τ=
∆t