PHYSICS 1210 Exam 1
University of Wyoming
14 February 2013
150 points
This test is open-note and closed-book. Calculators are permitted but computers are not.
No collaboration, consultation, or communication with other people (other than the
administrator) is allowed by any means, including but not limited to verbal, written, or
electronic methods. Sharing of materials including calculators is prohibited.
If you have a question about the test, please raise your hand. Please do not open this test
booklet until everyone has received a booklet and the test administrator has indicated for
you to begin. While you are waiting, make sure that your name is written clearly at the
top of the first page.
g = 9.80 m/s2 1 radian = 57.3°
1 mi = 1.609 km
v v
A B = ABcos = Ax Bx + Ay By + Az Bz
v
v
v
v
v
a = dv dt = d 2 r dt 2
v = dr dt
vx = v0x + ax t
x = x0 + v0x t + 1/2 ax t2
vx2 = v0x2 + 2ax (x – x0)
x – x0 = (v0x + vx) t/2
vx = v0x + ax dt
x = x0 + vx dt
v v v
v = v 0 + at
v v v
v
r = r0 + v 0 t + 1 2 at 2
x = (v0 cos0) t
y = (v0 sin0) t – 1/2 g t2
vx = v0 cos0
vy = v0 sin0 – g t
2
ar = v /R
ar = 42R/T2 vT = 2R/T
v
v
v
v P/A = v P/B + v B/A
A
1 m = 39.37 in = 3.28 ft
R = v02 sin(20)/g
Name:
Answer Key
1. (10 points) If the scalar product (dot product) of two vectors is positive, what does
that tell you about the vectors?
The angle between the two vectors is acute, so they point mostly in the
same direction.
2. (10 points) A classmate tells you that the formula for the volume of a cone of base
radius r and height h is r3h/3. Explain why that formula cannot be correct.
That formula would give units of (distance)4. Volume units must be
(distance)3.
3. (15 points) The diagrams to the right show five pairs of acceleration and velocity
vectors. (The outlined arrow is the acceleration vector and the simple arrow is the
velocity vector.) For each verbal description of motion, identify which numbered pair of
vectors corresponds to the motion described.
a.
3
speeding up without changing direction
v
1
a
b.
1
slowing down without changing direction
2
v
a
c.
5
changing direction at a constant speed
v
3
a
a
d.
e.
2
4
speeding up while changing direction
slowing down while changing direction
PHYS 1210-02 Exam 1
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4
5
a
v
v
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4. (10 points) A stone is thrown into the air at an angle above the horizontal and feels
negligible air resistance. Which one graph best depicts the stone’s speed as a function of
time while it is in the air?
5. (16 points) Each of the questions
below contains the same graph of position as a
x
function of time and a verbal description of some particular conditions of motion. Mark
on each graph the specific time, times, or time intervals that the conditions on the graph
match the description. If the conditions described are not met at any time on tthe graph,
indicate that.
x
a. speeding up
t
x
b. slowing down
t
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x
c. ax < 0
t
x
d. ax = 0
t
x
e. ax > 0
t
x
f. vx = 0
t
x
g. vx < 0
t
x
h. vx > 0
t
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6. (10 points) A sled starts from rest and slides down a frictionless air track at constant
acceleration. In 5.0s, it travels a distance of 50m down the track. What is its
acceleration?
The equation of motion is x = 1/2 at2. We know x = 50 m and t = 5.0 s;
so we can solve for a.
a = 2x/t2 = 2(50 m)/(5.0 s)2 = 100/25 m/s2 = 4 m/s2.
7. A dart is launched at an angle of below horizontal with initial speed v0 at time t = 0.
The launch position is a height H above and a distance L to the right of the origin. In
describing the situation, use the coordinate system in which up is the +y direction, right is
the +x direction, and the launch position is (L, H).
a. (6
r points) Fill
r in the table with
r formulas for the x- and y-components of acceleration
a , velocity v , and position r as functions of time t. Express them in terms of only
the constants v0, , H, L, and g, and the variable t.
x-component
y-component
r
a
0
–g
r
v
v0 cos()
–v0 sin() – gt
r
r
L + v0 cos() t
H – v0 sin() t – 1/2 gt2
b. (8 points) At what time will the dart land on the ground at height y = 0? Express as a
formula again in terms of some or all of the quantities v0, , H, L, and g.
y = H – v0 sin() t – 1/2 gt2
0 = H – v0 sin() t – 1/2 gt2
0 = 1/2 gt2 + v0 sin() t – H
– v0 sin() ± (v0 sin)2 + 2gH
t=
g
– v0 sin() + (v0 sin)2 + 2gH
The positive solution is t =
g
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c. (8 points) At what horizontal distance x from the origin will the dart land on the
ground? Express as a formula again in terms of some or all of the quantities v0, , H,
L, and g.
– v0 sin() + (v0 sin)2 + 2gH
x = L + v0 cos() t = L + v0 cos()
g
2
– v0 sin() + (v0 sin) + 2gH
= L + v0 cos()
g
Notice that this reduces to the range equation if L = 0 and H = 0.
d. (3 points) There should be two solutions for the time and horizontal position of the
dart reaching the ground. Only one is physically appropriate. What is the meaning of
the other solution?
If the equation of motion had been valid before the launch, the dart
would have traveled upward, reached the top of its arc, and then begun
traveling downward before t = 0. The other solution would have been
when the dart was traveling upward at ground level.
e. (4 points) What is the minimum speed of the dart during the time interval starting
immediately after it is launched and ending immediately before it lands?
v0
f. (3 points) As the dart falls, does the rate at which its speed changes increase,
decrease, or remain constant?
It increases. The rate of change of speed is the component of
acceleration parallel to the velocity. Since the velocity becomes more
downward as the dart falls, and since the acceleration is g downward,
the parallel component increases with time.
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8. A ball is dropped from rest from the top of a building that is 176.4 m high. At the
same instant, time t = 0, a second ball is projected vertically upward from the ground,
such that its speed is zero when it reaches a height of 176.4 m.
a. (8 points) At what time do the balls cross?
Projected ball:
Dropped ball:
yD = H – 1/2gt
2
yP = v0t – 1/2 gt2 = H at time tH
v = v0 – gt = 0 at time tH
tH = v0/g
H = v02/g – 1/2 g(v0/g)2 = 1/2 v02/g
v0 = 2gH = (19.6m/s2)(176.4 m)
= 58.8 m/s
When they cross:
yD = yP
H – 1/2 gt2 = 2gH t – 1/2gt2
t = H/2g = (176.4 m)/(19.6 m/s2) = 9 s = 3.0 s
b. (8 points) At what height do the balls cross?
Dropped ball:
Projected ball:
yD = H – 1/2gt2
= 176.4 m – (4.9 m/s2)(9 s2)
= 176.4 m – 44.1 m
= 132.3 m
yP = v0 t – 1/2 gt2
= (58.8 m/s)(3 s) – (4.9 m/s2)(9 s2)
= 176.4 m – 44.1 m
= 132.3 m
c. (6 points) When they cross, what are the speeds of the two balls? Be sure to give the
speed of each one.
Dropped ball:
Projected ball:
2
v = –gt = –(9.8 m/s )(3 s)
= –29.4 m/s
v = v0 – gt = 58.8 m/s –(9.8 m/s2)(3 s)
= 58.8 m/s – 29.4 m/s = +29.4 m/s
For both balls, the speed is 29.4 m/s.
d. (4 points) What is the acceleration of the balls when they cross? Be sure to give the
acceleration of each one.
For both balls, the acceleration is g downward.
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9. In an episode of the British television show Top Gear, one of the hosts drove a car on
a helical path (like a corkscrew) in a tunnel so that it briefly traveled upside down on the
roof of the tunnel. The radius of the tunnel was 5.0 meters.
a. (8 points) How quickly would the car need to make one loop (what is its period of
rotation) so that its centripetal acceleration was at least g?
a = 42R/T2
T = 2R/a = 2R/g = 2(5.0 m)/(9.8 m/s2) = 2 (0.71429 s) = 4.49 s
b. (8 points) If the car’s path was angled 20° to the roadway, how fast was the car
traveling to execute the loop?
vx
v
vx = 2R/T
vx/v = sin()
v = vx/sin() = 2R/(T sin) =
2
= (5 m)(9.8 m/s ) /sin(20°)
= 7/(0.34202) m/s = 20.47 m/s
PHYS 1210-02 Exam 1
2 R
2R/g sin()
A
= Rg /sin()
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