PHYS 2013, General Physics I, Spring 2015
Name:
ID:
First Exam
Problem Max. Points
1
5
2
5
3
5
4
5
5
5
6
10
7
15
8
25
9
25
10 (bonus)
10
Score
Total:
Comments:
* You must clearly show your work in order to receive full credit on any
problem.
* You are allowed to use a graphing calculator on this exam. You may NOT
use a smartphone or any other device with internet capabilities.
* Turn in your equation sheet with your exam.
* If you need more space for calculations, a blank page is included at the
end of the exam.
* You will lose points for numerical answers with missing or incorrect units.
* You will lose points for inappropriate use of sig-figs.
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Problem 1 (5 points): Consider a graph of an object’s velocity along the x-axis as a
function of time:
Which of the acceleration-vs-time graphs below could possibly correspond to the above
graph? Correct answer: D.
(A)
(B)
(C)
(D)
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Problem 2 (5 points): A ball is thrown straight up. Which of the following statements
are true at the instant when the ball reaches its maximum height? Ignore air resistance.
Correct answer: D.
(A)
The ball’s velocity and acceleration are both zero.
(B)
The ball’s velocity and acceleration are both non-zero.
(C)
The ball’s acceleration is zero, but its velocity is non-zero.
(D)
The ball’s velocity is zero, but its acceleration is non-zero.
(E)
The ball’s speed is zero, but it’s velocity is not zero.
Problem 3 (5 points): A car travels at a constant speed of 20 m/s around a circular track
with radius 100 m. What is the magnitude of the car’s acceleration? Correct answer: E.
(A)
0.0 m/s2
(B)
0.5 m/s2
(C)
1.0 m/s2
(D)
2.0 m/s2
(E)
4.0 m/s2
4
Problem 4 (5 points): The figure below represents the velocity and acceleration vectors
of a bumble bee at some instant in time during its flight as viewed from above (assume that
both vectors lie in the plane of the paper). Correct answer: A.
At the instant shown, which of the following statements best describe the bee’s motion?
(A)
The bee is turning to its left and slowing down. (Correct Answer!)
(B)
The bee is turning to its right and speeding up.
(C)
The bee is moving in a straight line with constant speed.
(D)
The bee is moving in a straight line and slowing down.
(E)
None of the above.
Problem 5 (5 points): A 100 kg crate is placed on a scale inside an elevator. When the
scale reads 1200 N, what can you say about the motion of the elevator? Correct answers
B and D.
(A)
Moving upwards at constant speed.
(B)
Moving downwards while slowing down.
(C)
Moving upwards while slowing down.
(D)
Moving upwards while speeding up.
(E)
None of the above.
5
~ = −2.0î + 5.0ĵ and B
~ = 4.0ĵ − 3.0k̂.
Problem 6 (10 points): Consider two vectors A
Calculate the following:
~ −B
~
(a) 2A
~ −B
~ = 2 −2.0î + 5.0ĵ − 4.0ĵ − 3.0k̂ = −8.0î + 6.0ĵ + 3.0k̂
2A
~ ·B
~
(b) A
~ ·B
~ = −2.0î + 5.0ĵ · 4.0ĵ − 3.0k̂ = 0 + 20.0 + 0 = 20.0
A
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Problem 7 (15 points): A particle undergoes motion in a straight line, starting at the
origin, with velocity given by At2 +Bt3 , where t is time, A = −0.12 m/s3 and B = 0.40 m/s4 .
(a) What is the particle’s acceleration at t = 1.0 s ?
From definitions of acceleration
d 2
At + Bt3
dt
= 2At + 3Bt2
a(t) =
and thus
a(1.0 s) = 2 −0.12 m/s3 (1.0 s) + 3 0.40 m/s4 (1.0 s)2
= 0.96 m/s2
(b) What is the particle’s position at t = 1.0 s?
Position is given by
ˆ
x(t) =
t
v(τ )dτ =
0
and thus
At3 Bt4
+
3
4
−0.12 m/s3 (1.0 s)3 0.4 m/s4 (1.0 s)4
x(1.0 s) =
+
= .06 m
3
4
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Problem 8 (25 points): Two blocks A and B are pushed along a frictionless horizontal
table by a force of magnitude F , as shown. Block A has mass of 3m, block B has mass of
5m.
NOTE: in your answers to the following, “minus” and “plus” (or similarly “negative”
and “positive”) do not indicate direction on their own. More is need to indicate a direction.
(a) Draw a free-body diagram for each block with unique labels for each force that
appears.
(b) What is the magnitude and direction of the acceleration of the system? Express the
magnitude in terms of F and m.
Let
~ AB = f î
F
(1)
~ BA = −F
~ AB = −f î
F
(2)
then from third law
and from second laws applied to each object
−f = 3ma and
− F + f = 5ma
(3)
−F
î
8m
(4)
or
−F − 3ma = 5ma
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⇒
~a =
(d) What is the magnitude and direction of the force that block A exerts on block B.
Express the magnitude in terms of F and m.
From (3) and (4) we get
~ AB = f î = −3maî = 3 F î
F
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(e) At the instant shown (which you just solved for), which of the following descriptions
of motion are possible. Circle ALL that apply. Correct answers are in bold.
• The blocks are moving to the right with constant speed.
• The blocks are moving to the right and speeding up.
• The blocks are moving to the right and slowing down.
• The blocks are moving to the left with constant speed.
• The blocks are moving to the left and speeding up.
• The blocks are moving to the left and slowing down.
• The blocks are at rest.
• It is impossible to tell without numbers.
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Problem 9 (25 points): Alice and Bob are on the roof tops of adjacent buildings. Alice’s
building has a height of 98 m, Bob’s building has a height of 125 m. The two building are
20 m apart. Alice throws a baseball with at a speed of 40 m/s, at an angle of 60◦ above
the horizontal. She releases the ball at the edge of her building from a height 100 m (2 m
above the roof of her building). Ignoring air resistance, answer the following questions.
(a) Does the ball reach Bob’s roof? (assume that the roof is at lest 200 m wide.) Justify
your answer with a quantitative calculation.
Initial velocity is v(0) = (20 m/s, 35 m/s) and thus at time T =
20 m
20 m/s
= 1 s the base ball will
either hit the wall of Bob’s building or will be on its way to land on the roof. But since
1
ry (1 s) = 98 m + 35m/s · 1 s − 9.8 m/s2 · (1 s)2 ≈ 128 m > 125 m
2
the answer is yes.
(b) What is the ball’s velocity when it reaches the height of Bob’s roof?
When the ball reaches the height of Bob’s roof the y-component of velocity is
vy2
r
= ± (35 m/s)2 − 2 · 9.8 m/s2 (125 m − 98 m) = ±26
and thus
v = (20 m/s, 26 m/s) .
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(Bonus) Problem 10 (10 points): Alice and Bob are standing on opposite sides of a
river that is 30 m wide and flows due west with a constant speed of 4 m/s.
Bob wishes to swim directly across the river, along the straight line connecting him and
Alice. He swims at his maximum speed of 5 m/s relative to the water.
(a) As he swims along the straight line, what is Bob’s speed relative to Alice?
Bob’s speed must be
q
vx2 + vy2 = 5 m/s
and in order to swing along a straight line the x-component must be
vx = 4 m/s
and thus
vy = ±3 m/s.
Therefore in order to swim along a straight line from B to A the velocity must be
v = (4 m/s, 3 m/s) .
(b) How long does it take Bob to cross the river?
The time it would take to cross the river is then
t=
30 m
= 10 s.
3 m/s
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