Instructor: Hacker
Course: Physics 121
Sample Exam 1 Solutions (Mathematical Prerequisites)
Print your name neatly. If you forget to write your name, or if the grader can’t read your
writing, you can lose up to 100 points. Answer all the questions that you can.
This exam will consist of 21 multiple-choice problems. You may not use calculators or
other electronic devices on this exam. The use of such a device will be regarded as an
attempt to cheat, and will be pursued accordingly. All diagrams and figures on this exam
are rough sketches: they are not generally drawn to scale.
No partial credit will be given for these problems. However, you can miss one of the 21
problems without penalty. Your grade will be based on your best 20 problems. You will
not receive extra credit for getting all 21 right.
Your grade on the exam will be based entirely on the answers that you circle on this
sheet. If you have no answer or a wrong answer there, the grader will not look at the
page with the problem to see if the right answer appears there. Illegible or ambiguous
answers will be graded as wrong. You are responsible for copying your answers clearly,
correctly, and in the right place.
Although there is no partial credit on this exam, you must show your work in the space
provided on the exam. There is additional scratch paper at the end of the exam: do
not use it unless you have filled all the scratch space provided on the page with the
problem. If you answer a difficult problem without doing any written work, the grader
will assume that you got the answer by guessing or by copying from someone else, and
will not give you credit for the problem even though you’ve indicated the correct solution
on the answer sheet.
Circle your answers here. Do not detach this sheet from the test.
1.
a
b c
○
d
e
8.
a
b c
○
d
e
a b
15. ○
2.
a
b
c
d e
○
9.
a b
○
c
d
e
16.
3.
a
b
c
d e
○
a b
10. ○
c
d
e
4.
a
b
c
d e
○
a b
11. ○
c
d
5.
a
b
c
d e
○
12.
a
b
c d
○
6.
a
b
c
d e
○
13.
a
b c
○
7.
a
b
c
d e
○
a b
14. ○
c
c
d
e
a
b c
○
d
e
17.
a
b
c d
○
e
e
18.
a
b
c d
○
e
e
19.
a
b
c
d e
○
d
e
20.
a
b
c
d e
○
d
e
21.
a
b c
○
d
e
physics 121 sample exam 1 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 2
Geometry
Problem 1. If the area of a circle is A, what is its circumference?
√
√
(a) 2π A
*(b) 2 πA
r
r
A
2A
(c)
(d)
π
π
(e)
None of these
r
Solution: A = πR2 and C = 2πR; so R =
A
π
r
⇒
C = 2π
√
A
= 2 πA
π
Problem 2. What is the area of a rectangular windowpane measuring 20 cm wide by
25 cm high?
(a) 50 cm2
(c) 125 cm2
(e) None of these
(b) 80 cm2
*(d) 500 cm2
Solution: A = lw = (20 cm)(25 cm) = 500 cm2
Problem 3. Find r in the right triangle at right.
(a) √
5.5
(c)
11
(e) None of these
(b)
*(d)
r 11
√
61
5
6
Solution: r is a length, so it must be positive. Hence
√
2
2
r =x +y
2
−−→ r =
p
√
√
√
x2 + y 2 = 62 + 52 = 36 + 25 = 61
Problem 4. An airplane flies a route involving three cities. From Rio Cordaro, it flies
100 miles straight east to Hackerville. From Hackerville, it flies 200 miles straight north
to San Pitucco. How far does it fly from San Pitucco to Rio Cordaro?
√
(a) 225 miles
(b) 100√3 miles
(c) 250 miles
*(d) 100 5 miles
(e) None of these
Solution: Sketch a diagram, as at right. The route forms a
right triangle, with the legs from Rio Cordaro to Hackerville and
from Hackerville to San Pitucco on either side of the right angle.
The leg from San Pitucco to Rio Cordaro is the hypotenuse. Use
the theorem of Pythagoras:
r 2 = x2 + y 2
p
p
−−→ r = x2 + y 2 = (100 mi)2 + (200 mi)2
q
√
= 50, 000 mi2 = 100 5 mi
SP
r
r
200 mi
√
r
RC
100 mi
r
HV
physics 121 sample exam 1 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 3
Problem 5. A vertical telephone pole is stabilized by a
diagonal guy wire anchored in the ground, as shown at right.
The wire is 10 m long; it is attached to the pole 7 m above
ground level. How far from the base of the pole does the wire
meet the ground?
(a) 3√
m
(c) 2 15 m
(e) None of these
(b) 9√m
*(d)
51 m
@
7m
@
@ 10 m
@
x @@
Solution: We have a sketch already; we can label the sides that we know and the side
that we want to find. This is a right triangle where the hypotenuse is 10 m and the
vertical side is 7 m; we want to know the horizontal side x. Use Pythagoras:
r 2 = x2 + y 2
−y 2
−−−→ x2 = r2 − y 2
√
p
p
√
√
−−→ x = r2 − y 2 = (10 m)2 − (7 m)2 = 100 − 49 m = 51 m
Algebra
Problem 6. If f (x) = x2 − 5, find f (5).
(a) −5
(c) 9
(e) None of these
(b)
*(d)
0
20
Solution: f (5) = 52 − 5 = 25 − 5 = 20
Problem 7. If f (x) = 3x + 2, find f (a + 1).
(a) 3a + 1
(c) 3a + 3
(e) None of these
(b) 3a + 2
*(d) 3a + 5
Solution: f (a + 1) = 3(a + 1) + 2 = 3a + 3 + 2 = 3a + 5
Problem 8. If f (x) =
(a)
(c)
(e)
a − 23
4
a + 23
4
x−3
, find f (a − 5).
4
a−8
*(b)
4
a+8
(d)
4
None of these
Solution: f (a − 5) =
(a − 5) − 3
a−8
=
4
4
physics 121 sample exam 1 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 4
Problem 9. If√an object is dropped from a height h, the speed with which it hits the
ground is: v = 2gh, where g is the gravitational acceleration. (If you don’t know what
gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If
an object is dropped from a height of 20, and the gravitational acceleration is 10, what
is the speed with which the object hits the ground? Round your answer to the nearest
integer.
*(a)
(c)
(e)
20√
(b) 40√
10 2
(d) 20 2
None of these
p
p
√
Solution: v = 2gh = 2(10)(20) = 400 = 20
Problem 10. The force exerted by a stretched spring is given by the formula:
F = k(x − x0 ), where k is the spring constant, x is the stretched length of the spring,
and x0 is its unstretched length. (You will learn about forces and spring constants
during this course.) What is the force F if k = 100, x = 6, and x0 = 5? Round your
answer to the nearest integer.
*(a)
(c)
(e)
100
250
3
None of these
(b)
120
(d)
595
Solution: F = k(x − x0 ) = (100)(6 − 5) = 100 · 1 = 100
Problem 11. If 3x − 2 = −9, find x. Which of the following statements is true?
*(a) x < −1
(c)
0≤x<1
(e)
None of these
Solution: 3x − 2 = −9
Hence x < −1
(b) −1 ≤ x < 0
(d) x ≥ 1
+2
−−→
3x = −7
÷3
−−→
x=−
7
3
Problem 12. If x + 7 = −4x + 9, find x. Which of the following statements is true?
(a) x < −1
*(c) 0 ≤ x < 1
(e) None of these
(b) −1 ≤ x < 0
(d) x ≥ 1
Solution: We need to collect all the terms with x on one side of the equation, and all
of the terms with no x on the other.
x + 7 = −4x + 9
Hence 0 ≤ x < 1
+4x
−−−→
5x + 7 = 9
−7
−−→
5x = 2
÷5
−−→
x=
2
5
Copyright ©Wayne Hacker 2010. All rights reserved. 5
physics 121 sample exam 1 solns
Problem 13. If P V = nRT , find P .
nRT
V
*(b) P =
(a) P =
nRT
V
(c)
P = V − nRT
(e)
None of these
Solution: P V = nRT
P = nRT − V
(d)
÷V
−−→ P =
nRT
V
Problem 14. If v = v0 + at, find a.
v − v0
v
*(a) a =
(b) a = − v0
t
t
vt − v0
v0
(c)
a=
(d) a = v −
t
t
(e)
None of these
Solution: v0 + at = v
−v
0
−−−→
at = v − v0
÷t
−−→
a=
v − v0
t
Problem 15. For the following pair of equations, find x and y:
2x + y = 4
3x + 2y = 12
What is the product xy?
*(a) xy = −48
(c)
xy = 48
(e)
None of these
(b) xy = −60
(d) xy = 60
Copyright ©Wayne Hacker 2010. All rights reserved. 6
physics 121 sample exam 1 solns
Solution: There are two different approaches that we can use to solve this. We’ll go
through both.
The first approach is to solve one of the equations for one of the variables, then to
substitute that expression into the other equation. In this case, since y has a coefficient
of 1 in the first equation, it’s easy to solve it for y:
−2x
−−→
2x + y = 4
y = −2x + 4
Now substitute this into the second equation:
3x + 2y = 12
y=−2x+4
−−−−−→
3x + 2(−2x + 4) = 12
−8
−→
−x = 4
×(−1)
−−−→
simplify
−−−−→
−x + 8 = 12
x = −4
Now that we know x, we can substitute into the expression for y:
y = −2x + 4 = −2(−4) + 4 = 12
It’s always a good idea to check your solutions in the original equations:
2x + y = 2(−4) + 12 = 4
3x + 2y = 3(−4) + 2(12) = 12
Then xy = (−4)(12) = −48.
The second approach is to multiply one or both of the equations by constants so that the
coefficient of one of the variables is the same in both equations (up to sign); then add
or subtract the two equations to eliminate that variable. In this case, if we multiply the
first equation by 2, we can eliminate y.
×2
−→
−→
2x + y = 4
3x + 2y = 12
subtract
−−−−→
4x + 2y = 8
3x + 2y = 12
x + 0y = −4
⇒
x = −4
Now we can substitute this value into either of the original two equations and solve for
y. Since the coefficient of y in the first equation is 1, it’s easiest to use that one:
2x + y = 4
−2x
−−→
y = −2x + 4 = −2(−4) + 4 = 12
physics 121 sample exam 1 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 7
Problem 16. Solve the equation: x2 + 8x + 15 = 0. There are two solutions, x1 and x2 ,
with x1 ≥ x2 . (It is possible that x1 = x2 .) What is the difference x1 − x2 ?
(a) x1 − x2 = 0
(c) x1 − x2 = 7
(e) None of these
*(b) x1 − x2 = 2
(d) x1 − x2 = 8
Solution: We can solve this by factoring, then by setting each factor equal to zero.
Factor:
Set factors equal to zero:
Solve:
Answer:
x2 + 8x + 15 = 0
(x + 3)(x + 5) = 0
x + 3 = 0 or x + 5 = 0
x1 = −3 and x2 = −5
x1 − x2 = −3 − (−5) = 2
Problem 17. Solve the equation: 3x2 = 4x + 1
1
1
(a) x = or x = 1
(b) x = − or x = −1
3
3
√
√
2± 7
−4 ± 2
*(c) x =
(d) x =
3
6
(e) No real solution
Solution: Before we can apply the quadratic formula, we need the equation to be in the
form: ax2 + bx + c = 0.
3x2 = 4x + 1
−4x−1
−−−−→
3x2 − 4x − 1 = 0
Then the quadratic formula gives
p
√
√
√
4 ± (−4)2 − 4(3)(−1)
−b ± b2 − 4ac
4 ± 28
2± 7
x=
=
=
=
2a
2(3)
6
3
physics 121 sample exam 1 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 8
Problem 18. An airplane flies at a speed of 80 miles per hour in still air. On a day
when the wind is blowing from the north at 20 miles per hour, the airplane flies 200 miles
straight north, then turns around and returns to its starting point. What is its average
speed on the round trip?
(a)
64 miles per hour
(b)
66 23 miles per hour
*(c)
75 miles per hour
(d)
80 miles per hour
(e)
None of these
Solution: The average speed for the round trip is the total distance divided by the total
time. The total distance is 2 × 200 = 400 miles. The total time is
200
200 200
10
16
200
+
=
+
=
+2=
hours
80 − 20 80 + 20
60
100
3
3
Hence the average speed is
400 miles
3
= 75 miles/hour
= 400 ·
16
16
hours
3
Graphs
Problem 19. The graph at right shows four points
labelled with letters. The points are
y
6
Ar
rB
(3, 3), (2, −5), (−6, 5), and (−4, −2).
-
Which of the four points is C?
(a) (3, 3)
(c) (−6, 5)
(b) (2, −5)
*(d) (−4, −2)
C
x
r
r
D
Solution: Each of the four points is in a different quadrant, so we can use that to
identify them. Point A is in the second quadrant, with a negative x-coordinate and a
positive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- and
y-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; its
x- and y-coordinates are both negative. It must be (−4, −2). Point D is in the fourth
quadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2, −5).
Copyright ©Wayne Hacker 2010. All rights reserved. 9
physics 121 sample exam 1 solns
Problem 20. The four graphs (a), (b), (c), and (d) below represent four different equations:
y =x−1
y =x+1
y = −x + 1
y = −x − 1
Which of the four graphs represents y = x + 1?
(a) @ y 6
(b)
y6
(c)
y6
*(d)
y6
@
@
@
@
-
@x
@
@
@
@
-
x
@
@
-
-
x
x
@
@
@
@
Solution: The equation y = x + 1 is written in the form: y = mx + b. Here m = 1 > 0
and b = 1 > 0. Since m > 0, the graph should have a positive slope (as x increases, y
increases). Since b > 0, the graph should have a positive y-intercept (when x = 0, y > 0).
Only graph (d) slopes upward and crosses the y-axis above the x-axis.
Problem 21. Which equation is shown on the graph
at right?
(a)
y6
y = x2 + 1
*(b) y = x2 − 1
-
x
2
(c)
y = −x + 1
(d)
y = −x2 − 1
Solution: The graph has a negative y-intercept: it crosses the y-axis below the x-axis.
This means that when x = 0, y < 0. That allows us to rule out equations (a) and (c).
In the graph, when x has large absolute values, y > 0. This is consistent with (b), where
the coefficient of x2 is positive; but not with (d), where the coefficient of x2 is negative.