Exam 1
Instructions
Student ID Number: ______________________
Section Number: ______________________
TA Name: ______________________
Please fill in all the information above.
Please write and bubble your Name and Student Id number on your scatron. Also fill in your section number
under special codes. Finally question 1 asks you to fill in your test ID code which is necessary to correctly
grade your exam.
The exam has 20 questions (questions 2-21) and you have 1 hour and 15 minutes to complete the exam.
Please use 9.8 m/s^2 for the acceleration of gravity on the Earth.
.
____
1. Please fill in your test ID code
a. A
b. B
c. C
d. D
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
2. One number has three significant figures and another number has four significant figures. If these numbers are
added, subtracted, multiplied, or divided, which operation can produce the greatest number of significant
figures?
a. the addition
b. the subtraction
c. the multiplication
d. the division
e. All the operations result in the same number of significant figures.
à a: the addition and b: subtraction
Example: 100.1+10.2 = 110.3
Example: 100.1-(-10.2) = 110.3
Multiplication and division would result in three significant figures.
____
3. Vx is the velocity of a particle moving along the x axis as shown. If x = 2.0 m at t = 1.0 s, what is the position
of the particle at t = 6.0 s?
a.
b.
c.
d.
e.
-2.0 m
+2.0 m
+1.0 m
-1.0 m
6.0 m
à d: -1.0 m
Average velocity from 1.0-3.0s is 0.0 m/s
Average velocity from 3.0-6.0 seconds -1.0m/s
Thus 2.0 + 0.0*2.0 + -1.0*3.0 = -1.0 m
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4. A rocket, initially at rest, is fired vertically with an upward acceleration of 10.0 m/s2. At an altitude of 0.50
km, the engine of the rocket cuts off. What is the maximum altitude it achieves?
a. 1.9 km
b. 1.3 km
c. 1.6 km
d. 1.0 km
e. 2.1 km
à d 1.0 km
Using:
y(t) = y0 +viy*t +(½)at2
500 = 0.0 + 0.0*t + ½ 10.0 * t2
t = 10 s
v(t=10) = 100m/s
Using:
vf2 = vi2 +2*a*y
0 = 10000+2(-9.8)y
y = 510 m
total 1010m = 1.0 km
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5. Two identical balls are at rest side by side at the bottom of a hill. Some time after ball A is kicked up the hill,
ball B is given a kick up the hill. Ball A is headed downhill when it passes ball B headed up the hill. At the
instant when ball A passes ball B,
a. it has the same position and velocity as ball B.
b. it has the same position and acceleration as ball B.
c. it has the same velocity and acceleration as ball B.
d. it has the same displacement and velocity as ball B.
e. it has the same position, displacement and velocity as ball B.
àb
It’s position, displacement and acceleration are the same.
____
6. A juggler throws two balls to the same height so that one is at the halfway point going up when the other is at
the halfway point coming down. At that point:
a. Their velocities and accelerations are equal.
b. Their velocities are equal but their accelerations are equal and opposite.
c. Their accelerations are equal but their velocities are equal and opposite.
d. Their velocities and accelerations are both equal and opposite.
e. Their velocities are equal to their accelerations.
àc
Because the motion is symmetric the velocities will be equal (but opposite) when they pass each other.
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7. A particle is moving at constant velocity. Its position at t = 1.0 s is 3.0 m and its position at t = 4.0 s is 15.0 m.
What is the slope of the position-time graph for this particle?
a. 0, since this is a constant velocity situation.
b. 4.0 m/s
c. 4.0 m/s2
d. 9.0 m/s
e. 12 m/s2
à b: 4.0 m/s
The slope gives average velocity (15.0-3.0)/(4.0-1.0) = 4.0 m/s
Instructions: On occasion, the notation = [A, q] will be a shorthand notation for .
Exhibit 3-1
The three forces shown act on a particle.
Use this exhibit to answer the following question(s).
____
8. Refer to Exhibit 3-1. What is the magnitude of the resultant of these three forces? The angles are also known
to three digits of precision.
a. 27.0 N
b. 33.2 N
c. 36.3 N
d. 23.8 N
e. 105 N
à d 23.8 N
Convert to Cartesian coordinates to add
[65*cos(30),65*sin(30)] + [30*cos(180),30*sin(180)] + [20*cos(250),20*sin(250)]
[19.451,13.706]
r = sqrt(19.4512 + 13.7062) = 23.794 = 23.8 N
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9. At t = 0, a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction. Its acceleration is
given by (3.0i - 2.0j) m/s2. At the instant the particle reaches its maximum y coordinate how far is the
particle from the origin?
a. 11 m
b. 16 m
c. 22 m
d. 29 m
e. 19 m
à a 11 m
The maximum y coordinate will occur after 2.5 s when the particle reverses direction.
x(t) = x0 +vix*t +(½)at2
x(t) = 0.0 +0.0*2.5 +(½)(3.0)2.52
x(t) = 9.375m
y(t) = y0 +viy*t +(½)at2
y(t) = 0.0 +5.0*2.5 +(½)(-2.0)*2.52
y(t) = 6.25 m
r = sqrt(9.375*9.375 + 6.25*6.25)
r = 11.267 m
r = 11 m
____ 10. A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of
53.0° above the horizontal. The rock strikes the ground a horizontal distance of 25.0 m from the base of the
building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?
a. 25.3 m
b. 29.6 m
c. 27.4 m
d. 23.6 m
e. 18.9 m
à d: 23.6 m
v0x = 12.2 cos(53) = 7.3421 m/s
v0y = 12.2 sin(53) = 9.7434 m/s
t = d/v = 25.0/12.2 cos(53) = 3.4050 s
y(t) = y0 +viy*t +(½)at2
0.0 = y0 +12.2*sin(53)*3.4050 +(½)(-9.8)3.40502
y0 = 23.635 m = 23.6 m
____ 11. A particle moves along a circular path having a radius of 2.0 m. At an instant when the speed of the particle is
equal to 3.0 m/s and changing at the rate of 5.0 m/s2, what is the magnitude of the total acceleration of the
particle?
a. 7.5 m/s2
b. 6.0 m/s2
c. 5.4 m/s2
d. 6.7 m/s2
e. 4.5 m/s2
à d: 6.7 m/s2
Find the centripetal acceleration
ac = v2/r = 3.02/2.0 = 4.5 m/s2
Total acceleration
a = sqrt(4.5*4.5 + 5.0*5.0) = 6.7268 = 6.7 m/s2
____ 12. A 0.14-km wide river flows with a uniform speed of 4.0 m/s toward the east. It takes 20.0 s for a boat to cross
the river to a point directly north of its departure point on the south bank. What is the speed of the boat
relative to the water?
a. 5.7 m/s
b. 8.5 m/s
c. 8.1 m/s
d. 7.0 m/s
e. 6.4 m/s
à c: 8.1 m/s
To move directly a point north of the departure point the swimmer must orient themselves partially up river to the north.
west component: 4m/s
north component: v = d/t = 140/20 = 7 m/s
s= |v| = sqrt (4*4+7*7) = 8.0623 = 8.1 m/s
____ 13. A river has a steady speed of 0.30 m/s. A student swims downstream a distance of 1.2 km and returns to the
starting point. If the student swims with respect to the water at a constant speed and the downstream portion
of the swim requires 20 minutes, how much time is required for the entire swim?
a. 50 minutes
b. 80 minutes
c. 90 minutes
d. 70 minutes
e. 60 minutes
à d: 70 minutes
velocity with respect to stationary observer downstream: v = 1200/(20*60) = 1.0 m/s
velocity with respect to river v = 1.0- 0.30 = 0.7
velocity with respect to stationary observer upstream: v = 0.70-0.30 = 0.4
time to swim upriver t = 1200/0.4 = 3000 s = 50 minutes
total time 70 minute
____ 14. A tennis player wants to slam a serve at O so that the ball lands just inside the opposite corner of the court.
What should the ratio v0y/v0x be for the initial velocity ? The time t = 0 is the time when the ball is hit by the
racket.
a.
b.
c.
d.
e.
W/L
L/W
½ gt2/L
½ gt2/W
½ gt2/sqrt(L2+W2)
à b L/W
You have to read the question carefully!. x and y coordinates are along the two sides of the court. You would also have
to hit the ball at the correct downward velocity to hit the opposite side of the court but you are not asked about
that coordinate.
v0y = L/t
v0x = W/t
v0y/v0x = L/W
____ 15. If a = 40.0° and the tension in string 2 is 30 N, determine M.
a.
b.
c.
d.
e.
3.4 kg
3.6 kg
2.6 kg
4.9 kg
7.5 kg
à c 2.6 kg
If string 2 has a 30N tension then both strings 1 and 3 have a 30 N tension in the x direction
Tan(theta) = y/x
y = tan(theta)x = tan(40)*30 = 25.172
times two since both strings 1 and 3 pull up
a = F/g = 50.345/9.8 = 5.1373
divided by two since there are two weights
2.5686 = 2.6 kg
____ 16. The only two forces acting on a body have magnitudes of 20.0 N and 35 N and directions that differ by 80.0°.
The resulting acceleration has a magnitude of 20.0 m/s2. What is the mass of the body?
a. 2.4 kg
b. 2.2 kg
c. 2.7 kg
d. 3.1 kg
e. 1.5 kg
à b 2.2 kg
Assume one force at 0 degrees and the other at 80
F = (20,0) + (35 cos(80),35 sin(80)) = (20,0) + (6.0777, 34.468) = (26.078, 34.468)
|F| = sqrt(26.078^2+34.468^2) = 43.222 N
m = |F|/a = 2.1611 = 2.2 kg
____ 17. A 5.0-kg mass is suspended by a string from the ceiling of an elevator that is moving upward with a speed
which is decreasing at a constant rate of 2.0 m/s in each second. What is the tension in the string supporting
the mass?
a. 49 N
b. 39 N
c. 59 N
d. 10 N
e. 42 N
à b: 39 N
Upward acceleration which reduces tension -2.0 m/s2
Downward acceleration which increases tension 9.8 m/s2
9.8 – 2.0 = 7.8 m/s2
T = F = ma = 7.8 * 5.0 = 39 N
____ 18. The total force needed to drag a box at constant speed across a surface with coefficient of kinetic friction µk is
least when the force is applied at an angle q such that
a. sinq = µk.
b. cosq = µk.
c. tanq = µk.
d. cotq = µk.
e. secq = µk.
à c: tanq = µk.
F = µk N
Fcos(theta) = µk(mg-Fsin(theta))
Fcos(theta) + µk Fsin(theta) = µkmg
F = µkmg/( cos(theta)+ µk sin(theta))
Take the derivative with respect to theta and set to zero to fine the extreemum
µkmg(-sin(theta) + µk cos(theta))/ ( cos(theta)+ µk sin(theta))2 = 0
(-sin(theta) + µk cos(theta)) = 0
µk cos(theta) = sin(theta)
tan(theta) = µk
____ 19. A 4.0-kg block slides down a 35° incline at a constant speed when a 16-N force is applied acting up and
parallel to the incline. What is the coefficient of kinetic friction between the block and the surface of the
incline?
a. 0.20
b. 0.23
c. 0.26
d. 0.33
e. 0.41
à a: 0.20
The forces up and down the incline must be equal
FNET = 0 = F - mgsin(theta) + µkmgcos(theta)
µk = ((mgsin(theta) –F)/( mgcos(theta)) = (4.0*9.8*sin(35)-16)/(4.0*9.8*cos(35)) = 0.20193 = 0.20
____ 20. In order to jump off the floor, the floor must exert a force on you
a. in the direction of and equal to your weight.
b. opposite to and equal to your weight.
c. in the direction of and less than your weight.
d. opposite to and less than your weight.
e. opposite to and greater than your weight.
à e: opposite to and greater than your weight
larger than your weight to accelerate upwards
____ 21. As in Lab M 4, Acceleration in free fall, if you dropped a tape through a spark machine. If the tape falls 1.0m
and the spark machine sparks at a rate of 30.0Hz about how many spark marks could there be on the tape.
a. 6
d. 30
b. 27
e. 20
c. 13
à c 13
To fall 1 m
y(t) = y0 +viy*t +(½)at2
-1 = (½)(-9.8)t2
t = 0.45175 sec, times 30 Hz (1/s): 13.553. In that time there may be 13 or 14 marks made depending on where first
spark occurs compared to t0
Exam 1
Answer Section
MULTIPLE RESPONSE
1. ANS: A
MULTIPLE CHOICE
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