Practice Exam 1 (with solutions)
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. The quantity with the same units as force times time, Ft, with dimensions MLT-1 is
a. mv
b. mvr
c. mv2r
d. ma
e. mv2/r
à a mv
Ft: ma*t : kg*(m/s2)*s = kg*m/s : ML/T
mv: kg*m/s : ML/T
____
2. A particle confined to motion along the x axis moves with constant acceleration from x = 2.0 m to x = 8.0 m
during a 2.5-s time interval. The velocity of the particle at x = 8.0 m is 2.8 m/s. What is the acceleration
during this time interval?
a. 0.48 m/s2
b. 0.32 m/s2
c. 0.64 m/s2
d. 0.80 m/s2
e. 0.57 m/s2
à b 0.32 m/s2
x(t) = x0 +vi*t +(½)at2
8.0 = 2.0 +vi*2.5 +(½)a*2.52
vf = vi + at
2.8 = vi + a*2.5
vi = 8.0 – a*2.5
8.0 = 2.0 +(2.8 – a*2.5)*2.5 + (½)a*2.52
a*2.52 - (½)a*2.52 = 2.0 + 2.8*2.5 - 8.0
(½)a*2.52 = 1.0
a = 0.32 m/s2
____
3. A particle starts from rest at xi = 0 and moves for 10.0 s with an acceleration of +2.0 cm/s2. For the next 20.0
s, the acceleration of the particle is -1.0 cm/s2. What is the position of the particle at the end of this motion?
a. Zero
b. +3.0 m
c. -1.0 m
d. +2.0 m
e. -3.0 m
à b 3.0 m
x(t) = x0 +vi*t +(½)at2
x(10) = 0.0 +0.0*10 +(½)2.0*102 = 100 cm
vf = vi + at = 0.0 + 10*2.0 = 20 cm s
Starting from the above position
x(t) = x0 +vi*t +(½)at2
x(20) = 100 +20*20 +(½)(-1.0)*202 = 300 cm = 3.0 m
____
4. A rock is thrown downward from an unknown height above the ground with an initial speed of 10.0 m/s. It
strikes the ground 3.0 s later. Determine the initial height of the rock above the ground.
a. 44 m
b. 14 m
c. 74 m
d. 30 m
e. 60 m
à c 74 m
y(t) = y0 +vi*t +(½)at2
0 = y0 + -10.0*3.0 + (1/2) (-9.8)*3.02
y0 = 74.1 m
y0 = 74 m
____
5. The velocity at the midway point of a ball able to reach a height y when thrown with velocity vi at the origin
is:
a. vi/2
b. sqrt(vi2*2*gy)
c. sqrt(vi2/2)
d. Sqrt(vi2+gy)
e. gy
à c: sqrt(vi2/2)
The initial velocity is found using:
vf2 = vi2 +2*a*y
0 = vi2 – 2*g*y
vi = sqrt(2gy)
To get to ½y
vf2 = vi2 +2*a*½y
vf2 = 2gy - 2*g*½y = gy = ½ vi2
vf = sqrt(vi2/2)
____
6. Two children start at one end of a street, the origin, run to the other end, then head back. On the way back
Joan is ahead of Mike. Which statement is correct about the distances run and the displacements from the
origin?
a. Joan has run a greater distance and her displacement is greater than Mike's.
b. Mike has run a greater distance and his displacement is greater than Joan's.
c. Joan has run a greater distance, but her displacement is less than Mike's.
d. Mike has run a greater distance, but his displacement is less than Joan's.
e. Mike has run a shorter distance, and his displacement is less than Joan's.
à c: Joan has run a greater distance, but her displacement is less than Mike's.
Instructions: On occasion, the notation = [A, q] will be a shorthand notation for .
____
7. If
a.
b.
c.
d.
e.
= [10 m, 30°] and = [25 m, 130°], what is the direction of the sum of these two vectors?
17°
73°
107°
163°
100°
à c: 107°
Convert to Cartesian to add
[10*cos(30),10*sin(30)] + [25*cos(130),25*sin(130)]
tan(theta) = ((10*sin(30)+ 25*sin(130))/( ((10*cos(30)+ 25*cos(130)))
tan(theta) = 25.151/(-7.4094)
Be careful of the quadrant
theta = 107.06 degrees
theta = 107 degrees
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 direction of the resultant of these three forces?
a. 35°
b. 45°
c. 65°
d. 55°
e. 85°
à a: 35°
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]
tan(theta) = 13.706/19.451
theta = 35.170
theta = 35 degrees
____
9. A particle starts from the origin at t = 0 with a velocity of 6.0i m/s and moves in the xy plane with a constant
acceleration of (-2.0i + 4.0j) m/s2. At the instant the particle achieves its maximum positive x coordinate, how
far is it from the origin?
a. 36 m
b. 20 m
c. 45 m
d. 27 m
e. 37 m
à b 20 m
The maximum x coordinate will occur after 3 seconds when the particle reverses direction in x.
x(t) = x0 +vix*t +(½)at2
x(t) = 0.0 +6.0*3.0 +(½)(-2.0)3.02
x(t) = 18.0-9.0 = 9.0m
y(t) = y0 +viy*t +(½)at2
y(t) = 0.0 +0.0*3.0 +(½)4.0*3.02
y(t) = 18.0 m
r = sqrt(9.0*9.0 + 18.0*18.)
r = 20.125 m
r = 20 m
____ 10. A particle leaves the origin with a velocity of 7.2 m/s in the positive y direction and moves in the xy plane
with a constant acceleration of (3.0i - 2.0j) m/s2. At the instant the particle moves back across the x axis (y =
0), what is the value of its x coordinate?
a. 65 m
b. 91 m
c. 54 m
d. 78 m
e. 86 m
à d 78 m
Solve for time using the motion in the y coordinate
y(t) = y0 +viy*t +(½)at2
0.0 = 0.0 + 7.2*t + (½)(-2.0)*t2
7.2*t = t2
t = 7.2 s
then
x(t) = x0 +vix*t +(½)at2
x(t) = 0.0 +0.0*t +(½)(3.0)t2
x(t) = 77.76 m
x(t) = 78 m
____ 11. A rifle is aimed horizontally at the center of a large target 60 m away. The initial speed of the bullet is 240
m/s. What is the distance from the center of the target to the point where the bullet strikes the target?
a. 48 cm
b. 17 cm
c. 31 cm
d. 69 cm
e. 52 cm
à c 31 cm
Calculate the time to the target in the x dimension
t = d/v = 60/240 = 0.25s
then
y(t) = y0 +viy*t +(½)at2
y(t) = 0.0+0.0*t +(½)(-9.8)t2
y(t) = (½)(-9.8)0.252
y(t) = 0.30625 m
y(t) = 31 cm
____ 12. The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an
instant when the magnitude of the total acceleration is 6.0 m/s2, what is the speed of the particle?
a. 3.9 m/s
b. 2.9 m/s
c. 3.5 m/s
d. 3.0 m/s
e. 1.4 m/s
à b: 2.9 m/s
Using the sum of the tangential and centripetal acceleration
a c2 + a T 2 = a 2
ac = sqrt(6.02 – 4.42)
ac =4.0792 m/s2
then
ac = v2/r
v = sqrt(4.0792*2.0) = 2.8663 m/s
v = 2.9 m/s
____ 13. A 0.20-km wide river has a uniform flow speed of 4.0 m/s toward the east. It takes 20 s for a boat to cross the
river to a point directly north of its departure point on the south bank. In what direction must the boat be
pointed in order to accomplish this?
a. 23° west of north
b. 20° west of north
c. 24° west of north
d. 22° west of north
e. 17° west of north
à d 22° west of north
The velocity vector north, across the river is:
v = 200/20 = 10 m/s
The boat must be pointed 4 m/s west and 10 m/s north
theta = tan-1(4/10) = 21.801 degrees
theta = 22 degrees
____ 14. A student in the front of a school bus tosses a ball to another student in the back of the bus while the bus is
moving forward at constant velocity. The speed of the ball as seen by a stationary observer in the street:
a. is less than that observed inside the bus.
b. is the same as that observed inside the bus.
c. is greater than that observed inside the bus.
d. may be either greater or smaller than that observed inside the bus.
e. may be either greater, smaller, or equal to that observed inside the bus.
à e : may be either greater, smaller, or equal to that observed inside the bus.
a) Possible: a ball thrown backwards at just less than the speed of the bus would move slowly in the stationary observer’s
frame.
b) Possible: a ball thrown with ½ of the speed of the bus would move forward with ½ of the speed of the bus in the
stationary observer’s frame.
c) Possible: a ball thrown with very at a very slow speed would move with almost the velocity of the bus in the stationary
observer’s frame.
Therefore e
____ 15. In the figure, if the tension in string 1 is 34 N and the tension in string 2 is 24 N, what is the mass of the
object shown?
a.
b.
c.
d.
e.
7.3 kg
5.5 kg
1.8 kg
3.7 kg
4.5 kg
à d : 3.7 kg
The left tension of string 1 needs to equal the right tension of string 2
sin(40)*34 = sin(theta)*24
theta = 65.591 degrees
Then add the two upward tensions
cos(40)*34 + cos(65.591)*24 = 35.963 = 36 N/9.8 = 3.6697 = 3.7 kg
____ 16. The tension in a string from which a 4.0-kg object is suspended in an elevator is equal to 44 N. What is the
acceleration of the elevator?
a. 11 m/s2 upward
b. 1.2 m/s2 upward
c. 1.2 m/s2 downward
d. 10 m/s2 upward
e. 2.4 m/s2 downward
à b 1.2 m/s2 upward
Consider the total force
44 N = 9.8*4.0 + a*4.0
a = 1.2 m/s2 upward
____ 17. A 2.0-kg block slides on a frictionless 15° inclined plane. A force acting parallel to the incline is applied to
the block. The acceleration of the block is 1.5 m/s2 down the incline. What is the applied force?
a. 8.1 N down the incline
b. 3.0 N down the incline
c. 2.1 N up the incline
d. 3.0 N up the incline
e. 8.1 N up the incline
à c 2.1 N up the include
a = FNet /m
-1.5 = (-9.8*2.0*sin(15) + F)/2.0
F = -1.5*2.0 + +9.8*2.0*sin(15) = 2.0729 N (opposite sign of gravity) = 2.1 N up the include
____ 18. A constant force F is applied to a body of mass m that initially is headed east at velocity v0 until its velocity
becomes -v0. The total time of travel is 2t. The total distance the body travels in that time is
a. ½ Ft2/m.
b. Ft2/m.
c. v0t- ½ Ft2/m.
d. v0t- ½ Ft2/m.
e. 2(v0t- ½ Ft2/m).
à b Ft2/m
The object is going to travel in one direction reverse direction and come back to the starting place. Calculating half the
distance.
Using
x(t) = x0 +vix*t +(½)at2
x(t) = 0.0 +v0*t +(½)at2
x(t) = 0.0 +v0*t - (½)(Ft2/m)
but
x(2t) 0.0 = 0.0 +v0*2t - (½)(4Ft2/m)
so v0*t = Ft2/m
d(2t) = 2(v0*t - (½)(Ft2/m))
d(2t) = 2(Ft2/m - (½)(Ft2/m))
d(2t) = Ft2/m
____ 19. The block shown is pulled across the horizontal surface at a constant speed by the force shown. If M = 5.0 kg,
F = 14 N and q = 35°, what is the coefficient of kinetic friction between the block and the horizontal surface?
a.
b.
c.
d.
e.
0.44
0.33
0.38
0.28
0.17
à d: 0.28
The force to the right and the friction force must equal
F cos(theta) = mu*N = mu*(g*m-Fsin(theta))
mu = F cos(theta)/ (g*m-Fsin(theta)) = 14*cos(35)/(9.8*5.0-14*sin(35))
mu = 0.27991 = 0.28
____ 20. In the figure, the coefficient of kinetic friction between the surface and the larger block is 0.20, and the
coefficient of kinetic friction between the surface and the smaller block is 0.30. If F = 14 N and M = 1.0 kg,
what is the magnitude of the acceleration of either block?
a.
b.
c.
d.
e.
2.0 m/s2
1.3 m/s2
1.5 m/s2
1.8 m/s2
3.5 m/s2
à b 1.3 m/s2
Using a = FNET/m
a = (F – FfM –Ff3M)
a = (14 – 0.3*9.8*1.0 – 0.2*9.8*3.0)/4.0
a = 1.295 = 1.3 m/s2
Practice Exam 1
Answer Section
MULTIPLE CHOICE
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
A
B
B
C
C
C
C
A
B
D
C
B
D
E
D
B
C
B
D
B