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1. The brightest star in the night sky is α Canis Majoris, also known as Sirius. It lies 8.8 light-years away. Express this distance in meters. (A light-year is the distance covered by light in one year. Light travels in vacuum at 3 × 10
8
m/s). a. 2.6 × 10
9 b. 7.2 × 10
m
10
m c. 7.1 × 10 d. 8.3 × 10
12
16
m
m e. 3.8 × 10
20
m
8.8 light years = 8.3 10 m
1 year 1 day 1 h 1 s
2. Let
G
A = + 5 k
ˆ and
G
B = − + 2 ˆ j + α k
ˆ
. Find the value of scalar α that makes these two vectors
α perpendicular to each other. a. α = 0.3 b. α = 0.6 c. α = 0.9 d. α = 1.2 e. α = 1.7
Perpendicular
G G
A B 3 0 5
⇒
α = 0
A B 0
= 0.6
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3. A box has an initial speed v
0
up a frictionless incline. The box moves up the incline and then returns to its initial position, and continues to move down the incline. Which of the following velocity versus time graphs corresponds to this motion?
v
v v
A
v
t
B t
v t
C t t
D
E
The box goes from positive velocity to zero velocity (for an instant) and then to negative velocity. also, the slope of the v(t) graph should be constant (constant acceleration).
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The situation below refers to the next two questions:
A particle moves along the trajectory shown below at constant speed. The positions of the particle at t = 0 and at t = 1 s are shown. t = 1 s t = 0
4.
Which of these vectors best describes the direction of the average velocity of the particle between t = 0 and t = 1 s? a. / b. 2 c.
d.
e.
The average velocity points in the same direction as ∆ r : t = 1 s
∆ r t = 0
5.
Which of these vectors best describes the direction of the instantaneous acceleration of the particle at t = 0? a. / b. 2 c.
d.
e.
The instantaneous acceleration is the centripetal acceleration (speed is constant), so its direction is perpendicular to trajectory (or to the instantaneous velocity at that point) and pointing towards the center of curvature.
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The situation below refers to the next two questions:
You are standing on the edge of a cliff of height h . You throw a ball at an angle θ = 30° above the horizontal with an initial speed v
0
= 54 m/s. Using a stop watch, you determine that the ball strikes the ground at the base of the cliff 7 s after release.
θ h
D
6. What is the horizontal distance D covered by the ball? a. D = 203 m b. D = 282 m c. D = 327 m d. D = 482 m e. D = 564 m
=
0 x
=
0 cos θ t =
7.
At what time t after the release does the ball reach its maximum height? a. t = 1.9 s b. t = 2.8 s c. t = 3.5 s d. t = 4.1 s e. t = 4.8 s
Condition for maximum height: v y
= 0
0 = v
0 y
+ a t t
0 = v
0 sin θ
= v
0 sin θ g
− gt
=
(54 m/s)sin 30 D
9.8 m/s 2
= 2.8 s https://webct.ait.iastate.edu/PHYS221/Exam%20solutions%20Summer%202003/Exam1a...
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The situation below refers to the next two questions:
A tennis ball of weight W = 0.5 N is attached to a massless rope and swung in a vertical circle. The rope is L = 2 m long.
L g
8 . What is the minimum speed of the ball at the highest point of the trajectory if the rope is to stay straight?
= 1.0 m/s a. v min b. v min
= 4.4 m/s
= 8.8 m/s c. v min d. v min
= 19.6 m/s
= 79.3 m/s e. v min
This is the free-body diagram of the ball on top of its trajectory:
T
W v 2
The minimum speed is given by the condition T = 0: v 2 min
L v min
= Lg =
L
4.4 m/s https://webct.ait.iastate.edu/PHYS221/Exam%20solutions%20Summer%202003/Exam1a...
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9. Which of the following is the correct expression for the tension in the string T when the ball is at the lowest point? a. T = 0 b. c. T = 2 W d.

1 + v 2 gL

 e.


1 − v 2 gL


This is the free-body diagram at the bottom of the trajectory:
T
W
T W = m v 2
L
T = m v
L
2
+ W = m

 v 2 gL
+ W

 https://webct.ait.iastate.edu/PHYS221/Exam%20solutions%20Summer%202003/Exam1a...
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10. Two spaceships A and B move toward one another as shown. At time t = 0, spaceship A launches a shuttle craft toward spaceship B. At time t = t f
, the shuttle reaches spaceship B. t = 0
A d
0
B
Shuttle t = t f
A d f
B
Shuttle
The average speed of the shuttle relative to spaceship A is: a.
v = d t f f b. v = d
0 t f
= d f
− d
0 t f c. v
= d f
+ d
0 t f d. v e. None of the above
At t =0, the shuttle is relative distance zero from spaceship A. At t = t f
.
, its distance to spaceship A is d f
.
So the displacement relative to A is ∆ x = d f v =
∆ x
∆ t
= d t f f https://webct.ait.iastate.edu/PHYS221/Exam%20solutions%20Summer%202003/Exam1a...
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The situation below refers to the next two questions:
Block 1 with mass m
1
= 1.0 kg and block 2 with mass m
2
= 1.6 kg sit atop a frictionless double inclined plane as shown. The two blocks are connected to one another through a massless string around an ideal pulley.
Direction of acceleration
N
1
T
T
N
2 m
1 m
2
W
1
W
2
θ
1
= 30° θ
2
= 60°
11.
What is the magnitude of the acceleration of the blocks? a. a = 0.35 m/s
2 b. a = 1.6 m/s
2 c. d. a a
= 2.5 m/s
= 3.0 m/s
2 e. a = 3.3 m/s
2
2
Use the free-body diagram above.
If the whole system m
1
+ m
2
is considered as one, internal forces can be ignored. The net force in the direction of motion is: sin θ
1
− sin θ
2
= ( m
1
+ ) m
1 sin θ
1
− m
2 m
1
+ m
2 sin θ
2 = 3.3 m/s 2
12.
Now block 1 is coated with a rough paint so the friction force between this block and the surface is not negligible (but not strong enough to prevent the blocks from moving). Compare the tension on the string before and after the coating. a. The tension stays the same. b. The tension increases only on the side of mass 1 (left side). c. The tension decreases only on the side of mass 1 (left side). d. The tension increases on both ends.
e. The tension decreases on both ends.
The system has now a smaller acceleration, so the net force on both blocks needs to be smaller. The forces on block 2 along the incline are the tension and the appropriate component of the weight: sin θ
2
− = . For a to be smaller, T must be larger. And the change applies to any point on the string, since this is a massless string.
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The situation below refers to the next two questions:
A box of weight W = 30 N is being pulled across a rough, horizontal surface with a force
T = 10 N exerted by a string that makes an angle θ = 45° with the horizontal. The box moves at a constant speed v = 3 m/s across the surface.
N
T
θ cos θ
13 . Which of the following expressions is correct for the magnitude of the normal force N exerted by the floor on the box? a. b. c.
d.
cos θ sin θ sin θ e.
W
Newton’s second law in the vertical direction, from the free-body diagram above: sin θ − W = ma y
= 0 sin θ
14.
What is the magnitude of the friction force between the box and the surface?
a. f
b. f k
= 3.4 N
c. f k d. f k
e. f k k
= 5.2 N
= 6.9 N
= 7.1 N
= 10 N
Newton’s second law in the horizontal direction, from the free-body diagram above:
T cos θ − f k
= ma x
= 0 f k
= T cos θ = 7.1 N https://webct.ait.iastate.edu/PHYS221/Exam%20solutions%20Summer%202003/Exam1a...
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The situation below refers to the next two questions:
Blocks A and B accelerate on a frictionless horizontal surface while being pushed by your hand with a force F . The mass of block A is larger than the mass of block B.
F
A
B a. a
A
< a
B
F net,
15 . Compare the acceleration of each block and the net force on each block:
A
< F net, B b.
a
A
= a
B
F net, A
< F net, B c. a
A
= a
B
F net, A
= F net, B d. a
A
> a
B
F net, A
= F net, B e. a
A
= a
B
F net, A
> F net, B
The acceleration needs to be the same for both blocks, since they are moving together.
Net force is equal to mass times acceleration, so a greater net force is required for A.
16 . The force exerted by block B on block A is: a. 0 b. F = F c.
F = m
A m
A
+ m
B
F d.
F = m
B m
A
+ m
B
F e.
F = m m
B m
A
+ m
B
F
It is also equal (in magnitude) to the force by A on B, which is the only force on block B, so
F = F = F = = m
B m
A
+ m
B
F
(To find the acceleration, use the net force on the whole system: F = ( m
A
+ )
) https://webct.ait.iastate.edu/PHYS221/Exam%20solutions%20Summer%202003/Exam1a...
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17. Two boxes of masses m and 2 m are stacked as shown in the figure below on top of a horizontal frictionless table. The coefficient of static friction between the two boxes is µ s
. The bottom box is being pulled with a force F . Find the maximum value of F for which the boxes will move together.
µ s f s bottom by top m f s top by bottom
F
2 m frictionless a. F max
= 3 µ s mg b. F max
= µ s mg c.
F max d.
F max
=
=
3
2
µ s mg
1
2
µ s mg
F max
3 e.
F max
=
1
3
µ s mg
The only horizontal force on the top box is the static friction f s
, so by F = 3 ma , so the static friction on the top box needs to be f
S
=
F
3 f
S
= ma . The acceleration is given
Static friction has a maximum magnitude f
S
= µ
S
N = µ
S mg , so the maximum applied force is:
= µ
S mg https://webct.ait.iastate.edu/PHYS221/Exam%20solutions%20Summer%202003/Exam1a...
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