Physics 221
Name:
Quiz 1 — chapters 1–2, Form: A
Date:
G= 6.67 × 10−11 N m2 /kg2
c= 3 × 108 m/s
Mearth = 6.0 × 1024 kg
Rearth = 6.38 × 106 m
1. The NFL football season began last night with a game between the New England Patriots and Indianapolis Colts. It’s a good reminder that ALL motion is determined by the Momentum Principle (i.e.
Newton’s Second Law).
Figure 1:
The picture above shows a 0.4-kg football during a time interval of 0.2 s after it is thrown. The ball
is traveling from left to right in the picture. On the picture, sketch the initial position vector ~ri , the
final position vector ~rf and the displacement vector ∆~r for the football.
2. If the velocity of the football at the beginning of the time interval is < 20, 10, 0 > m/s and the velocity
of the football at the end of the time interval is < 19, 8, 0 > , what is the net force on the football
during this time interval? Calculate it, express it as a vector, and sketch the vector.
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3. If you travel around a loop on a roller coaster, you feel “heavy” at the bottom and you feel “light” at
the top. Consider the roller coaster shown below.
Figure 2:
Your momentum just before the dip and your momentum just after the dip are shown in the picture.
Sketch the direction of ∆~
p. (Note: you must construct it correctly using the vectors shown. You may
not simply guess and draw a vector.)
4. A space probe is moving in the −y direction when a thruster briefly fires and exerts a force on the
probe in the +x direction. After a time interval ∆t, the thruster stops firing. Which path below is
the path of the space probe before, during, and after the thruster fired. (Circle or otherwise mark the
correct path.)
Figure 3:
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5. At a certain instant of time t, a star is located at < 6 × 1010 , 9 × 1010 , 8 × 1010 > m and a planet is
located at < −8 × 1010 , −4 × 1010 , 8 × 1010 > m. What is the vector pointing from the planet to the
star?
6. What is the distance |~r| from the star to the planet?
7. What is the unit vector from the planet to the star, r̂?
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8. Suppose that the star is 1000 times more massive than the planet and at this distance, the magnitude
of the force of the star on the planet is |F~ |. What is the magnitude of the force of the planet on the
star?
(a)
|F~ |
(b)
(c)
1000|F~ |
|F~ |/1000
(d)
|F~ | + 1000
(e)
none of the above
9. Suppose that the planet is in an elliptical orbit around the star. When it is furthest from the star,
the gravitational force on the planet is |F~ | and its distance from the star is |~r|. When the planet is
nearest to the star, it is a distance 1/3|~r| from the star. What is the gravitational force of the star on
the planet when it is nearest to the star, compared to when it is furthest from the star?
(a)
(b)
|F~ |
3|F~ |
(d)
9|F~ |
|F~ |/3
(e)
|F~ |/9
(c)
10. GPS satellites are used to precisely determine positions on the surface of the Earth. If a 2030-kg GPS
satellite is in orbit at an altitude of 20,000 km (convert this to m!), what is the gravitational force of
the Earth on the satellite at this altitude?
11. When the GPS satellite was sitting on Earth, before it was launched into orbit, what was the gravitational force on the satellite?
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Answer Key for Exam A
Figure 4:
1.
2.
∆~
p
F~net =
∆t
0.4 ∗ (< 19, 8, 0 > − < 20, 10, 0 >)
=< −2, −4, 0 > N
F~net =
0.2
Figure 5:
3.
4. see Figure 6
5.
~r = ~rs −~rp =< 6×1010 , 9×1010 , 8×1010 > − < −8×1010 , −4×1010 , 8×1010 >=< 14×1010 , 13×1010 , 0 > m
6.
|~r| = |~rp − ~rs =< 6 × 1010 , 9 × 1010 , 8 × 1010 > − < −8 × 1010 , −4 × 1010 , 8 × 1010 > |
p
= 14e102 + 13e102 = 1.91 × 1011 m
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Figure 6:
7.
r̂ = ~r/|~r|
= < 0.733, 0.681, 0 >
8. (a)
9. (c)
10.
r = Rearth + altitude = 6.38e6 + 2e7 = 26.38e6 m
Gm1 m2
r2
6.67e − 11(2030)(6e24)
=
= 1170 N
(26.38e6)2
|F~grav | =
11.
r = Rearth = 6.38e6 m
Gm1 m2
r2
6.67e − 11(2030)(6e24)
=
= 2.0e4 N
(6.38e6)2
|F~grav | =
Note that |F~grav | = m|~g | and near the surface of the earth |~g | = 10 N/kg. Thus,
|F~grav | = 2030 kg(10 N/kg) = 2.0e4 N
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