Question

Section 12.3 Newton’s Third Law of
Motion and Momentum
(pages 372–377)
This section describes action-reaction forces and how the momentum of
objects is determined.
Reading Strategy (page 372)
Summarizing As you read about momentum in this section, complete
the concept map to organize what you learn. For more information on
this Reading Strategy, see the Reading and Study Skills in the Skills
and Reference Handbook at the end of your textbook.
Newton’s Third Law (page 373)
1. According to Newton’s third law of motion, what happens whenever
one object exerts a force on a second object?
2. The equal and opposite forces described by Newton’s third law are
called
and
forces.
3. Circle the letters that identify each sentence that is true about
action-reaction forces.
a. Newton’s second law describes action-reaction forces.
b. Forces always exist in pairs.
c. Action-reaction forces never cancel.
d. All action-reaction forces produce motion.
4. Is the following statement true or false? Action-reaction forces do
not cancel each other because the action force is always greater than
the reaction force.
Momentum (pages 374–375)
5. Circle the letter of each factor that affects the momentum of a
moving object.
a. mass
b. volume
c. shape
d. velocity
6. If two identical objects are moving at different velocities, the object
that is moving faster will have
momentum.
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Physical Science Guided Reading and Study Workbook
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Chapter 12 107
Momentum
is measured
in
is calculated by
multiplying

7. Your in-line skates are sitting in a box on a shelf in the closet. What
is their momentum?
8. Is the following sentence true or false? An object with a small mass
can have a large momentum if the object is traveling at a high
speed.
9. Write the momentum formula, including the correct units.
10. Circle the letter of the object that has the greatest momentum.
a. a 700-gram bird flying at a velocity of 2.5 m/s
b. a 1000-kilogram car traveling at 5 m/s
c. a 40-kilogram shopping cart rolling along at 0.5 m/s
d. a 300-kilogram roller coaster car traveling at 25 m/s
Conservation of Momentum (pages 376–377)
11. What does conservation of momentum mean?
12. Is the following sentence true or false? Objects within a closed
system can exert forces on one another, but other objects and forces
cannot leave or enter the system.
13. According to the law of conservation of momentum, what happens
to the total momentum of a system if no net force acts on the system?
14. Is the following sentence true or false? In a closed system with two
objects, the loss of momentum of one object equals the gain in
momentum of the other object.
For questions 15 and 16, refer to the graph below.
15. The momentum of the ball at one second is
.
16. What is the speed of the ball at 0.5 seconds? Show your
calculation. Hint: Solve the momentum formula for velocity.
Name ___________________________
Class ___________________
Date _____________
Chapter 12
Forces and Motion
108
Physical Science Guided Reading and Study Workbook
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Chapter 12
Time (s)
Momentum (kg m/s)
4
3
2
1
0
0
0.5
1.0
1.5

Answer 1

Newton’s Third Law (page 373)
1. Whenever one object exerts a force on a second object, the second object exerts an equal-in-magnitude and opposite-in-direction force on the first object.
2. action and reaction
3. b and c
- b. Forces always exist in pairs. (true)
- c. Action–reaction forces never cancel. (true — they act on different objects)
- a and d are false.
4. False. Action–reaction forces do not cancel because they act on different objects, not because one is larger.

Momentum (pages 374–375)
5. a and d (mass and velocity)
6. greater
7. 0 kg·m/s (they are at rest)
8. True.
9. p = m v ; units: kg·m/s
10. d (300 kg × 25 m/s = 7500 kg·m/s)

Conservation of Momentum (pages 376–377)
11. The total momentum of a closed (isolated) system remains constant (is conserved) unless an external net force acts on the system.
12. True.
13. The total momentum remains the same (is conserved).
14. True.

Questions 15–16 (from the graph)
15. The momentum at 1.0 s is 2 kg·m/s.
16. At 0.5 s the momentum is 1 kg·m/s. Use v = p / m.
Example: if the ball’s mass is 0.50 kg, v = (1 kg·m/s) / (0.50 kg) = 2.0 m/s.