= ′
= −
=
2 cart unknown mass
But m = m +m , therefore
unknown mass 2 cart
unknown mass
0.75 kg 0.50 kg 0.25 kg
m m m
m
= −
= −
=
Evaluation
(i) The predictions and hypothesis were correct to within experimental errors.
(j) Sources of error during the experiment include:
•
There was friction between the cart and the track.•
There was friction between the tape and the timing recorder.•
It is difficult to measure the exact distance between the ticker tape timer dots because the distance is so small.Synthesis
(k) It is important to distinguish between scalar and vector quantities because a system momentum of zero may still involve dangerous speeds or motions.
(l) Using electronic values, the refuse bag could be tossed to a stationary astronaut. After catching the bag, the velocities and known mass of the astronaut could be used to calculate the mass of the refuse bag.
(m) The friction pads would provide a source of external force t o the system. This would increase the loss of kinetic energy before and after the collision and change the total momentum before and after as well.
Investigation 5.3.1: Analyzing Two-Dimensional Collisions
(Pages 262–265)
Question
The laws of conservation of momentum and kinetic energy can be verified through two-dimensional collisions.
Prediction
(a) The kinetic energy and the total momentum will be the same before and after the collisions.
(b) The change in momentum for each puck will be the same as the change in momentum of the other puck.
Hypothesis
(c) I expect that the two pucks will stick together and move along a straight path that will bisect the initial angle of collision.
(d) The change in momentum of puck A will be equal to the change in momentum of puck B. The velocities can me measured, to the mass can be calculated.
(e) Category I and Category II
A B A B
A B A B
A B A B
p p p p
mv mv mv mv v v v v
′ ′
+ = +
′ ′
+ = +
′ ′
+ = +
Category III
(f) Mark off the four markers to get a known distance over a known time (the distance between each pair of dots should be approximately the same). Measure the angles. The sum of the two velocity vectors before should be equal to the sum of the velocity vectors after for Category I.
(g) For category II, the change in each vector should be the same magnitude and opposite direction.
(h) Category III should have the two velocity vectors before be equal to the twice the combined after velocity.
(i) For Category IV, subtract the two velocity vectors for each puck and substitute the values into the above equation.
(j) Measure each length of velocity vectors before and after. Substitute with the mass into the kinetic energy equation and compare the total values before and after. Any lost energy went into thermal energy.
Evaluation
(l) Some possible sources of error:
•
The table surface may not be flat.•
There may be some frictional or air current forces acting on the pucks.•
The air supply line to the pucks may exert some lateral forces on the pucks.•
Masses of the pucks may not be identical.•
Frequency of the spark timer may not be perfectly uniform.Synthesis
(m) This would apply to all of the categories because all of them have the total momentum conserved.
(n) The momentum would only be conserved if you considered both pucks.
(o) It is wish to avoid dots during the impulse and collision because there is a changing speed during these interactions. The analysis of this experiment is not designed to include a puck that is accelerating.
(p) It is better to have steel barriers because they will absorb some energy and prevent the cars from bouncing off as much as a rubber barrier would.
CHAPTER 5 SELF QUIZ
(Pages 267–268)
True/False
1. F The impulse is the same in magnitude and is in the same direction.
2. T
3. F You have increased the amount of time the same force is applied.
4. F The total momentum before and after is the same.
5. T 6. T
7. F The kinetic energies will be different.
8. F The momentum is conserved in the snowball-earth system, even though the collision is completely inelastic.
9. T
10. F The final kinetic energy can be greater (e.g., an exploding cart system).
Multiple Choice
11. (e)
12. (d) Momentum will increase by 2 × 2, and kinetic enegy by 2 × 22. 13. (d)
14. (d) 15. (e)
16. (d) Some energy is lost to thermal energy of the collision.
17. (b) The momentum is p
=
2 mE K so p∝
E K .18. (c) M only stops, while T must continue to act on R to give it momentum in the opposite direction.
19. (a) They have both stopped, so the total momentum is zero, but the initial kinetic energy is stored as electric potential.
20. (d)
21. (a) 1
(2 ) ( ) ( 2 )
mv
+
m
2− = +
v
m m v′
3 0
mv mv
v m
v
′ = −
′ =
22.(d) 0
=
mv M m V−
(−
) V mvM m
= −
CHAPTER 5 REVIEW
(Pages 269–271)
Understanding Concepts
1. Impulse is the product of the force and the time. A smaller force exerted over a long period of time can impart a larger impulse than a large force for a short time.
2. As the meteor comes to a stop, the kinetic energy is converted into thermal energy which melts the material at the impact site.
3. This observation does not contradict the law of conservation of momentum. The momentum of the earth increases toward the falling object. The amount of change in the earth’s velocity is so small it is imperceptible, so it appears the law of conservation is being contradicted, but it is not.
4. The change in momentum of the ball that bounces is greater than the putty that sticks to the floor. Assuming both have the same mass and are dropped from the same height:
For the putty, For the ball
2 1
1
1
( )
(0 ) p m v v
m v
p mv
∆ = −
= −
∆ = −
2 1
1 1
1
( )
(( ) )
2 p m v v
m v v p mv
∆ = −
= − −
∆ = −
5.
2
K 2
E
=
mv2 2
2 K
2
2 m v
m E p
m
=
=
The momentum of the two players is the same, but the person with the larger mass will have a smaller kinetic energy. It would be better to avoid the faster moving lighter player, p2 = m2v2.
6. A car crashing into a tree is an example. Momentum is not conserved in the tree-car system because the tree is attached to the earth. The earth supplies a net external force to the system of objects so that momentum is not conserved.
7. Diagram is not to scale.
2
N
E
1.3 10 kg
8.7 m/s [44 E of N]
?
? m v p p
= ×
= °
=
=
(1.3 10 kg)(8.7 m/s)2
1131 kg m/s [44 E of N]
p mv
p
=
= ×
= ⋅ °
N
2 N
cos44
(1131 kg m/s)(cos44 ) 8.1 10 kg m/s p p
p
= °
= ⋅ °
= × ⋅
E
2 E
sin44
(1131 kg m/s)(sin 44 ) 7.9 10 kg m/s p p
p
= °
= ⋅ °
= × ⋅
The northward component of the boat’s momentum is 8.1 10 kg m/s
×
2⋅
. The eastward component is 7.9 10 kg m/s×
2⋅
. 8. Diagram is not to scale.4 E
3
2.6 10 kg m/s 1.1 10 kg 22
? p
m
v
θ
= × ⋅
= ×
= °
=
E
The car is travelling at a speed of 25 m/s.
9. Choose west as positive.
3
The velocity of the ball just as it leaves the racket is 28 m/s [fwd].
11. m1
=
0.112 kg1 1 2 2
(0.112 kg)(1.38 m/s) 0.154 kg
The speed of the second car is 1.00 m/s.
12. mP
=
1.67 10 kg×
−27The speed of the alpha particle is 0.620 km/s.
13. m1
=
2.67 kg1.70 10 m/s [toward Jupiter]
185 m/s [toward Jupiter]
183 m/s [toward Jupiter]
?
Choose the direction toward Jupiter as positive.
1 1 2 2 1 1 2 2
1.90 10 m/s [toward Jupiter]
m v m v m v m v
The velocity of the more massive rock is 1.90 10 m/s [toward Jupiter]
×
2 .14. (a) v1i
=
6.0 m/sThe two objects have the same final velocity. Having the same final velocity is an indication of a ‘hit-and-stick’
collision. All collisions that have the objects with the same speed (in the same direction) after are completely inelastic collisions.
There is some loss of kinetic energy. This is an inelastic collision.
(c) v1i
=
12 m/sThere is no loss of kinetic energy, so this collision is elastic.
15. Choose north as positive.
Conservation of Energy Conservation of Momentum
2 2 2
0.253(1.80) 0.253 0.232
3.24 0.917 (Equation 1)
(0.253)(1.80) 0.253 0.232
1.80 0.917
Substitute Equation 2 into Equation 1:
2 2
Substitute back into Equation 2:
1
1
1.80 0.917(1.88) 0.08 m/s [N]
v v
′ = −
′ =
The velocity of the first cart is 0.08 m/s [N]. The velocity of the second cart is 1.88 m/s [N].
16. (a) Diagram not to scale.
(b)
B A
B A
B A
B A
B A
0 0
sin 46 sin 33 sin33 sin 46
y y
y y
y y
mv mv v v v v
v v
v v
= −
= −
=
° = °
= °
°
A1 A B
A1 A B
A B
A A
A
A
5.4 cos 33 cos 46 sin33
5.4 cos 33 cos 46
sin46 5.4 1.365
3.957 m/s, or 4.0 m/s
x x
x x
mv mv mv
v v v
v v
v v
v v
= +
= +
= ° + °
°
= ° + ° °
=
=
Substitute back into equation for vB :
B A
B
sin33 sin 46 sin33
(3.957 m/s) sin 46
3.0 m/s
v v
v
= °
°
= °
°
=
The speed of the first puck is 4.0 m/s after the collision. The speed of the second puck is 3.0 m/s after the collision.
17. θ A
=
67.8°
B
B A
B A
30.0 3.30
? v v m m
θ
= °
=
=
A B
A A B B
B B A A
B A
A B
B A
B A
0
0 sin 67.8 sin 30.0
sin 30.0 sin 67.8 sin 67.8 sin30.0 sin 67.8
(sin30.0 ) sin 67.8 (3.30)(sin30.0 ) 0.561
y y
p p
m v m v
m v m v
m v
m v
v v
m m
′ ′
= −
= ° − °
° = °
= °
°
= °
°
= °
°
=
The ratio of the masses of the particles is 0.561.
18. Diagrams not to scale.
Conserve momentum in the y direction.
sin 69.2 sin 62.8
(2.37 m/s)(sin 69.2 ) (2.49 m/s)(sin 62.8 )
Conserve momentum in the x direction.
1 2 1 2
cos 69.2 cos 62.8
(2.37 m/s)(cos69.2 ) (2.49 m/s)(cos62.8 ) 2.70 m/s 3.00 m/s [E]
There is no y component for velocity, so the unknown initial velocity is 3.00 m/s [W]
19. (a) g
=
9.8 m/s2Use conservation of mechanical energy for the swing.
2
The speed of the ball just before the collision is 2.3 m/s.
(b) Conservation of Energy Conservation of Momentum
2 2 2
0.25(2.2574) 0.25 0.21
5.096 0.84 (Equation 1)
(0.25)(2.2574) 0.25 0.21
2.2574 0.84
Substitute Equation 2 into Equation 1:
2 2
5.096 (2.2574 0.84 ) 0.84
5.096 (5.096 3.792 0.7056 ) 0.84 0 3.792 1.5456
0 ( 3.792 1.5456 )
Since v2
′ =
0 is not valid (no change in speed), then:The speed of the 0.21-kg ball just after the collision is 2.5 m/s [fwd].
20. (a) m1
=
0.45 kg(0.45 kg)(2.2 m/s) 0.45 kg 0.79 kg
The speed of the ball and the box just after the collision is 0.80 m/s.
(b)
∆ =
d 0.051 mThe magnitude of the friction force acting on the ball and the box is 7.8 N.
21. m1
= ×
1.9 10 kg4Using the sine law:
22. The total momentum before is zero, so the total momentum after must also be zero.
Diagram not to scale.
1
The final velocity of the third piece is 2.0 m/s [22º S of W].
Applying Inquiry Skills
23.
24.
25. (a) The coefficient of restitution is a measure of how quickly an object returns to its shape after it has been compressed.
The apparatus shown could help determine the coefficient of restitution because the speed the sphere returns to its original shape corresponds to the height the ball bounces to.
26. (a) The bed sheet spreads the force applied over a greater distance (and therefore time) than if the egg was to strike the ground directly. The same impulse is imparted to the egg (i.e. it is brought to rest), so the force required is much smaller.
(b) This can be used to cushion the landing of a person who was falling from a high position by reducing the average force needed to stop them. You could test several heights with a sack of potatoes and plot the results on a graph. For a
human, you could extrapolate from the graph.
Making Connections
27. It is better for safety to have telephone poles that collapse upon impact. In this way, they absorb some energy and increase the time of collision. Both of these reduce the force imparted to the vehicle and occupants in a collision.
28. High speed photography and spark timers both show the location of an object at fixed time intervals. High speed photography gives more thorough information because you can see the state of the object s during each stage of the
collision.
29. Many arrester cables are connected to a water squeezer that dampens the motion and converts the kinetic energy of the aircraft to kinetic energy of ejected water.
30. The main purpose of the ablation shield is to protect the shuttle during re-entry into the earth’s atmosphere.
Extension
31. (a) Conservation of Momentum Conservation of Energy
1 1 2 2 1 1 2 2 2
Divide Equation 2 by Equation 1:
2 2 2
Substitute Equations 3 and Equation 4 back into the original conservation of momentum equation:
1 1 1 1 2 2
For the collision:
Calculate the frictional force on the plane and on the barge:
For the plane For the barge
K
2450 N [backward]
F mg
F
=
= ×
=
By Newton’s third law, F K = 2450 N [forward]
Acceleration of the plane Acceleration of the barge
3
Distance plane will travel during collision Distance barge will travel during collision
2 2
CHAPTER 6 GRAVITATION AND CELESTIAL MECHANICS Reflect on Your Learning
(Page 272)
1. (a) The force of gravity exerted by Saturn keeps the rings in their orbit around Saturn.
(b) The force of gravity exerted by Earth keeps the Hubble Space Telescope in a stable orbit around Earth. The force exerted by the rockets of the telescope counteracts the small amount of friction of the upper atmosphere.
2. (a) The two probes would have the same minimum speed (escape speed) even though the probes have different masses.
Students will discover that the escape speed depends on the mass of Earth, not on the mass of the probe, according to the equation: 2GM E
v
= r .
(b) The minimum kinetic energy of the probe of mass 2m would be twice that of the probe of massm.The minimum kinetic energy of the probe is equal to the gravitational potential energy, which is proportional to the mass of the probe.
K g
K
E E
E GMm r
= −
∴ =
3. Since the gravitational potential energy E g is inversely proportional to the distance from the main body, E gwill approach zero as the distance increases, according to the equation presented later in the chapter:
g
E GMm
= − r
4. Students should be able to answer this question based on their study of the law of universal gravitation in Section 3.3.
5. As students have seen on page 147 of the text, a black hole is an extremely dense celestial body. The gravitational field of a black hole is so strong that nothing, including electromagnetic radiation, can escape from its vicinity. Since light can enter, but not escape, a black hole appears totally black. The surface of the body is called its event horizon because no event can be observed from outside the surface. At the core of a black hole is a dense centre called a singularity. The distance from the centre of the singularity to the event horizon is the Schwartzschild radius.
Black holes are thought to form during the course of stellar evolution. When the nuclear fuels are exhausted in the core of the star, the star collapses. If the mass of the core is greater than a critical value that is almost twice as great as the mass of the Sun, the core may collapse into a black hole.