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```					                                    Solutions to Homework #3

Chapter 4
8.   Picture the Problem: Two divers run horizontally off the edge of a low cliff.
Strategy: Use a separate analysis of the horizontal and vertical motions of the divers to answer the conceptual
question.
Solution: 1. (a) As long as air friction is neglected there is no acceleration of either diver in the horizontal direction.
The divers will continue moving horizontally at the same speed with which they left the cliff. However, the time of
flight for each diver will be identical because they fall the same vertical distance. Therefore, diver 2 will travel twice
as much horizontal distance as diver 1.
2. (b) The best explanation (see above) is I. The drop time is the same for both divers. Statement II is true but not
relevant. Statement III is false because the total distance covered depends upon the horizontal speed.
Insight: If air friction is taken into account diver 2 will travel less than twice the horizontal distance as diver 1. This
is because air friction is proportional to speed, so diver 2, traveling at a higher speed, will experience a larger force.

20. Picture the Problem: The pumpkin’s trajectory is depicted in the
figure at right.
Strategy: The horizontal velocity remains 3.3 m/s throughout the
flight of the pumpkin, but the vertical velocity continually increases in
the downward direction due to gravity. Use the given time
information to find the vertical component of the velocity, and use it
together with the horizontal component to find the direction and
magnitude of the velocity vector. Equation 4-7 apply because the
initial velocity of the pumpkin is horizontal.

Solution: 1. (a) Determine the                  v y   gt
   9.81 m/s 2   0.75 s 
vertical component of the velocity:

v y  7.4 m/s

 7.4 m/s 
2. Find the direction of the velocity vector:     tan 1                66 or 66 below horizontal
 3.3 m/s 

3. Find the magnitude of the velocity vector: v      3.3 m/s          7.4 m/s   8.1 m/s
2               2

2y   2   9.0 m 
4. (b) Find the time of flight from equation 4- y   1 gt         t                             1.35 s
2

7:
2
g    9.81 m/s2 
5. Determine the vertical component of v:       v y   gt    9.81 m/s 2  1.35 s   13.2 m/s

 13.2 m/s 
6. Find the direction of the velocity vector:     tan 1                76 or 76 below horizontal
 3.3 m/s 

7. Find the magnitude of the velocity vector: v      3.3 m/s          –13.2 m/s   14 m/s
2                   2
Insight: When we study energy in Chapter 8 we’ll learn a simple way to find the magnitude of the velocity as a
function of altitude. Finding the direction, however, requires an approach similar to the one outlined above.

34. Picture the Problem: The trajectory of the ball is
shown at right.

Strategy: Use vy  v0 y  2 g y to find v y , then use v y ,
2    2

v0 y and the acceleration of gravity to find the time of
flight and hence the horizontal distance traveled.

Solution: 1. Find v y using equation 4-       v y   v0 sin 2   2 g y
2

 4.3 m/s  sin 2 15   2  9.81 m/s 2    0.80 m 
10:

2

  4.1 m/s  The ball is traveling downward 

v y  v0 y        4.1   4.3sin15 m/s 
2. Find the time of flight from equation      t                                                0.53 s
4-6:                                                  g                   9.81 m/s 2

3. Find the horizontal distance traveled: x  v0 x t   v0 cos   t   4.3 m/s  cos 15  0.53 s   2.2 m

Insight: Another way to solve this problem is to use y   v0 sin   t  1 gt 2 to find the time the ball is in the air. Then
2

use the known initial speed and angle to find the horizontal speed of the ball, which together with the time of flight can
be used to find the horizontal distance traveled. This method saves a step but requires you to use the quadratic
formula.

54. Picture the Problem: The soccer ball travels along a parabolic arc, maintaining its horizontal velocity but changing
its vertical speed due to the constant downward acceleration of gravity.
Strategy: Use equation 4-10 to find the initial vertical velocity of the ball from the time interval between the kick
and the peak of the flight. Then use the known magnitude of the initial velocity together with its vertical component
in order to find the direction of the kick.
Solution: 1. When the ball is at the                     vy  0  v0 sin   gt
peak
of flight its vertical speed is zero:

 gt             9.81 m/s 2   0.750 s  
2. Solve the equation for θ:                               sin 1          sin 1                               46.2
 v0           

10.2 m/s            

Insight: You can verify for yourself that with the new angle of the kick the maximum height is 2.76 m and the
downfield range is 10.6 m.
Chapter 5
12. Picture the Problem: The 747 is accelerated horizontally in the direction opposite its motion in order to slow it
down from its initial speed of 27.0 m/s.
Strategy: Use equation 5-1 to find the acceleration from the known force and mass, then use equations 2-7 and 2-10
to find the speed and distance traveled.

F  – 4.30  10  x N
ˆ
5

Solution: 1. (a) Use equation 5-1 to find a :         a                          1.23 m/s 2  x
ˆ
m    3.50  10 kg
5

2. Use equation 2-7 to find the final speed:                                                       
v  v0  at  27.0 m/s  1.23 m/s2  7.50 s   17.8 m/s

3. (b) Use equation 2-10 to find the distance         x    1
2    v0  v  t  1  27.0  17.8 m/s  7.50 s   168 m
2
traveled by the 747 as it slows down:
Insight: The landing speed of a Boeing 747-200 is 71.9 m/s (161 mi/h) and it has a specified landing roll distance of
2,121 m, requiring an average landing acceleration of −1.22 m/s2.

14. Picture the Problem: A small car collides with a large truck.
Strategy: Consider Newton’s Laws of motion when answering the conceptual question.
Solution: 1. (a) Newton’s Third Law states that when the car exerts a force on the truck, the truck exerts an equal and
opposite force on the car. We conclude that the magnitude of the force experienced by the car is equal to the
magnitude of the force experienced by the truck.
2. (b) The best explanation is I. Action-reaction forces always have equal magnitude. Statements II and III are both
false.
Insight: The force has a larger effect (produces a larger acceleration) on the smaller car due to its smaller inertia.

20. Picture the Problem: The force pushes on box 1 in the
manner indicated by the figure at right.
Strategy: The boxes must each have the same acceleration,
but because they have different masses the net force on each
must be different. These observations allow you to use
Newton’s Second Law for each individual box to determine
the magnitudes of the contact forces. First find the
acceleration of all the boxes and then apply equation 5-1 to
find the contact forces.
F
Solution: 1. (a) The 7.50 N force                      F  (m1  m2  m3 )a  a 
accelerates all the boxes together:                                                           m1  m2  m3
7.50 N
a                                 0.798 m/s 2
1.30 kg  3.20 kg  4.90 kg

2. Write Newton’s Second Law for the first box:        F  F  F          c12    m1 a
Fc12  F  m1 a  7.50 N  1.30 kg   0.798 m/s 2   6.46 N

3. (b) Write Newton’s Second Law for the third         F  F       c23    m3 a   4.90 kg   0.798 m/s 2   3.91 N
box:
Insight: Another way to solve the problem is to note that Fc12 accelerates boxes 2 and 3:
Fc12   3.20  4.90 kg   0.798 m/s 2   6.46 N , the same result as in step 2.
28. Picture the Problem: The two men pull on the barge in the
directions indicated by the figure at right.
Strategy: Place the x-axis along the forward direction of the
boat. Use the vector sum of the forces to find the force F such
that the net force in the
y-direction is zero.

Solution: Set the sum of  Fy   130 N  sin 34  F sin 45  0
the forces in the y                   sin 34
direction                F  130 N           100 N  0.10 kN
sin 45
equal to zero:

Insight: The second crewman doesn’t have to pull as hard as the first because a larger component of his force is
pulling in the y direction. However, his force in the forward direction (73 N) is not as large as the first crewman
(110 N).

34. Picture the Problem: The uncoupled train car coasts up the incline, slowing down under the influence of gravity,
and briefly comes to rest before accelerating backwards down the incline.
Strategy: After the last car breaks free, it will continue coasting up the incline until the force of gravity brings it
ˆ
momentarily to rest. Let the x direction point uphill and parallel to the incline. The only force acting on the
train car is the parallel component of its weight, Wx  mg sin  . Use the known acceleration ax   g sin  to
find the time required to bring the train car to a stop, and then use equation 2-10 to determine the distance it
travels along the incline while slowing down.

Solution: 1. (a) Find the time                                   v  v0    0  v0          3.25 m/s
t                                              5.09 s
required to stop the train car:                                    ax      g sin   9.81 m/s 2  sin 3.73

2. (b) Use eq. 2-10 to find the                             x   1
2    v0  v  t  1  3.25  0 m/s  5.09 s  
2
8.27 m
distance traveled while stopping:
Insight: In most cases when motion occurs along an inclined surface it is much simpler to tilt the coordinate axes
until they are parallel and perpendicular to the incline, as it was in this case.

54. Picture the Problem: Various objects either accelerate or move with constant velocity.
Strategy: Consider Newton’s Second Law as a vector equation when answering the conceptual questions.
Solution: 1. (a) “A car accelerating northward from a stoplight.” The acceleration vector points northward and
therefore the net force points northward, due to the vector nature of Newton’s Second Law.
2. (b) “A car traveling southward and slowing down.” In this case the acceleration points opposite to the direction of
motion, and so the acceleration and the net force point northward.
3. (c) “A car traveling westward with constant speed.” In this case the acceleration and the net force are zero.
4. (d) “A skydiver parachuting downward with constant speed.” In this case the acceleration and the net force are zero.
5. (e) “A baseball during its flight from pitcher to catcher (ignoring air resistance).” In this case the acceleration points
downward (due to gravity) and therefore the net force points downward.
Insight: The acceleration of an object must always point in the same direction as the net force on it.

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