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Recit. 9, week of Mar. 2125: Ch. 7 # 14, 32, 40, 45
14.
An 85kg jogger is heading due east at a speed of 2.0 m/s. A 55kg jogger is heading 32
o
north of east at a speed of 3.0 m/s. Find the magnitude and direction of the sum of the momenta
of the two joggers.
REASONING
This is a problem in vector addition, and we will use the component method for vector
addition. Using this method, we will add the components of the individual momenta in the direction
due north to obtain the component of the vector sum in the direction due north. We will obtain the
component of the vector sum in the direction due east in a similar fashion from the individual
components in that direction. For each jogger the momentum is the mass times the velocity.
SOLUTION
Assuming that the directions north and east are positive, the components of the
joggers’ momenta are as shown in the following table:
Direction due east
Direction due north
85 kg jogger
85
2 0
170
kg
m / s
kg m / s
b gb
g
.
=
⋅
0 kg
⋅
m/s
55 kg jogger
55
3 0
32
140
kg
m / s
kg m / s
b gb
g
.
cos
°
=
⋅
55
3 0
32
87
kg
m / s
kg m / s
b gb
g
.
sin
°
=
⋅
Total
310 kg
⋅
m /s
87 kg
⋅
m/s
Using the Pythagorean theorem, we find that the magnitude of the total momentum is
310
87
322
2
2
kg m / s
kg m / s
kg m / s
⋅
+
⋅
=
⋅
b
g b
g
The total momentum vector points north of east by an angle
θ
, which is given by
=
⋅
⋅
F
H
G
I
K
J
=
°

tan
1
87
310
16
kg m / s
kg m / s
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View Full Document32.
A car (mass = 1100 kg) is traveling at 32 m/s when it collides headon with a sport utility
vehicle (mass = 2500 kg) traveling in the opposite direction. In the collision, the two vehicles
come to a halt. At what speed was the sport utility vehicle traveling?
REASONING
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 Spring '08
 Staff

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