Game Development Reference
In-Depth Information
(b) First, we compute the duration of the journey by considering the man's motion in
the coordinate space fixed to the car, as
∆t = ∆x/v = (20.0 m)/(1.25 m/s) = 16.0 s.
We then multiply the velocities of the car and man by this duration to obtain their
displacements.
∆x
m
= v
m
∆t = (1.16 m/s)(16.0 s) = 18.6 m
∆x
c
= v
c
∆t = (−0.0869 m/s)(16.0 s) = −1.39 m
An alternate approach is to recognize that the center of mass of the system does
not move, since there are no external forces, and treat the man and the car as point
masses. Since the man walked the length of the car,
∆x
m
= ∆x
c
+ 20.0 m.
Now the movement of the man must be offset by the movement of the car, such that
the center of gravity does not shift.
∆x
m
m
m
+ ∆x
c
m
c
= 0
One again, the system of equations is solved by plugging the first equation into the
second.
(∆x
c
+ 20.0 m)m
m
+ ∆x
c
m
c
= 0
(∆x
c
+ 20.0 m)(75.0 kg) + ∆x
c
(1.00×10
3
kg) = 0
∆x
c
(75.0 kg) + (1.50×10
3
kg m) + ∆x
c
(1.00×10
3
kg) = 0
∆x
c
(1.08×10
3
kg) = −1.50×10
3
kg m
∆x
c
= −1.39 m
(c) The car's velocity would also increase in proportion to the man's. At all times, the
total momentum and total displacement of the center of mass would be zero. The
ending configuration of the car and the man would be the same as before.
(d) Here all we must do is add +5.00 m/s to our earlier results.
v
c
= −0.0869 m/s + 5.00 m/s = 4.91 m/s
v
m
= 1.16 m/s + 5.00 m/s = 6.16 m/s
(e) ∆x
m
= v
m
∆t = (6.16 m/s)(16.0 s) = 98.6 m
∆x
c
= v
c
∆t = (4.91 m/s)(16.0 s) = 78.6 m
9. First, we must compute the contact normal n as
cos−110
o
sin−110
o
−0.342
−0.940
n =
≈
,
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