Game Development Reference
In-Depth Information
1
3
(152 g m/s)
2
=
v
2
+
k
n
m
2
−4 m/s
−4 m/s
′
v
=
+
Calculating the
post-impulse velocity for
the SuperBall
50 g
−4 m/s
−4 m/s
3.04 m/s
9.12 m/s
−0.96 m/s
5.12 m/s
=
+
=
.
Before we show this velocity graphically, let's look at the beanbag. We
treat the beanbag collision as almost completely inelastic and use e = 0.01.
Other than the change to e, the procedure is the same as for the SuperBall:
(1/m
1
+ 1/m
2
)
n
n
=
1.01 (16 m/s)
(e + 1)
v
rel
n
k =
= 80.8 g m/s,
Calculating the impulse
multiplier k and
post-impulse velocity for
the beanbag
10/(50 g)
1
3
(80.8 g m/s)
2
=
v
2
+
k
n
m
2
−4 m/s
−4 m/s
−2.38 m/s
0.85 m/s
′
v
=
+
=
.
50 g
Notice that the reduced coe
cient of restitution caused the beanbag to
receive a smaller impulse scale, and the resulting bounce velocity was also
lower. This difference is shown graphically in Figure 12.15. For comparison,
we've also included a perfectly inelastic collision (e = 1). The perfectly
inelastic collision (e = 0) is very close to the beanbag result and is not
depicted.
E
=1
E
=0.9
N
=[1,3]
E
=0.01
M
2
=50 G
Figure 12.15
Post-impact
velocities for
different values of
e
,
the coefficient of
restitution
M
1
=
V
2
=[4,4]
V
1
=
0
It is obvious from the beanbag trajectory in Figure 12.15 that an impor-
tant aspect of collisions is not captured by the model: friction. The velocity
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