Game Development Reference
InDepth Information
m
1
v
1n
+ m
2
v
2n
= m
1
v
1n+
+ m
2
v
2n+
Noting that
m
1
equals
m
2
since the balls are identical, and that
v
2n
is 0, and then solving
for
v
1n+
yields:
v
1n+
= v
1n
 v
2n+
To actually solve for these velocities, you need to use the equation for coefficient of
restitution and make the substitution for
v
1n+
. Then, you'll be able to solve for
v
2n+
. Here's
how to proceed:
e = (v
1n+
+ v
2n+
) / (v
1n
 v
2n
)
e v
1n
= (v
1n
 v
2n+
) + v
2n+
v
2n+
= v
1n
(e + 1) / 2
v
2n+
= (5.18 m/s)(1.9) / 2 = 4.92 m/s
Using this result and the formula for
v
1n+
yields:
v
1n+
= 5.18 m/s  4.92 m/s = 0.26 m/s
Since the collision is frictionless, there is no impulse acting in the tangential direction.
This means that momentum is conserved in that direction too and that the final tan‐
gential speed of ball 1 is equal to its initial tangential speed, which in this case is equal
to 3 m/s (this equals (6m/s) sin 30°). Since ball 2 had no initial tangential speed, its
velocity after impact is solely in the normal direction. Converting these results back to
x

y
coordinates instead of normal and tangential coordinates yields the following ve‐
locities for each ball after impact:
v
2+
= (4.92 m/s) sin 60°
i
 (4.92 m/s) cos 60 °
j
v
1+
= [(0.26 m/s) cos 30° + (3 m/s) sin 30°]
i
+
+ [(−0.26 m/s) sin 30° + (3 m/s) cos 30°)]
j
v
1+
= (1.72 m/s)
i
+ (2.47 m/s)
j
To further illustrate the application of these collision response principles, consider an‐
other example, this time the collision between a baseball bat and baseball (as shown in
Figure 53
). We are looking at a side view, staring down the barrel of the bat.