Mechanics 2: Linear and Rotational Dynamics - 3D Math Primer for Graphics and Game Development - page 599

Let's say we have one SuperBall and one beanbag, each weighing 50 grams.
To avoid making the solution to our example completely obvious without
Equation (12.23), we will arrange for the point of impact to occur on an
inclined surface, so that n is not a trivial cardinal axis. Since we are not
bound to use a unit vector, let's say that n = [1,3]. In our imagination, we
throw both objects in exactly the same way, such that the impact velocity
in both cases is v 2 = [−4,−4]. This is illustrated in Figure 12.14.

N =[1,3]

M 2 =50 G

M 1 =

V 2 =[4,4] V 1 = 0

Figure 12.14
Bouncing a beanbag
or SuperBall off of an
inclined surface

To determine the resulting velocity, we first solve for k, the scale factor
for the impulse. To do this, we must choose a coe cient of restitution.
For the SuperBall collision, we'll use e = 0.9, which is near the advertised
value. Solving for k (remember that v rel = v 1 − v 2 ) we get

Calculating the impulse
multiplier k for the
SuperBall

n
(1/m 1 + 1/m 2 ) n n

(e + 1) v rel

k =

4 m/s
4 m/s

1
3

(0.9 + 1)

= 1.9 (16 m/s)
10/(50 g)

= = 152 g m/s.

1
3

1
3

(0 + 1/(50 g))

To compute the post-collision velocity, we add an impulse of k n to the
momentum of the SuperBall. Since momentum is mass times velocity, the
change in velocity is equal to this impulse divided by m 2 , the mass of the
SuperBall: