The superball
Since we are told that |v_y| is unchanged by the bounce, we can ignore it when applying conservation of energy. And since the vertical impulse from the floor provides no torque around the ball's CM, we can completely ignore the y motion in this problem. The horizontal impulse from the floor is responsible for changing both v_x and ω.
With positive directions defined as in the statement of the problem, Eq. (8.61) gives
ΔL = RΔp ⇒ I(ω′ − ω) = Rm(v_x′ − v_x). (8.130)
And conservation of energy gives
1/2 mv_x′² + 1/2 Iω′² = 1/2 mv_x² + 1/2 Iω²
⇒ I(ω′² − ω²) = m(v_x² − v_x′²). (8.131)
Dividing this equation by Eq. (8.130) gives¹⁵
R(ω′ + ω) = −(v_x′ + v_x). (8.132)
We can now combine this equation with Eq. (8.130) which can be rewritten as, using I = (2/5)mR²,
2/5 R(ω′ − ω) = v_x′ − v_x. (8.133)
Given v_x and ω, the previous two equations are two linear equations in the two unknowns, v_x′ and ω′. Solving for v_x′ and ω′, and writing the result in matrix notation, gives
[v_x′] 1 [ 3 -10] [v_x ]
[Rω′] = 7 [-4 -3 ] [Rω], (8.134)
as desired. Note that Eq. (8.132), when written in the form of v_x + Rω = −(v_x′ + Rω′), says that the relative velocity of the ball's contact point and the ground simply changes sign during the bounce.
Remark: For a ball with a general moment of inertia I = βmR², you can use the above procedure to show that the matrix in Eq. (8.134) takes the general form,
1 [ 1-β -2β ]
——— [ ]. (8.135)
1+β [ -2 -(1-β)]
¹⁵ We have divided out the trivial ω′ = ω and v_x′ = v_x solution, which corresponds to slipping motion on a frictionless plane. The nontrivial solution we will find shortly is the nonslipping one. Basically, to conserve energy, there must be no work done by friction. And since work is force times distance, this means that either (1) the plane is frictionless, so that the force is zero, or (2) there is no relative motion between the ball's contact point and the plane, so that the distance is zero. The latter case is the one we are concerned with here.
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Angular momentum, Part I (Constant L^)
For β = 1 (a hoop), this becomes
[ 0 -1]
[-1 0], (8.136)
which means that the bounce simply interchanges and negates the values of v_x and Rω. In particular, if you throw a "super-hoop" sideways with no spin (that is, Rω = 0), then it will bounce straight up in the air (that is, v_x′ = 0) while spinning. ♣