I used to work with SI units:
m1 = 20 lb = 9.07 kg,
m2 = 10 lb = 4.536 kg,
R = 1ft = 0.305 m,
s =2 ft = 0.61 m.
k =400 lb/ft = 5837 N/m,
μ = 0.3,
coefficient of restitution e = 0.8.
For the ball:
n1 = 300 rpm = 5 rev/s =>
ω1 =2 πn(b1) = 10π =>
v1 = ω(b1) R = 10π0.305 =9.58 m/s.
The kinetic energy of the rolling ball at its initial position is
KE1 =mv1^2/2 +I ω1^2/2 = mv1^2/2 +(2mR^2/5) (v1^2/2r^2) = 0.7m1v1^2 =0.79.07(9.58)^2 = 583 J.
KE1 KE2 =W(friction)
W(friction) = μmgs =0.3 9.079.80.61 = 16.27 J.
KE2 = KE1 - W(friction) = 583 - 16.27 = 566.73 J.
v2 = sqrt(2KE2/m1) = sqrt(2566.73/9.07) = 11.18 m/s.
Law of conservation of linear momentum:
m1v2 = m1v3 +m2u
According to the definition the coefficient of restitution is
e = (u v3)/v2,
ev2 = u v3
v3 = u - ev2.
m1v2 =m1 (u - ev2) + m2u = m1 u - m1 ev2 + m2u.
u (m1+m2) = m1v2 (e + 1 )
u = m1v2(e + 1 )/ (m1+m2) =
9.0711.181.8/(9.07 + 4.536) = 13.42 m/s.
kx^2/2 = m2u^2/2,
x =usqrt(m2/k) = 13.42sqrt(4.536/5837) = 0.37 cm ≈ 1.2 ft.
Maybe Ive mistaken in calculations (especially due to the units transformation), but the general form is correct. If you know the answer, post it
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