high level physics

Let mass of small sphere be m. Since mass of sphere is proportional to the cube of its radius, mass of large sphere = m(2a/a)^2 = 8m

(a) Vertical distance fell by each sphere when the large sphere touches the ground
= (h - 2a)
Hence, speed of each sphere u when the large sphere touches the ground is
u^2 = 2g(h-2a) where g is the acceleration due to gravity

Because the large sphere rebounds elastically, the speed after rebound = u
When the large sphere collides with the small sphere, momentum is conserved,
hence, (8m)u + m(-u) = (8m)(v1) + m(v2)
where v1 and v2 are the velocities of the large and small spheres respectively.
i.e. 7u = 8(v1) + v2 ------------- (1)

By conservation of kinetic energy,
(1/2)(8m)u^2 + (1/2)mu^2 = (1/2)(8m)(v1)^2 + (1/2)(m)(v2)^2
i.e. 9u^2 = 8(v1)^2 + (v2)^2 ------------- (2)

After solving (1) and (2) for v1 and v2 and simplify, we have
v1 = 5u/9
v2 = 23u/9

Height reached by small sphere (measured from collision point)
= (v2)^2/2g = (23u/9)^2/2g = (23/9)^2.[2g(h-2a)]/2g
= (23/9)^2(h-2a)
Thus, height reached by small sphere measured from ground
= (23/9)^2(h-2a) + 5a

(b) If the collision between the spheres is perfectly elastic, the spheres have a common velocity v after collision. By momentum conservation,
Hence, (8m)u + m(-u) = (8m+m)v
i.e. v = 7u/9

Height reached by small sphere (measured from collision point)
= (7u/9)^2/2g = (49/81)u^2/2g = (49/81)[2g(h-2a)]/2g
= (49/81)(h-2a)
Thus, height reached by small sphere measured from ground
= (49/81)(h-2a) + 5a