hard physics, nobody here has been able to get the right answer for this one........?

Question

A small ball of mass m is aligned above a larger ball of mass M = 0.77 kg (with a slight separation, as with the baseball and basketball), and the two are dropped simultaneously from height h = 2.1 m. (Assume the radius of each ball is negligible relative to h.)



(a) If the larger ball rebounds elastically from the floor and then the small ball rebounds elastically from the larger ball, what value of m results in the larger ball stopping when it collides with the small ball?
kg
(b) What is the maximum height the small ball then reaches?
m

actually i already figured it out, but if somebody can give me the explanations i guess u can get the best answer points.....

a) 0.257kg
b) 8.4m

Answer 1

♠ kin energy of M-ball is the same as its pot energy, that is
0.5*M*V^2 = Mgh, hence V^2=2gh;
since V is not dependent of mass the speed of m-ball is also V;
♣ now all this bound-rebound stuff is just lyrics to muddle our brain; and we can re-word the problem like this;
“M-ball and m-ball collide head-to-head, both having speed V;
what is mass m if M-ball stops after collision?”
♦ then momentum conserv law says:
M*V - m*V = M*0+m*u,
where u is new speed of m-ball after collision;
thus u=(M-m)*V/m;
And energy conserv law says:
0.5m*V^2 +0.5M*V^2 = 0.5M*0^2 +0.5m*u^2;
thence m*V^2+M*V^2 = m*{(M-m)*V /m}^2; or deleting V^2;
m^2 +m*M =M^2 –2m*M +m^2, hence m=M/3 =0.77/3 =0.257kg;
and u =(2M/3)*V/(M/3) =2V= 2*sqrt(2gh) – (see ♠);
♥ now forget M-ball;
here m-ball with kin energy
0.5m*u^2 =0.5m*(2V)^2 =4mgh
will stop with pot energy mgH reaching altitude H;
thus mgH=4mgh, hence H=4h =8.4 m;

Answer 2

Not so hard, the smaller ball is equal to the mass of the larger ball divided by 3.


Proof:

conservation of momentum with the large mass having zero velocity after collision. Note that the two masses have the same speed with opposite direction under the assumptions you stated.

so
(ml-ms)*V=ms*vs
where ml is the large, ms the small
V is the velocity at collision (sqrt(2*g*h))
and vs is the velocity of the small mass after collision

solving for vs,
vs=(ml-ms)*V/ms

Conservation of energy with the large mass having zero kinetic after collision:
I'll skip a bunch of algebra

(ml+ms)*V^2=ms*vs^2

plugging in vs from the top

ml*ms+ms^2=ml^2-2*ml*ms+ms^2

note that the answer is independent of the height of the drop

simplifying
3*ml*ms=ml^2
ms=ml/3

The return height is simply the translation of all of the potential energy of both balls converted into potential energy of the small ball (yawn)

g*h*(ml+ms)=g*d*ms
recall
ml=3*ms
so
d=h*4
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