Answer:
New height reached = (M²/m²)h
Explanation:
As M>>m you can see the final height is very large compared to the initial height, which agrees the experiment.
<em>Theory</em>
<u>1.The principle of conservation of Mechanical energy</u>
In a system which the only forces are acting are associating with potential energy, the sum of kinetic energy and potential energies is constant.
<u>2. The law of conservation of linear momentum
</u>
The sum of linear momentum of a closed system under no external unbalance force remains the same.
⇒
m ≪ M suggest you the mass of small ball is negligible compared to the mass of super ball.
1.Put the principle of conservation of Mechanical energy ( or principle of conservation of energy),
(K.E. of two balls at height h)+(P.E. of two balls at height h) =
(K.E. of two balls just before bounce)+(P.E. of two balls just before bounce)
We consider gravitational potential energy at ground level as 0.
We get,
0 + (M+m)gh = (M+m)
u = velocity of the composite body.
u = -------------------(1)
2.As two balls collide superball get highly compressed and transfer all its momentum to the ball.
Now apply the law of conservation of linear momentum
, it is given that collisions with the superball are elastic,
(linear momentum of super ball + small ball before collision) = (linear momentum of super ball after collision) + (linear momentum of small ball after collision)
(M+m)u + 0 = 0 + mv but m ≪ M
Mu + 0 = 0 + mv assuming all the momentum is transferred to the small ball from superball
from (1) v = mu/M ------------------(2)
3. Now apply the principle of conservation of Mechanical energy to the small ball,
(K.E. of small ball just after bounce)+(P.E. of small ball just after bounce) =
(K.E. of small ball at maximum height)+(P.E. of small ball at maximum height)
m +0 = 0 + mgH
H = new height attained,
From (2)
H = (M²/m²)h