Question

A small ball of mass m is placed on top of a "superball" of mass M, and the two balls are dropped to the floor from height h. How high does the small ball rise after the collision? Assume that col- lisions with the superball are elastic, and that m ≪ M. To help visualize the problem, assume that the balls are slightly separated when the superball hits the floor. (If you are surprised by the result, try demonstrating the problem with a marble and a superball.)

Answer 1

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.

Theory

1.The principle of conservation of Mechanical energy

In a system which the only forces are acting are associating with potential energy, the sum of kinetic energy and potential energies is constant.

2. The law of conservation of linear momentum

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 = $\frac{1}{2}$ (M+m) $u^{2}$

u = velocity of the composite body.

u = $\sqrt{2gh}$ -------------------(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)

$\frac{1}{2}$m$v^{2}$ +0 =  0 + mgH

H = new height attained,

From (2)

H = (M²/m²)h