Question

A small ball of mass m is dropped immediately behind a large one of mass M from a height h mach larger then the size of the balls. What is the relationship between m and M if the large ball stops at the floor? Under this condition, how high does the small ball rise? Assume the balls are perfectly elastic and use an independent collision model in which the large ball collides elastically with the floor and returns to strike the small ball in a second collision that is elastic and independent from the first.

Answer 1

Answer:

a)   (M / m -1), b)   $h_{f}$ = h (M / m -1)²

Explanation:

For this exercise we will begin by looking for the speed of the balls when they reach the floor, for this we use the concepts of energy conservation

Initial. Highest point

      Em₀ = U = m g h

Final. Floor

      $Em_{f} }$ = K = ½ m v²

      Em₀ =  $Em_{f} }$

      m gh = ½ m v²

      v = √ 2g h

The speed is the same for the two balls since it does not depend on the mass.

First shock

The heaviest ball (M) with the floor, as the floor does not move the ball bounces with the same speed as it arrives, but in the opposite direction

Second shock

Between the large ball with velocity upwards of value v and the small ball of mass (m) with velocity v downwards. Let's use the moment

Initial. Before the crash

         p₀ = Mv - m v

After the crash

        $p_{f}$ = 0 + m $v_{f}$

        p₀ = $p_{f}$

        Mv - m v = m $v_{f}$

       $v_{f}$ = (M-m) / m v

        $v_{f}$= (M / m -1) v

Let's look with energy to where e raises the small ball with speed vf

      Em₀ = $Em_{f}$

       ½ m vf² = m g $h_{f}$

       $h_{f}$ = ½ $v_{f}$² / g

       $h_{f}$ = ½ v² / g (M / m-1)²

We substitute in value of v

       $h_{f}$ = ½ 2gh / g (M / m -1)²

       $h_{f}$ = h (M / m -1)²