Question

A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball? b) What final height, in terms of H, does the small ball reach?

Answer 1

Answer:

a) the large sphere has 3 times the mass of the small sphere

b)     y = 4H

Explanation:

We must start this problem by calculating the speed with which the spheres reach the floor

       vf² = vo² - 2g y

As the spheres are released v₀ = 0

      vf² = - 2g H

      vf = √ (2g H)

The two spheres arrive with the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the  spheres

     V₁₀ = √2gH

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call vh = √2gH

Small sphere m₂ and v₂o = - √2gH = -vh

Large sphere m₁ and v₁o = √ 2gh = vh

Before crash

        p₀ = m₁ v₁₀ + m₂ v₂₀

After the crash

      pf = m₁ v₁f + m₂ v₂f

      po = pf

      m₁ v₁₀ + m₂ v₂₀ = m₁ v₁f + m₂ v₂f

The conservation of kinetic energy

      Ko = ½ m₁ v₁₀² + ½ m₂ v₂₀²

      Kf = ½ m₁ v₁f² + ½ m₂ v₂f²

      Ko = KF

      ½ m₁ v1₁₀² + ½ m₂ v₂₀² = ½ m₁ v₁f² + ½ m₂ v₂f²

Let's write the values

      -m₁ vh + m₂ vh = m₁ v₁f + m₂ v₂f

       m₁ vh² + m₂ vh² = m₁ v₁f² + m₂ v₂f²

The solution to this system of equations is

       mt = m₁ + m₂

       v1f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂

       v₂f = 2m₁ /mt v₁₀ + (m₂-m₁) / mt v₂

The large sphere is labeled 1, we are asked for the mass so that v1f = 0, let's clear the equation

       v₁f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂₀

        0 = (m₁-m₂) / mt vh + 2 m₂ / mt (-vh)

       (m₁-m₂) / mt vh = 2 m₂ / mt vh

       (m₁-m₂) = 2m₂

        m₁ = 3 m₂

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

        v₂f = 2m₁ / mt v₁₀ + (m₂-m₁) / mt v₂₀

        v₂f = 2 m₁ / mt vh + (m₂-m₁) mt (-vh)

In addition, we know that m₁ = 3 m₂

        mt = 3m2 + m2

         mt= 4m2

        v₂f = 2 3m₂ / 4m₂ vh - (m₂-3m₂) 4m₂ vh

        v₂f = 3/2 vh +1/2 vh

        v₂f = 2 vh

        v₂f = 2 √ 2gh

This is the rate of rise of sphere 2 (small. At the highest point its zero velocity vf = 0

        V² = v₂f² - 2 g Y

          0 = (2√2gh)² - 2gy

        2gy = 4 (2gH)

         y = 4H

Answer 2

a) The large sphere has 3 times the mass of the small sphere

b)   The final height at which small ball reach y = 4H

What will be the mass of the sphere and height covered by the small ball?

We must start this problem by calculating the speed with which the spheres reach the floor

$V_f^2=V_o^2-2gy$      

As the spheres are released v₀ = 0

$V_f^2=2gH$      

$V_f=\sqrt{2gH}$      

The two spheres arrive at the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the  spheres

$V_{10}=\sqrt2gH$    

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call

$V_h=\sqrt2gH$

Small sphere m₂ and  $V_{20}=-\sqrt2gH=-V_h$

Large sphere m₁ and  $V_{10}=\sqrt{2gH}=V_h$  

Before crash

$P_o=m_1V_{10}+m_2V_{20}$      

After the crash

$P_f=m_1V_{1f}+m_2V_{2f}$    

   

$P_o=P_f$  

   

$m_1V_{10}+m_2V_{20}=m_1V_{1f}+m_2V_{2f}$  

The conservation of kinetic energy

$K_o=\dfrac{1}{2} m_1V_{10}^2+\dfrac{1}{2} m_2V_{20}^2$      

$K_f=\dfrac{1}{2} m_1V_{1f^2}+\dfrac{1}{2} m_2V_{2f}^2$    

$K_o=K_f$  

$K_o=\dfrac{1}{2} m_1V_{10}^2+\dfrac{1}{2} m_2V_{20}^2$  $=$  $K_f=\dfrac{1}{2} m_1V_{1f^2}+\dfrac{1}{2} m_2V_{2f}^2$    

Let's write the values

$-m_1V_h+m_2V_h=m_1V_{1f}+m_2V_{2f}$    

 $m_1V_h^2+m_2V_h^2=m_1V_{1f}^2+m_2V_{2f}^2$

 

The solution to this system of equations is

$m_t=m_1+m_2$    

$V_{1f}=\dfrac{(m_1-m_2)}{m_tV_{10}} +\dfrac{2m_2}{m_tV_2}$      

$V_{2f}=\dfrac{2m_1}{m_tV_{10}} +\dfrac{m_2-m_1}{m_tV_2}$        

The large sphere is labeled 1, we are asked for the mass so that  $V_{1f}$ = 0, let's clear the equation

$V_{1f}=\dfrac{m_1-m_2}{m_tV_{10}} +\dfrac{2m_2}{m_tV_{20}}$      

$0=\dfrac{m_1-m_2}{m_tV_h} +\dfrac{2m_2}{m_t(-V_h)}$      

 

$\dfrac{m_1-m_2}{m_tV_h} =\dfrac{2m_2}{m_tV_h}$      

$(m_1-m_2)=2m_2$    

$m_1=3m_2$      

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

  $V_{2f}=\dfrac{2m_1}{m_tV_{10}} +\dfrac{m_2-m_1}{m_tV_{20}}$    

 

$V_{2f}=\dfrac{2m_1}{m_tV_h} +\dfrac{m_2-m_1}{m_t(-V_h)}$

In addition, we know that m₁ = 3 m₂

$m_t=3m_2+m_2$  mt = 3m2 + m2

$m_t=4m_2$    

     

$V_{2f}=\dfrac{2\times 3m_2}{4m_2V_h-(m_2-3m_2)4m_2V_h}$  

   

$V_{2f}=\dfrac{3}{2} V_h+\dfrac{1}{2} V_h$  

       

$V_{2f}=2V_h$

$V_{2f}=2\sqrt{2gh$        

This is the rate of rising of sphere 2 (small. At the highest point, it zeroes velocity $V_f$= 0

$V^2=V_{2f}^2-2gy$  

$0=(2\sqrt{2gh})^2-2gy$        

       

$y=4H$

Thus

a) The large sphere has 3 times the mass of the small sphere

b)   The final height at which small ball reach y = 4H

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