Question

Suppose you drop a superball of massMand a marble of mass m(both treated as point masses) from a heighthwith the marble just on top of the superball. A superbowl has essentially elastic collisions with hard objects. Thesuperball hits the floor, rebounds and then collides with the marble. How highdoes the marble go if all the motion is vertical

Answer 1

Answer:

 h '= [ ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]²  h

Explanation:

Let's analyze this problem, first the two bodies travel together, second the superball bounces, third it collides with the marble and fourth the marble rises to a height h ’

let's start by finding the velocity of the two bodies just before the collision, let's use the concepts of energy

starting point. Starting point

         Em₀ = U = m g h

final point. Just before the crash

         Em_f = K = ½ m v²

as there is no friction the mechanical energy is conserved

         Em₀ = Em_f

         mg h = ½ m v²

         v = √2gh

this speed is the same for the two bodies.

Second point. The superball collides with the ground, this process is very fast, so we will assume that the marble has not collided, let's use the concept of conservation of moment

initial instant. Just when the superball starts contacting the ground

      p₀ = M v

this velocity is negative because it points down

final instant. Just as the superball comes up from the floor

      p_f = M v '

the other body does not move

      p₀ = p_f

     - m v = M v '

       v ’= -v

Therefore, the speed of the asuperbola is the same speed with which it arrived, but in the opposite direction, that is, upwards.

Let's use the subscript 1 for the marble and the subscript 2 for the superball

Third part. The superball and the marble collide

the system is formed by the two bodies, so that the forces during the collision are internal and the moment is conserved

initial instant. Moment of shock

        p₀ = M $v_{1'}$+ m v_2

final instant. When the marble shoots out.

        P_f = Mv_{1f'}+ m v_{2f}

        p₀ = p_

        M v_{1'}+ m v_2 = M v_{1f'} + m v_{2f}

        M (v_{1'} - v_{1f'}) = -m (v_2 - v_{2f})

in this expression we look for the exit velocity of the marble (v2f), as they indicate that the collision is elastic the kinetic nerve is also conserved

       K₀ = K_f

       ½ M v_{1'}² + m v₂² = M v_{1f'}²  + ½ m v_{2f}²

        M (v_{1'}² - v_{1f'}²) = - m (v₂² - v_{2f}²)

Let's set the relation  (a + b) (a-b) = a² - b²

      M (v_{1'} + v_{1f'})  (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})

let's write our two equations

           M ( v_{1'} - v_{1f'}) = -m (v₂ - v_{2f})                 (1)

           M (v_{1'} + v_{1f'})  (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})

       

if we divide these two expressions

           (v_{1'}+ v_{1f'}) = (v₂ + v_{2f} )

we substitute this result in equation 1 and solve

          v_{1f'}= (v₂ + v_{2f}) - v_{1'}

          M (v_{1'} - [(v₂ + v_{2f}) - v_{1'}] = -m (v₂ - v_{2f})

           -M v₂ - M v_{2f1'} + 2M v_{1'} = m v₂ - m v_{2f}

          -M v_{2f} -m v_{2f} = m v₂ -M v₂ + 2M v_{1'}

          v_{2f} (M + m) = - v₂ (M-m) + 2 M v_{1'}

           

          v_{2f} = - $( \frac{ M-m}{M +m } )$) v₂ + 2 $(\frac{M}{M+m})$ v_{1'}

now we can substitute the velocity values ​​found in the first two parts

          $v_{2f}$ = - ( \frac{ M-m}{M +m  } ) √2gh + 2(\frac{M}{M+m}) √2gh

we simplify

          v_{2f} = [( \frac{ M-m}{M +m  } ) + 2 (\frac{M}{M+m})] $\sqrt{2gh}$

let's call the quantity in brackets that only depends on the masses

          A = ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]

           

           v_{2f}= A \sqrt{2gh}

in general, the marble is much lighter than the superball, so its speed is much higher than the speed of the superball

finally with the conservation of energy we find the height that the marble reaches

       

Starting point

          Emo = K = ½ mv_{2f}²

Final point

          Emf = U = m g h'

          Em₀ = Em_f

          ½ m v_{2f}² = m g h ’

          h ’= ½ v_{2f}² / g

         h ’= ½ (A \sqrt{2gh})² / g

         h ’= A² h

         

         h '= [ ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]²  h