Question

Think of a Newtonian cradle made of only two marbles. These two marbles of masses m1 and m2 are each suspended by identical strings. The strings are initially both vertical so the marbles are touching each other when undisturbed. The marble with m1 is shifted away from the initial position until is reaching the height h1 while the length of the string does not change and then is set free to fall and collide with marble m2. What is the formula for the height which both marbles will reach after the collision if: a). The collision is elastic; b). The collision is inelastic; (c). What will be the heat developed during the collision in both cases

Answer 1

Answer:

a)  h₂ = (m₁ / (m₁ + m₂))² h₁

b)   h₂ = (2m₁ / (m₁ + m₂))² h₁

Explanation:

Let's analyze this exercise, we have an energy that is given to marble 1, then it collides with canine 2 and they rise to a new height; therefore the problem has to be solved in parts.

Let's start by using the energy concepts for the canine 1

Starting point. Highest point

           Em₀ = U = m₁ g h

Final point. Lowest point, just before touching the other marble

            $Em_{f}$ = K = ½ m₁ v²

            Emo = Em_{f}

            m₁ g h = ½ m₁ v²

            v₁ = √ 2gh

now let's analyze the clash of the two marbles

We define a system formed by the two marbles, so that the outside during the shock have been internal and the moment is preserved

initial. Just before the crash

           p₀ = m₁ v₁ + 0

finsl. Right after the crash

           $p_{f}$ = (m₁ + m₂) v

This case inelastic collisions

           p₀ = p_{f}

           m₁ v₁ = (m₁ + m₂) v

            v = m₁ / (m₁ + m₂) v₁

this is the speed of the set before starting to climb. Let's use energy conservation for these two marbles

      Starting point. Right after the crash

             Em₀ = K = ½ (m₁ + m₂) v

final point. At the highest point of the set

             Em_f = U = (m₁ + m₂) h₂

             Em₀ = Em_f

            ½ (m₁ + m₂) v² = (m₁ + m₂) gh

            we substitute

          ½ v² = gh

           h = v² / 2g

we substitute the equation for speed

          h = (m₁ / (m₁ + m₂))² (2gh₁) / 2g

          h₂ = (m₁ / (m₁ + m₂))² h₁

b) In the case of elastic collision

in this case the conservation of the moment changes

initial    p₀ = m₁ v₁

End       p_f = m₁ v₁ ’+ m₂ v₂’

            p₀ = p_f

            m₁ v₁ = m₁ v₁ ’+ m₂ v₂’

also kinetic energy is conserved

              K₀ = K_f

             ½ m₁ v₁² = ½ m₁ v₁’² + ½ m₂ v₂²

we write the two equations

          m₁ (v₁ - v₁ ’) = m₂ v₂²

          m1 (v₁² - v₁’²) = m₂ v₂²

solving this system of equations we are left with

        v₁ ’= (m₁-m₂) / (m₁ + m₂) v₁

        v₂ = 2m₁ / (m₁ + m₂) v₁

with this result marble 2 rises to height, let's use conservation energy

       Em₀ = Em_f

        ½ m₂ v₂² = m₂ g h₂

         h₂ = v₂² / 2g

          h₂ = (2m₁ / (m₁ + m₂))² 2gh₁ / 2g

          h₂ = (2m₁ / (m₁ + m₂))² h₁

C) in the elastic case there is no heat in the collision

in the inelastic case Q = ΔK