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(#24 on p. 292 of Ohanian’s Physics) Two small balls are suspended side by side from two strings of length L so that they touch when in their equilibrium position. Their masses are m and 2m, respectively. If the left ball (of mass m) is pulled aside and released from a height h, it will swing down and collide with the right ball (of mass 2m) at the lowest point. Assume the collision is elastic. (a) How high will each ball swing after the collision? (b) Both balls again swing down and collide once more at the lowest point. How high will each swing after the second collision?

Respuesta :

Answer:

a)  h3 = 4/9 gh

Explanation:

This problem must be worked in two separate parts: one with the conserving energy, another with the conservation of the moment during the crash.

Let's start looking for the lightest ball speed, let's use energy conservation, at the highest point and the lowest point of the trajectory

    Em₁ = ​​U = m g h

    Em₂ = K = ½ m v²

    Em₁ = ​​Em₂

    m g h = ½ m v²

    v² = 2gh

    v = √ 2gh

Now let's use moment conservation, the system is formed by the two balls before after the crash

Before the crash

    p₀ = m v

After the crash

    [tex]p_{f}[/tex] = m [tex]v_{f}[/tex] + m₂ v₂

    p₀ = [tex]p_{f}[/tex]

    m v = m [tex]v_{f}[/tex] + m₂ v₂

As the shock is elastic the kinetic energy is conserved

    K₀ = ½ m v²

    Kf = ½ m [tex]p_{f}[/tex]² + ½ m₂ v₂²

    K₀ = [tex]K_{f}[/tex]

     ½ m v² = ½ m [tex]v_{f}[/tex]² + ½ m₂ v₂²

Let's write the two equations together and solve the system

    mv = m [tex]v_{f}[/tex] + m₂ v₂

    m v² = m [tex]v_{f}[/tex]² + m₂ v₂²

Let's clear vf in the first equation and substitute in the second equation

    [tex]v_{f}[/tex] = (m v -m₂ v₂) / m

    m v² = m [(mv -m₂ v₂) / m]² + m₂ v₂²

    m v² = (m v -m₂ v₂)² / m + m2 v₂²

    m v² = [(m v)² - 2 m m₂ v v₂ + (m₂v₂)²2] /m + m₂ v₂²

    m v² = m v² - 2 m₂ v v₂ + m₂² /m v₂² + m₂ v₂²

    0 = - 2 m₂ v v₂ + v₂² (m₂² /m + m₂)

    2 m₂ v = v₂ (m₂² / m + m₂)

Let's replace the values

    m₂ = 2m

    2 2m v = v₂ (4m² / m + 2m)

    4m v = V₂ (4m + 2m)

     4 v = v₂ 6

.     v₂ = 2/3 v

     v₂ = 2/3 √2gh

Having the velocity of the second (heavier) ball we use energy conservation to find where it goes up

Lowest point

      Em₂2 = K = ½ (2m) v₂²

Highest point

      Em₃ = U = mgh₃

      Em₂ = Em₃

      ½ 2m v₂² = 2m g h₃

       h₃ = ½ v₂² / g

       h₃ = ½ (4/9 2gh)

       h₃ = 4/9 gh

b)    as the shock is elastic, the energy will not be lost at any point for which the balls must return to their highest points of height after each collision in succession,

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