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An object, with mass 64 kg and speed 14 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame

Sagot :

Answer:

 K_f = 1881.6 J

Explanation:

To solve this exercise, let's start by finding the velocities of the bodies.

We define a system formed by the initial object and its parts, with this the forces during the explosion are internal and the moment is conserved

initial instant. Before the explosion

        p₀ = M v₀

final instant. After the explosion

        p_f = m₁ v + m₂ 0

the moeoto is preserved

         p₀ = p_f

         M v₀ = m₁ v

         v = [tex]\frac{m_1}{M}[/tex]  v₀

in the exercise they indicate that the most massive part has twice the other part

         M = m₁ + m₂

         M = 2m₂ + m₂ = 3 m₂

         m₂ = M / 3

so the most massive part is worth

        m₁ = 2 M / 3

we substitute

        v = ⅔ v₀

with the speed of each element we can look for the kinetic energy

initial

         K₀ = ½ M v₀²

Final

         K_f = ½ m₁ v² + 0

         K_f = ½ (⅔ M) (⅔ v₀)²

         K_f = [tex]\frac{8}{27}[/tex] (½ M v₀²)

         K_f = [tex]\frac{8}{27}[/tex]  K₀

the energy added to the system is

         ΔK = Kf -K₀

         ΔK = (8/27 - 1) K₀

         ΔK = -0.7 K₀

         K_f = K₀ + ΔK

         K_f = K₀ (1 -0.7)

         K_f = 0.3 K₀

let's calculate

         K_f = 0.3 (½ 64 14²)

         K_f = 1881.6 J

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