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An object, with mass m and speed v relative to an observer, explodes into two pieces, one three 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? (Use any variable or symbol stated above as necessary.)

Respuesta :

Answer:

a) [tex]K_{f}[/tex] / K₀ = 27/64

Explanation:

We will solve this problem with the expressions of the moment. Let us form a system that contains the initial stone and the two final fragments, all the forces are internal and the moment is preserved.

Moment before the explosion

    p₀ = M v

After the explosion

    Mt = 3m + m = 4m

    [tex]p_{f}[/tex] = (3m) vf + m (0)

    p₀ =  [tex]p_{f}[/tex]

    M v = (3m)  [tex]v_{f}[/tex]

    [tex]v_{f}[/tex] = v 3m / M

     [tex]v_{f}[/tex] = v 3m / 4m

    [tex]v_{f}[/tex] = ¾ v

Having the speed of the stones before and the fragments we can calculate the kinetic energy

Before explosion

    K₀ = ½ M v²

    K₀ = ½ 4m v²

    K₀ = 2 mv²

After explosion

   [tex]K_{f}[/tex] = ½ (3m) [tex]v_{f}[/tex]² + 0

   [tex]K_{f}[/tex] = ½ (3m) (¾ v)²

  [tex]K_{f}[/tex] = 27/32 m v²

The relationship between energy is the division of them

   [tex]K_{f}[/tex] / K₀ = 27/32 mv2 / (2 mv2)

  [tex]K_{f}[/tex] / K₀ = 27/64

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