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An object, with mass 72 kg and speed 28 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?

Respuesta :

opudodennis

Answer:

14112 J

Explanation:

When the 72 Kg mass explodes into two, one mass is twice the other so 72/3=24 Kg

M1= 24 kg, M2= 72-24=48 kg

From law of conservation of linear momentum, the sum of initial and final momentum are equal. p=mv where p is momentum, m is mass and v is velocity. Fir this case, since the less massive piece stops, its final velocity is zero.

72*28=48v2

V2=72*28/48=42 m/s

Difference between initial and final kinetic energy will be

[tex]\triangle KE= 0.5(Mv^{2}-m2v2^{2})\\\triangle KE= 0.5(72*28^{2}-48*42^{2})=-14112 J[/tex]

Therefore, from observers reference, kinetic energy of 14112 J is added

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