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An object, with mass 32 kg and speed 26 m/s relative to an observer, explodes into two pieces, one 5 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?

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Answer:


\Delta K = 2164.053\,J

Step-by-step explanation:

Let consider the observer as an inertial reference frame. The object is modelled after the Principle of Momentum Conservation:


(32\,kg)\cdot (26\,(m)/(s) ) = (5.333\,kg)\cdot (0\,(m)/(s) )+(26.665\,kg )\cdot v

The speed of the more massive piece is:


v = 31.202\,(m)/(s)

The kinetic energy added to the system is:


\Delta K = (1)/(2)\cdot [(5.333\,kg)\cdot (0\,(m)/(s) )^(2)+(26.665\,kg )\cdot (31.202\,(m)/(s) )^(2)]-(1)/(2)\cdot (32\,kg)\cdot (26\,(m)/(s) )^(2)


\Delta K = 2164.053\,J

User Naman Kashyap
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