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smokeyd smokeyd
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11 years ago
In an explosion, a 9 kg object which was initially at rest, spilts into three parts which all have equal mass. Two of the parts move apart with equal speeds of 12m/s and an angle of 50 degrees between their directions of motion. How much kinetic energy is released in this explosion?
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wrote...
11 years ago
Write the total kinetic energy in terms of the momentums;
K = K1 + K2 + K3
K = P1^2/2m + P2^2/2m + P3^2/2m

Since the bomb was initially at rest you have the vector conservation of momentum eq;
P1 + P2 + P3 = 0
P3 = -P1 - P2

Square both sides (using dot product)
P3^2 = P1^2 + P2^2 - 2IP1IIP2ICos(50)

You have all the info. to calculate IP1I & IP2I and thus P1^2 & P2^2 and thus P3^2 which you can then put in the eq for K

EDIT_________________________ _____________________________ ____________
That last term on the right should be +2IP1IIP2ICos(50) , not minus. Sorry about that.
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