%3Cp%3EA%20fireworks%20rocket%20is%20fired%20vertically%20upward.%20At%20its%20m

Question

%3Cp%3EA%20fireworks%20rocket%20is%20fired%20vertically%20upward.%20At%20its%20maximum%0Aheight%20of%2090%20m%2C%20it%20explodes%20and%20breaks%20into%20two%20pieces%2C%20one%20with%20a%0Amass%20of%201.45%20kg%20and%20the%20other%20with%20a%20mass%20of%200.22%20kg.%20In%20the%0Aexplosion%2C%20930%20J%20of%20chemical%20energy%20is%20converted%20to%20kinetic%20energy%0Aof%20the%20two%20fragments.%3Cbr%20%2F%3E%0Aa)%20What%20is%20the%20speed%20of%20each%20fragment%20just%20after%20the%0Aexplosion%3F%3Cbr%20%2F%3E%0A%3Cbr%20%2F%3E%0Ab)%20It%20is%20observed%20that%20the%20two%20fragments%20him%20the%20ground%20at%20the%20same%0Atime.%20What%20is%20the%20distance%20between%20the%20points%20on%20the%20ground%20where%0Athey%20land%3F%20Assume%20that%20the%20ground%20is%20level%20and%20air%20resistance%20can%0Abe%20ignored.%3Cbr%20%2F%3E%0A%3Cbr%20%2F%3E%3C%2Fp%3E%0A

Explanation / Answer

a) Use Conservation of Energy. (1/2) { (m1 * v1*2 ) + (m2 * v2^2) = 930J

and Conservation of Momentum m1 * v1 = m2 * v2

To get we put in values of m1 = 1.45kg, m2 = 0.22kg

and substitute v2 = (1.45^2 / 0.22) v1

And get v1 = 13m/s, and v2 = 85.7m/s

b) Since they hit the ground at the same time, and by conservation of momentum, it is reasonable to assume that both the pieces flew horizontally. Therefore there initial velocity in y direction = 0

Therefore 1/2 * a * t^2 = 90m (height)

taking a = 9.8m/s2

We obtain t as 4.29s

Since Pieces flew in Opposite directions, relative velocity in x = (85.7 + 13) = 423m (Ans)

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