Asked by Rohan

An object of mass 9.0 kg is moving due west with a uniform speed of 4.0 m/s. It suddenly explodes into two parts having masses of 6.0 kg and 3.0 kg. Just after the explosion, the 3.0 kg mass has a velocity of 15 m/s due west relative to the Earth's surface. What is the speed of the 6.0 kg mass relative to the Earth's surface?

*If possible, please show the process in how you got the answer!*
Answered by Damon
the momentum after = the momentum before
momentum before = 9 * 4 West + 0 North
momentum after = 3 *15 West + 6 * v West + 0 North
so
36 = 45 + 6 v West
6 v West = - 9
v West = -9/6 = -1.5 West or in other words 1.5 East
Answered by Rohan
Thank you so much for your help!
Answered by Damon
You are welcome.
Answered by henry2,
Given: M1 = 9kg, V1 = -4m/s, M2 = 6kg, V2 = ?, M3 = 3kg, V3 = -15m/s.
Momentum before = Momentum after
M1*V1 = M2*V2 + M3*V3
9*(-4) = 6*V2 + 3*(-15)
-36 = 6V2-45
V2 = 1.5 m/s, East.

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