View Full Version : Momentum in explosions
mindhater
Dec3-03, 06:47 PM
Need help on this problem - exactly as stated
Homer the human powder keg, initially at rest, suddenly explodes into three pieces each with equal mass. One piece moves east at 30 m/s and a second piece moves at 30 m/s southeast.
Caculate the speed of the third piece and calculate the direction in which it travels
Thanks for the help
I'm going to go with 55.4m/s at 22.5degrees North of East.
Hopefully someone can verify that answer.
i agree with the direction but the speed wud be equal would it not
Adrian Baker
Dec4-03, 03:47 AM
Originally posted by FUNKER
i agree with the direction but the speed wud be equal would it not
Why? Momentum is conserved, not 'speed'. Before the explosion momentum was zero. It must be afterwards too.
selfAdjoint
Dec4-03, 08:17 AM
Since the three pieces had equal mass, you can factor the mass out of momentum and just use speed, if you want. The point is that the vector sum of the three vectors has to be 0, and so you have to do some trig.
whatgravity
Dec4-03, 09:51 AM
yeah, add the x components up = 8.8 m/s East
add the y components up = 21.2 m/s South
so 8.8 m/s West and 21.2 m/s North will give you the opposite vector, thereby conserving momentum
answer = 22.95 m/s @ 22.5 deg West of North [a)]
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.