The discussion focuses on calculating the speed of the third fragment resulting from the explosion of a 32.0-kg body moving at 215 m/s. The first fragment, weighing 7.0 kg, moves at 310 m/s along the positive y-axis, while the second fragment, weighing 4.5 kg, moves at 370 m/s along the negative x-axis. To find the speed of the third fragment, one must apply the conservation of momentum principle, calculating the momentum vectors of the first two fragments and subtracting their resultant from the initial momentum of the body.
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A 32.0-kg body is moving in the direction of the positive x-axis with a speed of 215 m/s when, owing to an internal explosion, it breaks into three pieces. One part, whose mass is 7.0 kg, moves away from the point of explosion with a speed of 310 m/s along the positive y axis. A second fragment, whose mass is 4.5 kg, moves away from the point of explosion with a speed of 370 m/s along the negative x axis. What is the speed of the third fragment? Ignore effects due to gravity.
You have to subtract the resultant from the original momentum vector of the bullet to get M3.Neerolyte said:Can't seem to get this right
M1 = 7.0kg
M2 = 4.5kg
M3 = Mtotal - (M1+M2)
Isn't it simply adding up momentum vector of M1, and momentum vector of M2 then the
resultant will be momentum of M3?
Neerolyte said:I'm not really sure what you mean by that, could you explain it again thanks
The total momentum which is equal to the momentum the bullet had just before the explosion is 32*215i as the piece moves along the positive x axis. The momentum of the 7 kg piece is 7*310j as the piece moves along the positive y axis. The momentum of the 4.5 kg piece is -4.5*370i.Neerolyte said:hm...Okay..i think this problem should be solve by components, but can i have some hints on how to use component methods?