Solving for the Third Fragment's Speed Using Momentum and Collision Equations

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The discussion focuses on solving for the speed of the third fragment after an explosion involving a 32.0-kg body. Participants emphasize the importance of using conservation of momentum, noting that the total momentum before the explosion must equal the sum of the momenta of the fragments afterward. The momentum equations are set up for both the x and y components, leading to the equations 32*215 = M3x - 4.5*370 and 0 = M3y + 7*310. By calculating the components of momentum for the third fragment and dividing by its mass, the speed can be determined. The conversation highlights the need for clarity in applying vector components to solve the problem correctly.
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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.

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?

If so, then you get a right angle triangle.

You know momentum of both sides and just by using pythagoras you can find
momentum of M3, then just divided by the mass then you can get the velocity right?
 
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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?
You have to subtract the resultant from the original momentum vector of the bullet to get M3.

ehild
 
I'm not really sure what you mean by that, could you explain it again thanks
 
Neerolyte said:
I'm not really sure what you mean by that, could you explain it again thanks

Sorry I did not notice your formula M3 = Mtotal - (M1+M2) before. I wanted to say the same - you should subtract the resultant of M1 and M2 from Mtotal to get M3, and I ment that Mtotal is momentum of the bullet before exploding.

M3 = Mtotal - (M1+M2).

ehild
 
hm...Okay..i think this problem should be solve by components, but can i have some hints on how to use component methods?
 
Use conservation of momentum...?
 
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?
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.
The sum of the x components is the same before and after explosion, and the same holds for the y components.
32*215=M3x-4.5*370
0=M3y+7*310.
Calculate the components of M3. Determine the magnitude and divide by the mass of the third piece to get the speed as you have written down in your first letter.
ehild
 
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