How Does Momentum Conservation Apply in Nuclear Disintegration?

AI Thread Summary
The discussion centers on applying the conservation of momentum to a nuclear disintegration scenario involving three particles. The initial calculations incorrectly assumed that the final velocity of the third particle could be derived by simply summing the components of the other two particles. Participants emphasize the need to treat velocity as a vector quantity, requiring proper vector addition to find the resultant velocity. The correct approach involves calculating the components separately and then using the Pythagorean theorem to determine the magnitude of the third particle's velocity. Ultimately, the correct magnitude of the third particle's velocity is identified as 1.27 x 10^7 m/s, highlighting the importance of vector analysis in momentum conservation problems.
riseofphoenix
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Conservation of Momentum?

An unstable nucleus of mass 2.7 x 10-26 kg, initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of m1 = 1.0 x 10-26 kg, moves in the positive y-direction with speed v1 = 4.8 x 106 m/s. Another particle, of mass m2 = 1.2 x 10-26 kg, moves in the positive x-direction with speed v2 = 3.6 x 106 m/s. Find the magnitude and direction of the velocity of the third particle.

Magnitude: ______ m/s
Direction: ______ ° counterclockwise from the +x-axis

This is what I did...

1) M = m1 + m2 + m3

(2.7 x 10-26) = (1.0 x 10-26) + (1.2 x 10-26) + m3
(2.7 x 10-26) - (1.0 x 10-26) - (1.2 x 10-26) = m3
5.0 x 10-27 = m3

2) Determine your systems

1st particle:
m1 = 1.0 x 10-25 kg
vinitial = 0 m/s
v1final = 4.8 x 106 m/s

2nd particle:
m2 = 1.2 x 10-26 kg
vinitial = 0 m/s
v2final = 3.6 x 106 m/s

3rd particle:
m3 = 5.0 x 10-27 kg
Vinitial = 0 m/s
V3final = ? m/s

3) Conservation of momentum equation

pinitial = pfinal

m1vinitial + m2vinitial + m3vinitial = m1v1final + m2v2final + m3v3final
(m1)(0) + (m2)(0) + (m3)(0) = (1.0 x 10-25)(4.8 x 106) + (1.2 x 10-26)(3.6 x 106) + (5.0 x 10-27)(v3final)
0 = (4.8 x 10-20) + (4.32 x 10-20) + (5.0 x 10-27)(v3final)
-(9.12 x 10-20) = (5.0 x 10-27)(v3final)
-18240000 = v3final
-1.82e+07 = v3final

"INCORRECT. CORRECT ANSWER IS: 1.27e+07"

I don't understand! How did they get 1.27e+07? Help!
 
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riseofphoenix said:
An unstable nucleus of mass 2.7 x 10-26 kg, initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of m1 = 1.0 x 10-26 kg, moves in the positive y-direction with speed v1 = 4.8 x 106 m/s. Another particle, of mass m2 = 1.2 x 10-26 kg, moves in the positive x-direction with speed v2 = 3.6 x 106 m/s. Find the magnitude and direction of the velocity of the third particle.

Magnitude: ______ m/s
Direction: ______ ° counterclockwise from the +x-axis

This is what I did...

1) M = m1 + m2 + m3

(2.7 x 10-26) = (1.0 x 10-26) + (1.2 x 10-26) + m3
(2.7 x 10-26) - (1.0 x 10-26) - (1.2 x 10-26) = m3
5.0 x 10-27 = m3

2) Determine your systems

1st particle:
m1 = 1.0 x 10-25 kg
vinitial = 0 m/s
v1final = 4.8 x 106 m/s

2nd particle:
m2 = 1.2 x 10-26 kg
vinitial = 0 m/s
v2final = 3.6 x 106 m/s

3rd particle:
m3 = 5.0 x 10-27 kg
Vinitial = 0 m/s
V3final = ? m/s

3) Conservation of momentum equation

pinitial = pfinal

m1vinitial + m2vinitial + m3vinitial = m1v1final + m2v2final + m3v3final
(m1)(0) + (m2)(0) + (m3)(0) = (1.0 x 10-25)(4.8 x 106) + (1.2 x 10-26)(3.6 x 106) + (5.0 x 10-27)(v3final)
0 = (4.8 x 10-20) + (4.32 x 10-20) + (5.0 x 10-27)(v3final)
-(9.12 x 10-20) = (5.0 x 10-27)(v3final)
-18240000 = v3final
-1.82e+07 = v3final

"INCORRECT. CORRECT ANSWER IS: 1.27e+07"

I don't understand! How did they get 1.27e+07? Help!
Velocity and momentum are vector quantities.

You need to keep track of vector components.
 


SammyS said:
Velocity and momentum are vector quantities.

You need to keep track of vector components.

I STILL get the same answer though...

0 = (4.8 x 10-20) + (4.32 x 10-20) + (5.0 x 10-27)(v3final)

x component


0 = [STRIKE](4.8 x 10-20)[/STRIKE] + (4.32 x 10-20) + (5.0 x 10-27)(v3final)
0 = (4.32 x 10-20) + (5.0 x 10-27)(v3final)
-(4.32 x 10-20) = (5.0 x 10-27)(v3final)
-(4.32 x 10-20)/(5.0 x 10-27) = v3final
-8.64 x 106= v3final in the x direction

y component


0 = (4.8 x 10-20) + [STRIKE](4.32 x 10-20)[/STRIKE] + (5.0 x 10-27)(v3final)
0 = (4.8 x 10-20) + (5.0 x 10-27)(v3final)
-(4.8 x 10-20) = (5.0 x 10-27)(v3final)
-(4.8 x 10-20)/(5.0 x 10-27) = v3final
-9.6 x 106 = v3final in the y direction

Add vectors

-8.64 x 106 - 9.6 x 106 = -18240000 = -1.82 x 10^7 m/s

-____-
 


riseofphoenix said:
Add vectors

-8.64 x 106 - 9.6 x 106 = -18240000 = -1.82 x 10^7 m/s
How do you find the magnitude of a vector given its components? (You don't just add the components!)
 


Doc Al said:
How do you find the magnitude of a vector given its components? (You don't just add the components!)

But I just did!
 


Yeah, he's saying it's wrong to just add the components.

Consider the vector and it's components as a triangle (I assume you're familiar with this representation). You're trying to the find the length of the hypotenuse.
 

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