How Does Momentum Conservation Apply in Nuclear Disintegration?

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SUMMARY

The discussion focuses on the application of momentum conservation in nuclear disintegration, specifically analyzing a nucleus of mass 2.7 x 10-26 kg that breaks into three particles. The first particle (mass 1.0 x 10-26 kg) moves at 4.8 x 106 m/s in the positive y-direction, while the second particle (mass 1.2 x 10-26 kg) moves at 3.6 x 106 m/s in the positive x-direction. The third particle's velocity was incorrectly calculated as -1.82 x 107 m/s, while the correct magnitude is 1.27 x 107 m/s, highlighting the importance of vector component analysis in momentum calculations.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Familiarity with vector quantities and their components
  • Basic knowledge of nuclear physics and particle disintegration
  • Ability to perform calculations involving mass and velocity
NEXT STEPS
  • Study vector addition and how to calculate the magnitude of a vector from its components
  • Learn about the principles of nuclear decay and particle interactions
  • Explore momentum conservation in multi-particle systems
  • Review examples of momentum conservation in different physical scenarios
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Physics students, educators, and professionals interested in nuclear physics, momentum conservation, and vector analysis in particle dynamics.

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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