How Does Momentum Conservation Apply in Nuclear Disintegration?

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Homework Help Overview

The discussion revolves around the application of momentum conservation in the context of nuclear disintegration. The original poster presents a scenario involving an unstable nucleus that disintegrates into three particles, with specific masses and velocities provided for two of the particles. The goal is to determine the magnitude and direction of the velocity of the third particle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of momentum and the need to account for vector components in the calculations. There are attempts to derive the velocity of the third particle using momentum equations, but discrepancies in results lead to questions about the methodology used.

Discussion Status

Some participants have provided guidance on the importance of treating velocity as a vector and not merely adding components. There is an ongoing exploration of how to correctly calculate the magnitude of the resultant vector from its components, indicating a productive direction in the discussion.

Contextual Notes

Participants are grappling with the implications of vector addition in the context of momentum conservation, and there is a noted lack of consensus on the correct approach to finding the velocity of the third particle. The original poster expresses confusion over the correct answer provided by others, highlighting the complexity of the problem.

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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