How Does Momentum Conservation Apply in Nuclear Disintegration?

In summary: So you have to use the Pythagorean theorem.a² + b² = c²Where c is the hypotenuse (the magnitude of the vector), and a and b are the components.Does that make sense?So in this case, we have:a = -1.82 x 10^7 m/s (x component)b = -1.82 x 10^7 m/s (y component)c = ? (magnitude of the vector)Using the Pythagorean theorem:a² + b² = c²(-1.82 x 10^7)² + (-1.82 x 10^7)² = c²6.6124 x 10^14
  • #1
riseofphoenix
295
2
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!
 
Physics news on Phys.org
  • #2


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.
 
  • #3


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

-____-
 
  • #4


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!)
 
  • #5


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!
 
  • #6


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.
 

FAQ: How Does Momentum Conservation Apply in Nuclear Disintegration?

What is the definition of conservation of momentum?

The conservation of momentum is a fundamental law in physics that states that the total momentum of a system remains constant unless acted upon by an external force. This means that in a closed system, the total momentum before and after an interaction will be the same.

Why is conservation of momentum important?

The conservation of momentum is important because it helps us understand and predict the behavior of objects in motion. It allows us to determine the outcomes of collisions and interactions between objects, which is crucial in many fields such as engineering, astronomy, and sports.

What are the two types of momentum that are conserved?

The two types of momentum that are conserved are linear momentum and angular momentum. Linear momentum refers to the motion of an object in a straight line, while angular momentum refers to the motion of an object around an axis.

How is conservation of momentum related to Newton's third law of motion?

Conservation of momentum is closely related to Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In other words, when two objects interact, their momentum changes in opposite directions but the total momentum remains constant.

Can the conservation of momentum be violated?

No, the conservation of momentum is a fundamental law of physics and has been extensively tested and proven to hold true in all physical interactions. However, in certain cases such as in quantum mechanics or at very high energies, the concept of momentum conservation may need to be modified, but it still holds true in these scenarios.

Similar threads

How much force does the 20 kg block exert on the 10 kg block?
Replies
17
Views
2K
Velocities of four masses | Conservation of Momentum
Replies
7
Views
527
Conservation of Momentum (elastic collision of masses)
Replies
15
Views
7K
Finding momentum of a photon/ Finding the mass of a object
Replies
8
Views
2K
Finding the Kinetic Energy using the mass, radius and initial KE
Replies
12
Views
2K
Conservation of momentum & energy
Replies
4
Views
2K
Elastic Collision of Hydrogen and Carbon Atoms
Replies
9
Views
2K
Two dimensional momentum problem
Replies
20
Views
3K
Conservation of momentum & energy
Replies
11
Views
1K
Collision Conservation of Energy
Replies
5
Views
1K

Hot Threads

Recent Insights

Back
Top