# Finding the Line of Motion of Two Particles

• tanaygupta2000
In summary, the equation of the line of motion of the two particles is given by the dot product of their respective center of mass coordinates. The angular momentum of the center of mass is found by adding the angular momentum of each particle in the frame of reference.
tanaygupta2000
Homework Statement
Two particles of masses m1 and m2, interact through a central force potential V(r). At t=0, their position and velocity vectors are given by r1=(0,0,a), r2=(a,0,0), v1=(b,2b,0) and v2=(0,0,3b), where a and b are constants. If m2=2m1, which of the following vectors is perpendicular to the plane of motion?
(a) (2,5,0)
(b) (2,-1,-1)
(c) (-1,2,1)
(d) (1,-3,2)
(e) (1,1,1)
Relevant Equations
Position of center of mass, R = (m1r1+m2r2)/(m1+m2)
Velocity of center of mass, V = (m1v1+m2v2)/(m1+m2)
I know that I need to find the equation of the line of motion of the two particles, the dot product of which with one of the options will give 0.
I began with founding the coordinates of center of mass:
R = (m1r1+m2r2)/(m1+m2) = (2a/3, 0, a/3)

and velocity of the center of mass:
V = (m1v1+m2v2)/(m1+m2) = (b/3, 2b/3, 2b)

the mass of the center of mass is given by, m* = m1m2/(m1+m2) = 2m1/3
Now applying law of conservation of momentum, m*V = m1v1 + m2v2
I am getting simply (2b/9, 4b/9, 4b/3) = (b, 2b, 6b)
which doesn't seem to be useful.

Kindly help me in solving this. Thanks !

Think about the orientation of the total angular momentum vector in the center-of-mass reference frame.

TSny said:
Think about the orientation of the total angular momentum vector in the center-of-mass reference frame.
If we consider m1 = m
m2 = 2m1 = 2m, then
For first particle, L1 = r1×m1v1 = -mab(2, -1, 0)
For second particle, L2 = r2×m2v2 = -mab(0, 6, 0)
So L(Total) = L1+L2 = -mab(2, 5, 0) which lies in the XY plane, while r1 and r2 lie in the XZ plane

So how should I approach to find values of a and b?

For center of mass,
L(c.m.) = R×m*V = -4mab/27 (1,3,-2)

tanaygupta2000 said:
If we consider m1 = m
m2 = 2m1 = 2m, then
For first particle, L1 = r1×m1v1 = -mab(2, -1, 0)
For second particle, L2 = r2×m2v2 = -mab(0, 6, 0)
So L(Total) = L1+L2 = -mab(2, 5, 0) which lies in the XY plane, while r1 and r2 lie in the XZ plane
These results look good

So how should I approach to find values of a and b?
You will not need to know the values of a and b.

tanaygupta2000 said:
For center of mass,
L(c.m.) = R×m*V = -4mab/27 (1,3,-2)
I get a different result for L(c.m.)

The two particles move in a plane when viewed from the center-of-mass reference frame.

How can you find the total angular momentum in the center-of-mass reference frame, L'(Total)?

How is L'(Total) oriented relative to the plane of motion?

TSny said:
I get a different result for L(c.m.)
My apologies, till now I was considering that the formula for mass of the C.M. is given by, m* = m1m2/(m1+m2)
while it was simply (m1 + m2)
With this in mind, I got the angular momentum of the center of mass as:
L(C.M.) = -mab/3 (2, 11, -4)

tanaygupta2000 said:
L(Total) = L1+L2 = -mab(2, 5, 0) which lies in the XY plane, while r1 and r2 lie in the XZ plane
I think the correct answer is option - (a)

tanaygupta2000 said:
I got the angular momentum of the center of mass as:
L(C.M.) = -mab/3 (2, 11, -4)
Looks good.

tanaygupta2000 said:
I think the correct answer is option - (a)
You need to consider the total angular momentum, L'(Total), in the center-of-mass frame of reference. In this frame, the particles move in a fixed plane. In the frame of reference for which the particles have the initial positions and velocities stated in the problem, the particles do not move in a plane.

There is an easy way to find L'(Total) from L(Total) and L(C.M.)

TSny said:
total angular momentum, L'(Total)
I guess L'(Total) = L(Total) + L(C.M.) ?

tanaygupta2000 said:
I guess L'(Total) = L(Total) + L(C.M.) ?
Not quite. The correct statement is L(Total) = L'(Total) + L(C.M.).

If you are not familiar with this theorem, then you can find L'(Total) by switching to the CM frame. That is, find the velocity of each particle in the CM frame and the position of each particle relative to the CM location. Then use these velocities and positions to calculate L'(Total).

So is the correct answer option - (e) ?

tanaygupta2000 said:
So is the correct answer option - (e) ?
Maybe. Why do you think the answer is (e)?

TSny said:
Maybe. Why do you think the answer is (e)?
As the Total angular momentum of the C.M. frame L'(Total) is a scalar multiple of (1, 1, 1) and being a cross product, it is perpendicular to the position vectors of the two particles and hence the line joining their centres (line of motion, since a central potential), as asked in the question (according to my understanding).

That sounds good.

tanaygupta2000
Thank You so much for help !

## 1. What is the line of motion of two particles?

The line of motion of two particles refers to the path that the particles take as they move through space. It is the trajectory that the particles follow as they move from one point to another.

## 2. How is the line of motion of two particles calculated?

The line of motion of two particles can be calculated using mathematical equations, such as the equations of motion or vector calculus. These equations take into account the initial positions, velocities, and accelerations of the particles to determine their path.

## 3. What factors can affect the line of motion of two particles?

The line of motion of two particles can be affected by various factors, such as the initial conditions of the particles (position, velocity, acceleration), external forces acting on the particles, and the properties of the medium in which the particles are moving (e.g. air resistance, friction).

## 4. Can the line of motion of two particles be predicted accurately?

The line of motion of two particles can be predicted accurately if all the relevant factors and initial conditions are known and accounted for in the calculations. However, uncertainties and external factors can also affect the accuracy of the prediction.

## 5. How is the line of motion of two particles used in scientific research?

The line of motion of two particles is used in various scientific fields, such as physics, chemistry, and biology, to study the behavior and interactions of particles. It can also be used to model and simulate real-world phenomena, such as the movement of planets in space or the behavior of molecules in a chemical reaction.

• Introductory Physics Homework Help
Replies
14
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
984
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
691
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
3K