Finding the Line of Motion of Two Particles

[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 !

 

[TSny]

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

 

[tanaygupta2000]

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?

 

[tanaygupta2000]

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

 

[TSny]

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.

 

[TSny]

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?

 

[tanaygupta2000]

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]

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)

 

[TSny]

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

Looks good.

 

[TSny]

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.)

 

[tanaygupta2000]

total angular momentum, L'(Total)

I guess L'(Total) = L(Total) + L(C.M.) ?

 

[TSny]

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).

 

[tanaygupta2000]

So is the correct answer option - (e) ?

 

[TSny]

So is the correct answer option - (e) ?

Maybe. Why do you think the answer is (e)?

 

[tanaygupta2000]

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).

 

[TSny]

That sounds good.

 

[tanaygupta2000]

Thank You so much for help !