Calculate the speed and angle of the center of mass before a collision

[rdemyan]

Homework Statement

center of mass problem

Relevant Equations

Provided in post

TL;DR Summary: Calculate the speed and angle of the center of mass before a collision

I want to calculate the speed and angle of the center of mass (CM) before a collision for a two dimensional problem.

The direction of the CM will be inferred from a drawing or from the calculated angle. So as shown in the drawing - two particles collide at an angle of $2\beta$. The x-axis exactly bisects this collision angle so the angle of interest is $\beta$. The masses of the particles are $m_1,m_2$ and the velocities are $u_1,u_2$.

So, the way I see calculating the speed of the CM is to calculate the momentum in the x direction and y direction. Then use the pythagorean theorem to get the momentum along the angle of the CM and divide by the sum of the masses to get the speed. So,
$$u_{cm} = \frac{\sqrt{(m_1u_1cos\beta + m_2u_2cos\beta)^2 +(m_1u_1sin\beta - m_2u_2sin\beta)^2}}{m_1+m_2}$$
And the angle of the CM relative to the x axis is given by
$$ \alpha = tan^{-1}(\frac{m_1u_1\sin\beta - m_2u_2\sin\beta}{m_1u_1\cos\beta + m_2u_2\cos\beta})$$
Am I looking at this correctly and are the equations correct for this particular situation?

View attachment 359396

 

[haruspex]

Homework Statement: center of mass problem
Relevant Equations: Provided in post

TL;DR Summary: Calculate the speed and angle of the center of mass before a collision

I want to calculate the speed and angle of the center of mass (CM) before a collision for a two dimensional problem.

The direction of the CM will be inferred from a drawing or from the calculated angle. So as shown in the drawing - two particles collide at an angle of $2\beta$. The x-axis exactly bisects this collision angle so the angle of interest is $\beta$. The masses of the particles are $m_1,m_2$ and the velocities are $u_1,u_2$.

So, the way I see calculating the speed of the CM is to calculate the momentum in the x direction and y direction. Then use the pythagorean theorem to get the momentum along the angle of the CM and divide by the sum of the masses to get the speed. So,
$$u_{cm} = \frac{\sqrt{(m_1u_1cos\beta + m_2u_2cos\beta)^2 +(m_1u_1sin\beta - m_2u_2sin\beta)^2}}{m_1+m_2}$$
And the angle of the CM relative to the x axis is given by
$$ \alpha = tan^{-1}(\frac{m_1u_1\sin\beta - m_2u_2\sin\beta}{m_1u_1\cos\beta + m_2u_2\cos\beta})$$
Am I looking at this correctly and are the equations correct for this particular situation?

View attachment 359396

Looks good to me.

 

[rdemyan]

Two other points I'd like to confirm.

1)The velocity of the center of mass is constant and the same regardless of the value of the coefficient of restitution.
2) Also, the angle at which the center of mass moves is a constant value for the system (assuming no external forces) regardless of the value of the coefficient of restitution.

Are those statements true?

 

[jbriggs444]

1)The velocity of the center of mass is constant and the same regardless of the value of the coefficient of restitution.
2) Also, the angle at which the center of mass moves is a constant value for the system (assuming no external forces) regardless of the value of the coefficient of restitution.

Are those statements true?

Yes and yes.

In the absence of any external forces or mass transfers, the velocity of the center of mass will be constant. This includes both its magnitude (speed) and its direction.

There is a tiny exception in the case of the zero velocity. The zero velocity has no angle.