Elastic Collisions in Two Dimensions

Elastic Collisions in Two Dimensions

In center of mass frame

Two dimensional collision in CM frame

Consider a system of two particles having masses\(m_{1}\) and\(m_{2}\) moving with velocities\(u_{1}\) and\(u_{2}\) respectively in lab frame. Thus, the velocity of CM in lab frame is,\begin{align*} u=\frac{m_{1}u_{1}+m_{2}u_{2}}{m_{1}+m_{2}}\end{align*} Let, the target is at rest ie. \(u_{2}=0\)\begin{align*} u=\frac{m_{1}u_{1}}{m_{1}+m_{2}}\end{align*} Thus, the CM moves with uniform velocity and can be considered as an inertial frame. The velocity of particles in CM frame are, \begin{align*} u_{1}^{'}=u_{1}-u \end{align*} and \begin{align*} u_{2}^{'}=u_{2}-u \end{align*} Momentum of centre of mass is, \begin{align*} Mu^{'}=m_{1}u'_{1}+m_{2}u'_{2}\end{align*}\begin{align*} or, Mu^{'}=m_{1}(u_{1}-u)+m_{2}(u_{2}-u)\end{align*} Since, \(u_{2}=0\), \begin{align*} Mu^{'}=m_{1}u_{1}-(m_{1}+m_{2})u\end{align*} But from above value of u, \begin{align*} Mu^{'}=m_{1}u_{1}-(m_{1}+m_{2})\frac{m_{1}u_{1}}{m_{1}+m_{2}} \end{align*} \begin{align*} or, Mu^{'}=0 \end{align*} Hence, the total momentum of the system is 0 when measured with respect to the centre of mass. Thus, it is also known as the zero momentum frame of reference.

Let, \(v'_{1}\) and \(v'_{2}\) be the velocities of the particle after collision, relative to the centre of mass. Then, it is true that, \begin{align*} m_{1}v'_{1}+m_{2}v'_{2}=0\end{align*} Also, from KE conservation,\begin{align*} \frac{1}{2}m_{1}{u'}_{1}^{2}+\frac{1}{2}m_{2}{u'}_{2}^{2}=\frac{1}{2}m_{1}{v'}_{1}^{2}+\frac{1}{2}m_{2}{v'}_{2}^{2}\end{align*} If we solve these equations, we get \(|u'_{1}|=|v'_{1}|\) and\(|u'_{2}|=|v'_{2}|\) Thus, in an ealstic collision in two dimensions in CM frame of reference, the magnitude of final velocity of the colliding bodies is same as their initial velocity.

In lab frame

Let, a body \(m_{1}\) moving with velocity\(u_{1}\) collide elastically with another body\(m_{2}\) at rest. After collision, let\(m_{1}\) move with velocity\(v_{1}\) making an angle\(\theta_{1}\) with the initial direction. Also,\(m_{2}\) gets scattered making an angle\(\theta_{2}\) with the initial direction of motion of\(m_{1}\) with velocity\(v_{2}\). Since, momentum is conserved,\begin{align*} m_{1}u_{1}=m_{1}v_{1}+m_{2}v_{2}\end{align*} Resolving into components, along X-axis, we get,\begin{align*} m_{1}u_{1}=m_{1}v_{1}\cos\theta_{1}+m_{2}v_{2}\cos\theta_{2}\end{align*} and along Y-axis, we get,\begin{align*} 0=m_{1}v_{1}\sin\theta_{1}-m_{2}v_{2}\sin\theta_{2}\end{align*} From KE conservation,\begin{align*} \frac{1}{2}m_{1}u_{1}^2=\frac{1}{2}m_{1}v_{1}^2+\frac{1}{2}m_{2}v_{2}^2\end{align*} Since, we have three equations and four unknowns, one of the four values must be provided to solve it. Thus in lab frame, solution of two dimensional collision is difficult. Thus, to describe elastic collisions in two dimensions, CM frame is used.

Value of Scattering Angle

In centre of mass frame of reference, there are absolutely no limitations on the value of scattering angle. It can have any possible values.

For lab frame of reference,

Consider, in CM frame, \(\theta\) be the scattering angle of \(m_{1}\) and\(\pi-\theta\) be he scattering angle of \(m_{2}\) while in lab frame their scattering angle be respectively \(\theta_{1}\) and\(\theta_{2}\). Now, velocity of CM in lab frame is,\begin{align*} v_{c}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}\end{align*} We have \(v_{1}^{'}=v_{1}-v_{c}\) and we can write \(\tan\theta_{1}=\frac{v_{1}\sin\theta_{1}}{v_{1}\cos\theta_{1}}\). Since there is no vertical component of \(v_{c}\), \(v'_{1}\sin\theta=v_{1}\sin\theta_{1}\). Also, we can write \(v_{1x}^{'}=v_{1x}-v_{c}\) which impies \(v_{1}\cos\theta_{1}=v_{1}^{'}\cos\theta+v_{c}\) So our equation becomes\(\tan\theta_{1}=\frac{\sin\theta}{\cos\theta+\frac{v_{c}}{v'_{1}}}\). Since, the velocity of CM does not change on collision, we get \(v_{c}=(\frac{m_{1}}{m_{2}})v'_{1}\). Then, we can get from the above relation,\(\tan\theta_{1}=\frac{\sin\theta}{\cos\theta+\frac{m_{1}}{m_{2}}}\).

Special Cases

If \(m_{1}>m_{2}\), \(\theta_{1}\) must be lesser than 90 degrees. Thus, in an elastic collision, if a massive body collides against a lighter one, it can never bounce back along its original path.

If\(m_{1}=m_{2}\),\(\theta_{1}\) can have any possible values upto the limiting value of 90 degrees. Thus, in an elastic collision between two bodies of equal mass, one of which is initially at rest, the two bodies can move at right angles to each other after collision.

If\(m_{1}<m_{2}\),\(\theta_{1}\) can have any possible values. Therefore, in an elastic collision between two bodies of which the heavier is at rest, the lighter body may bounce back.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.