Collisions in Two Dimensions

Fig. 1 also depicts the collision of a moving mass m1 with the stationary mass m2. Linear momentum is conserved in such a collision. Since momentum is a vector this implies three equations for the three directions {x, y, z}. Consider the plane determined by the final velocity directions of m1 and m2 and choose it to be the x-y plane. 

Fig.1: Collision of mass $m_1$, with a stationary mass $m_2$

The conservation of the z-component of the linear momentum implies that the entire collision is in the x-y plane. The x- and y-component equations are

$m_1v_{1i}=m_1v_{if} \ cos \ \theta_1+m_2v_{2f} \ cos \ \theta_2$

$0=m_1v_{if} \ sin \ \theta_1-m_2v_{2f} \ sin \ \theta_2$

One knows ($m_1$ , $m_2$ , $v_{1i}$) in most situations. There are thus four unknowns ($v_{1f}$ , $v_{2f}$ , θ$_1$ and θ$_2$), and only two equations. If θ$_1$ = θ$_2$ = 0, we regain Eq. (6.24) for one dimensional collision. If, further the collision is elastic,

$\frac{1}{2}m_1v_{1i}^2=\frac{1}{2}m_1v_{if}^2+\frac{1}{2}m_2v_{2f}^2$

We obtain an additional equation. That still leaves us one equation short. At least one of the four unknowns, say θ$_1$, must be made known for the problem to be solvable. For example, θ1 can be determined by moving a detector in an angular fashion from the x to the y axis. Given ($m_1$, $m_2$, $v_{1i}$, θ$_1$) we can determine ($v_{1f}$ , $v_{2f}$ , θ$_2$) from Eqs. (6.29)-(6.31).

Example 1

Consider the collision depicted in Fig. 6.10 to be between two billiard balls with equal masses $m_1 = m_2$. The first ball is called the cue while the second ball is called the target. The billiard player wants to ‘sink’ the target ball in a corner pocket, which is at an angle θ$_2$ = 37°. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain θ$_1$.

Answer 

From momentum conservation, since the masses are equal

$\mathbf{v_{1i}}=\mathbf{v_{1f}}+\mathbf{v_{2f}}$

or 

$v_{1i}^2=(\mathbf{v_{1f}}+\mathbf{v_{2f}}).(\mathbf{v_{1f}}+\mathbf{v_{2f}})$

      $=v_{1f}^2+v_{2f}^2+2\mathbf{v_{1f}}.\mathbf{v_{2f}}$

      $=v_{1f}^2+v_{2f}^2+2v_{1f}v_{2f} \cos \ (\theta_1+37^0)$

Since the collision is elastic and $m_1 = m_2$ it follows from conservation of kinetic energy that

$v_{1i}^2=v_{1f}^2+v_{2f}^2$

Comparing Eqs. (6.32) and (6.33), we get

$cos \ (\theta_1+37^0)=0$

or $\theta_1+37^0=90^0$

Thus, $\theta_1=53^0$

This proves the following result : when two equal masses undergo a glancing elastic collision with one of them at rest, after the collision, they will move at right angles to each other. 

The matter simplifies greatly if we consider spherical masses with smooth surfaces, and assume that collision takes place only when the bodies touch each other. This is what happens in the games of marbles, carrom and billiards.

In our everyday world, collisions take place only when two bodies touch each other. But consider a comet coming from far distances to the sun, or alpha particle coming towards a nucleus and going away in some direction. Here we have to deal with forces involving action at a distance. Such an event is called scattering. The velocities and directions in which the two particles go away depend on their initial velocities as well as the type of interaction between them, their masses, shapes and sizes.