ENERGY AND MOMENTUM: Subatomic Collisions, Billiard Balls, and the 90-Degree Rule ( Part 1 )

An elastic collision, within which kinetic energy ( KE ) and momentum ( p ) are conserved, is mathematically modeled in terms of momentum as follows: 

p_i = p_f , and m1v1_i + m2v2_i = m1v1_f + m2v2_f 

where p = mv, and m = mass in kilograms ( kg ), and velocity ( v ) is expressed in units of meters per second ( m/s ). In this system of two equal masses, mass ( m1 ) is initially moving with a constant velocity ( v1_i ), and it collides with a mass ( m2 ) that is originally at rest ( v2_i = 0 ). Therefore, the total initial momentum of the system before the collision occurs is as follows: 

p_i = m1v1_i + ( 0 )

IF BOTH MASSES ARE ASSUMED TO BE LOCATED ALONG THE X-AXIS OF A COORDINATE SYSTEM WITH ( m1 ) MOVING ALONG X, the initial system’s momentum ( p_i ) has an x-component, but its y-component = 0. Thus, the final system’s momentum p_f = p_{x initial} , and the sum of its p_y components must = 0.

If the collision in question is not a perfect head-on collision, the projectile and target masses will exit the collision with an angle between the paths they travel. Since the projectile initially travelled along the x-axis, we may plot the emerging ( m1 ) mass as travelling at some angle θ₁ above the x-axis, and we may plot the emerging ( m2 ) mass as travelling at another angle θ₂ below the x-axis. The x-components of motion for ( m1 ) and ( m2 ) = ( v cos θ ), and their y-components = ( v sin θ ). A review of the basic trigonometric functions is located here.

The initial and final kinetic energy of a system within which a collision occurs, KE_i = KE_f , may be represented as follows: 

½ m1v1_i² + ½ m2v2_i² = ½ m1v1_f² + ½ m2v2_f²  

When the target of mass ( m2 ) is stationary within the initial system ( v2_i = 0 ) , the conservation of momentum equation may be rewritten to represent the system’s energetics as follows: 

½ m1v1_i² = ½ m1v1_f² + ½ m2v2_f² + mv1_fv2_fcos( θ₁ – θ₂ )

Angle θ₁ is an angle that is positive with respect to the x-axis, and it is associated with a vector that is positioned diagonally within the first quadrant of a coordinate plane. The second angle, θ₂ , is negative with respect to the x-axis, and it is associated with a vector that is positioned diagonally within the fourth quadrant of a coordinate plane.  

Q: A particle accelerator is used to launch a projectile ( m1 ) with a constant velocity ( v1_i ) into a region within which a target of similar mass is at rest. If m1 emerges from the collision at +45 degrees to the x-axis, and m2 emerges at a -45 degree angle below the x-axis, what is the net angular separation of the two particles?

A: +45 degrees – ( – 45 degrees ) = 90 degrees

The 90-Degree Rule

Recall that KE_i and KE_f of an elastic collision is represented mathematically as follows:

½ m1v1_i² + ½ m2v2_i² = ½ m1v1_f² + ½ m2v2_f² 

Nonetheless, the momentum equation’s kinetic energy derivation of such a system is as follows:

½ m1v1_i² = ½ m1v1_f² + ½ m2v2_f² + mv1_fv2_fcos( θ₁ – θ₂ )

THERE ARE THREE CONDITIONS UNDER WHICH the mv1_fv2_fcos( θ₁ – θ₂ ) term is not applicable to the circumstance:

1. v1_f = 0. This occurs when there is a head-on collision between projectile and target. This would be represented by an initial condition in which both target and projectile lie along the x-axis, and after the collision occurs, the previously motionless target moves straight forward in the x-direction and the projectile stops moving and stays at rest on the x-axis.

2. v2_f = 0. No collision occurs. The projectile misses its mark or the target doesn’t exist. This scenario can occur when physics majors go to a bar and attempt to apply quantum tunneling to a friendly game of pool.

3. cos( θ₁ – θ₂ ) = 0. Under this circumstance, the difference in angles after the collision occurs is 90 degrees, and the previous equation reduces to ½ m1v1_i² = ½ m1v1_f² + ½ m2v2_f² .

The second part of this discussion will involve circumstances where a projectile emerges from a collision at an angle that is known. The masses of the emerging projectile ( m1 ) and target ( m2 ) differ, however. The angle and velocity with which the target emerges from the collision will be unknown, and we’ll use the aforementioned mathematical methods to solve for our unknown. The differences in mass would make measurements of the internal kinetic energy of the system slightly decrease, and therefore, our collision would be inelastic. Nonetheless, the observation that billiard ball and subatomic collisions oftentimes give rise to objects traveling with 90-degrees of separation validates the approximation that such collisions are indeed elastic.