The Problem:

A particle of mass 4.0 kg, initially moving with a velocity of 2.0 m/s collides elastically with a particle of mass 6.0 kg. initially moving with a velocity of -4.0 m/s. What are the velocities of the two particles after the collision?

- m
_{1 }= 4.0 kg - m
_{2}= 6.0 kg - v
_{1_before }= 2.0 m/s - v
_{2_before}= -4.0 m/s - v
_{1_after }= ? - v
_{2_after}= ?

Here, both m_{1} and m_{2} have initial
velocities. Three solutions are given below.

- Using the Conservation Laws
- Remembering the General Solution
- Using the Center of Mass Reference Frame

Look them over, and compare.

Solution 1: Using the Conservation Laws:

Although it does simplify the algebra somewhat, it isn't necessary to switch reference frames (as in solution 2 below) to solve the problem using the Conservation laws. Here, we don't.

Conservation of Momentum says:

Here, I've done a few of things that I hardly ever recommend. First, I have substituted numerical values very early in the solution process. In this case, if you want to solve the problem in general, you might as well do it once and remember the solution. Secondly, I have ignored units in equation 1. All of the units in the problem are standard MKS units, so that we know that the velocities will come out in meters/second, and deleting the units makes the equation look a little more "algebra-like" in order to facilitate the solution. Third, I've dropped the "after" part of the subscripts - I just don't want to have to write it over and over, so don't get confused.

Conservation of Kinetic Energy says:

Here again, values have been substituted and "after" and units have been dropped in order to make equation 2 look more "algebra-like".

Equations 1 and 2 are a set of 2 equations and 2 unknowns, and
there are many methods for solving for v_{1_after} and
v_{2_after}. Suppose we solve equation 1 for v_{1},
then substitute into equation 2.

This last equation is a quadratic in v_{2}, so put it into
standard form:

which can be solved routinely using the quadratic formula:

Physically, the solution v_{2_after} = -4.0 m/s is a
"miss" instead of a collision. The solution v_{2_after} = 0.8
m/s is the one we're looking for. Substituting this into equation 3
gives:

So, v_{1_after} = -5.2 m/s, and v_{2_after} = 0.8
m/s.

Solution 2: Remembering the General Solution:

- First, we must switch to a frame of reference in which one of
the particles is at rest. By subtracting -4.0 m/s from each
velocity we can bring m
_{2}to rest:

v = v_{1_before}- vv_{2_before}= 2.0 m/s - (-4.0 m/s) = 6.0 m/s - Now, calculate the velocities after the collision in this frame of reference:
- Now, return to the original frame of reference by adding -4.0
m/s to both velocities:

v_{1_afte}= v_{1}+ (-4.0 m/s) = -1.2 m/s + (-4.0 m/s) = - 5.2 m/s

v_{2_after}= v_{2}+ (-4.0 m/s) = 4.8 m/s + (-4.0 m/s) = 0.8 m/s

So, after the collision, m_{1} has a velocity of -5.2 m/s
and m_{2} has a velocity of 0.8 m/s.

Solution 3: Using the Center of Mass Reference Frame:

In this case, it is not necessary to switch to a reference frame in which one of the particles is at rest - instead, you switch to the center of mass reference frame.

- Find the velocity of the center of mass, v
_{cm}: - Calculate v
_{1_after}and v_{2_after}

v_{1_after}= 2v_{cm}- v_{1}= 2(-1.6 m/s) - 2.0 m/s = -5.2 m/s

v_{2_after}= 2v_{cm}- v_{2}_{ }= 2(-1.6 m/s) -(-4.0 m/s) = 0.8 m/s

So, after the collision, m_{1} has a velocity of -5.2 m/s
and m_{2} has a velocity of 0.8 m/s.

last update February 8, 2003 by JL Stanbrough