Physics Lab » One Dimensional Collisions

Jul 28 2008

One Dimensional Collisions

Collisions

A collision in physics is defined as two masses exerting equal forces on each other but opposite in direction over a period of time. A collision can be two balls striking each other, a ball striking a wall, two magnets repelling each other, or even a celestial body being diverted by the gravity of a planet or star. Collisions are categorized into three groups:

Elastic Collision - No energy is lost during the collision. In this case, the two masses will just bounce off each other and the total kinetic energy of the system is unchanged.
Inelastic Collision – Some energy is lost during the collision. This, like the elastic case, the two masses will bounch off each other, but the total kinetic energy of the system will be less than it was before the collision.
Totally Inelastic Collision – The maximum energy is lost during the collision. In this case, the two masses will stick together after the collision.

The Centre of Mass Reference System

When dealing with collisions, we usually know the initial velocities of the particles in the system, as well as their masses, and we want to determine the velocities of the particles after the collision. Consider two bodies of mass $m_1$ and $m_2$ moving with velocities $u_1$ and $u_2$ .

The Centre of Mass of the two particles is given by the following formula, where $x_1$ and $x_2$ are the positions of the particles in a coordinate system:

$\vec{X} = \frac{m_1x_1 + m_2x_2}{m_1+m_2}$

(1)

If we differentiate equation 1 in time, we get the velocity of the centre of mass.

$\vec{V} = \frac{m_1u_1 + m_2u_2}{m_1+m_2}$

(2)

Now consider a camera, placed directly on the center of mass and moves with the same velocity. If we look through this camera, then we are in the Centre of Mass Reference System (CMRS), when we are in the CMRS, we will see the two masses moving with different speeds, the speeds we would see them move at is denoted with ‘ over the variable and is given by the following formula:

$\vec{u_1}' = \vec{u_1}-\vec{V}$

(3a)

$\vec{u_2}' = \vec{u_2}-\vec{V}$

(3b)

We will be only working with the CMRS. To find velocities in the Lab Reference System (LRS), we can just use equation 3a and 3b. Note, in the CMRS, the two masses will either be moving towards each other, or away from each other. In the figure below, $u_1'$ will be positive, while $u_2'$ will be negative.

Now consider the total momentum in the CMRS. This is the sum of the momentum of the two CMRS masses.

$\vec{P}' =m_1 \vec{u_1}' + m_2\vec{u_2}'$

(4)

If we substitute equations 3a, and 3b along with equation 2 into equation 4, we get the following:

$\vec{P}' = m_1 (u_1 - \frac{m_1u_1 + m_2u_2}{m_1+m_2} ) + m_2 (u_2 - \frac{m_1u_1 + m_2u_2}{m_1+m_2} )$

(5)

$\vec{P}' = 0$

(6)

So we see that the total momentum of in the CMRS is 0. So, after the collision, the total momentum in the CMRS must also be 0 since momentum of a closed system is always conserved.

$\vec{P}'_f = 0 = m_1v'_1 + m_2v'_2$

(7)

$m_1v'_1 = -m_2v'_2$

(8)

So from equation 8, we know how to find the final velocity of one mass if we know the final velocity of the other mass (in the CMRS). Thanks to our good friend Isaac Newton, who, through lots of experimental work discovered that the ratio of the difference of the final velocities and initial velocities is constant. This is known as the restitution coefficient and depends on what the two masses are made out of.

$\epsilon = \frac{v_2 - v_1}{u_2 - u_1}$

(9)

This also holds true in the CMRS:

$\epsilon = \frac{v'_2 - v'_1}{u'_2 - u'_1}$

(10)

The value of restitution coefficient ranges from 0 to 1. A value of 1 represents an elastic collision. So now we can use equation 10 and equation 8 to solve for the two velocities. Note, that there are two solutions to the equations because of the absolute value in equations. Remember that if:

$x = |y|$
then
$y=x$ or $y=-x$

So, using equations 8 and 10, we get the following solutions to the system:

$v'_2 = \frac{\epsilon | u'_1 - u'_2| m_1}{m_1+m_2}$

(11a)

$v'_1 = -\frac{\epsilon | u'_1 - u'_2| m_2}{m_1+m_2}$

(11b)

$v'_2 = -\frac{\epsilon | u'_1 - u'_2| m_1}{m_1+m_2}$

(12a)

$v'_1 = \frac{\epsilon | u'_1 - u'_2| m_2}{m_1+m_2}$

(12b)

Now, consider equation 12a. We know that from equation 8 that:

$m_1v'_1 = -m_2v'_2$

(8)

If we substitute equation 7 into equation 12a, we get:

$v'_2 = -\frac{\epsilon | u'_1m_1 - u'_2m_1| }{m_1+m_2}$
$v'_2 = -\frac{\epsilon | -u'_2m_2 - u'_2m_1| }{m_1+m_2}$
$v'_2 = -\frac{\epsilon | -u'_2 | (m_1 + m_2) }{m_1+m_2}$
$v'_2 = \pm \epsilon u'_2$

(13a)

Following the same procedure using equation 12b we get:

$v'_1 = \pm \epsilon u'_1$

(13b)

Equation 13 has two solutions because equations 11 and 12 are both solutions to the collision problem. We must now determine which of these two cases is the correct one. In one of the cases the final velocity of the mass is in the same direction as the initial velocity of the mass. Since the two masses are travelling towards each other, this means that the two masses have some how travelled through each other. This case does not make any sense, so the only solution to the problem is:

$v'_1 = - \epsilon u'_1$

(14a)

$v'_2 = - \epsilon u'_2$

(14b)

So the final velocities of the masses in the CMRS is opposite in direction to the initial velocities, and proportional to the coefficient of restitution.

Energy after Collision

Now, consider the kinetic energy of the CMRS. The kinetic energy of a particle is given by the following formula:

$K = \frac{1}{2}mv^2$

(15)

The initial energy of the CMRS is just the sum of the energies of the two masses:

$K'_i = \frac{1}{2}m_1u'^2_1 + \frac{1}{2}m_2u'^2_2$

(16)

After the collision, the energy of the system is:

$K'_f = \frac{1}{2}m_1( -\epsilon u'_1)^2 + \frac{1}{2}m_2 (-\epsilon u'_2)^2$

(17)

Now, if we have an elastic collision, ( $\epsilon=1$ ), then the kinetic energy of the system after the collision is the same as the kinetic energy before the collision, which means the kinetic energy is conserved.

If $\epsilon = 0$ , then:

$K'_f = 0$

(18)

Have any questions about this information? Please contact me.

3 responses so far

3 Responses to “One Dimensional Collisions”

# Eltonon 03 Dec 2008 at 1:40 pm

Very cool. How about using text entry fields to enter values instead of the sliders? The sliders only aloow you to use certain values…
# lookangon 20 Feb 2009 at 2:12 am

My independently developed EJS java applet
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=831.msg3151#msg3151

interesting how we have the same idea to make simulations.

nice flash you have there!

keep up the good work
# Rosson 24 Mar 2009 at 6:38 am

Love it!!

One Dimensional Collisions

Collisions

The Centre of Mass Reference System

Energy after Collision

3 Responses to “One Dimensional Collisions”

Leave a Reply