To the average person in the street the term collision is likely to mean some sort of automotive disaster. We'll use it in that sense, but we'll also broaden the meaning to include any strong interaction between bodies that lasts a relative short time. So we cover not only car accidents but also balls colliding on a biliard table, etc….

If the forces between the bodies (or objects) are much larger than any external forces, as is the case in most collisions, we can neglect the external forces entirely and treat the bodies as an isolated system. Then momentum (if you haven't read the momentum tutorial, please do so) is conserved in the collision, and the total momentum of the system has the same value before and after the collision.

As mentioned above, the momentum of two colliding objects is conserved after the collision. Although we may know the masses and velocities of the bodies before the collision, we don't have enough information to find the velocities after the collision. For that, we use the kinetic energy (the measure of energy an object possesses because of its motion) of the objects to help solve their final velocities after collision. The formula used to calculate kinetic energy is the following:

Where K is the letter used for kinetic energy, m is the mass of the object, and V is the velocity of the object. J is the unit of energy and it is spelled Joules.

If the kinetic energy of a system is the same after the collision, we call this kind of collision elastic collision. A collision between two marbles or two billiard balls is almost completely elastic. The figure below shows a model for elastic collision.

A collision in which the total kinetic energy after the collision is less than that before the collision is called an inelastic collision. A bullet embedding itself in a block of wood is an example of inelastic collision. An inelastic collision in which the bodies stick together and move as one body after the collision is often called a completely inelastic collision.

We are now going to explain completely inelastic collisions followed by an example.

Completely Inelastic Collisions

Let's first look at what happens to momentum and kinetic energy in a completely inelastic collision of two bodies (A and B), as shown in the figure below

Because the two bodies stick together after the collision, their final velocities must be equal:

Conservation of momentum gives the relation

If we know the masses and initial velocities, we can compute the common final velocity V₂.

Suppose, for example, that a body with mass m_A and initial x−component of velocity V₁ collides inelastically with a body with mass m_B that is initially at rest (V_B1 = 0). From the equation for completely inelastic collisions, the commone x−component of velocity V₂ of both bodies after the collisions is

Let's verify that the total kinetic energy after this completely inelastic collision is less than before collision. The kinetic energies K₁ and K₂ before and after the collision, respectively, are

The ratio of final to initial kinetic energy is

The right side is always less than unity because the denominator is always greater than the numerator. Even when the initial velocity m_B is not zero, it is not hard to verify that the kinetic energy after a completely inelastic collision is always less than before.

To better understand this, let's look at an example.

Example 1: Suppose that two blocks move toward each other on a frictionless linear air track as shown in the figure below. After the collision, the two blocks stick together. The masses and initial velocities are m_A = 0.50 kg, V_A1 = 2.0 m⁄s, m_B = 0.30 kg, V_B1 = -2.0 m⁄s. We now want to find the common velocity V₂, and compare the initial and final kinetic energies.

Solution: Take the x−axis to lie along the direction of motion, as before. From conservation of the x−component of momentum,

m_AV_A1 + m_BV_B1 = (m_A + m_B)V₂,
V₂ = (m_A1 + m_B1) ⁄ (m_A + m_B),
V₂ = (0.50 kg)(2.0 m⁄s)(-2.0 m⁄s) ⁄ (0.50 kg + 0.30 kg)
V₂ = 0.50 m⁄s.

Because V₂ is positive, the blocks move together to the right (the +x−direction) after the collision.

Before the collision, the kinetic energies of the blocks A and B are

K_A = 1⁄2m_AV_A1² = 1⁄2(0.50 kg)(2.0 m⁄s)² = 1.0 J,
K_B = 1⁄2m_BV_B1² = 1⁄2(0.30 kg)(−2.0 m⁄s)² = 0.60 J.

Note that the kinetic energy of block B is positive, even though the x−component of its velocity V_B1 and momentum mV_B1 are both negative. The total kinetic energy before the collision is 1.6J (K₁ + K₂). The kinetic energy after the collision is

1⁄2(m_A + m_B)V₂² = 1⁄2(0.50 kg + 0.30 kg)(0.50 m⁄s)² = 0.10 J.

We are now going to describe elastic collisions and also give an example.

Elastic Collisions

As aforementioned, an elastic collision in an isolated system is a collision in which kinetic energy (as well as momentum) is conserved. Elastic collisions occur when forces between the colliding bodies are conservative. For example, when two steel balls collide, they squash a little near the surface of contact, but then they spring back.

If we compute the kinetic energy of each object before the collision as we did for the inelastic collisions, and use that along with the rule of conservation of kinetic energy (as well as conservation of momentum), we can derive the equations that solve for the final velocities of the objects. From conservation of kinetic energy we have

and conservation of momentum gives

If the masses m_A and m_B and the initial velocities V_A1 and V_B are known, these two equations can be solved simultaneously to find the two final velocities V_A2 and V_B2.

The general solution is a little complicated, so we will concentrate on the particular case in which body B is at rest before the collision. Think of it as a target for body A to hit. We can then simplify the velocity notation; we let V be the initial x−component of velocity of A, and V_A and V_B be teh final x−comonents for A and B. Then the kinetic energy and momentum conservation equations are, respectively.

1⁄2m_AV² = 1⁄2m_AV_A² + 1⁄2m_BV_B²,

m_AV = m_AV_A + m_B.

We may solve for V_A and V_B in terms of the masses and the initial velocity V. We are going to skip all the algebra involved and go straight to the formulas. The velocities for A and B after collision can be found with the following formulas

V_A = (m_A − m_B) * V ⁄ (m_A + m_B).

V_B = 2m_AV ⁄ (m_A + m_B).

Let's now look at an example.

Example 2: The figure below shows an elastic collision of two pucks on a frictionless air table. Puck A has mass m_A = 0.500 kg, and puck B has mass m_B = 0.300 kg. Puck A has an initial velocity of 4.00 m⁄s in the positive x−direction and a final velocity of 2.00 m⁄ in an unknown direction. Puck B is initially at rest. We want to find the final spee V_B2 of puck B and the angles α and β as illustrated in the figure.

Solution: Because the collision is elastic, the initial and final kinetic energies are equal:

1⁄2m_AV_A1² = 1⁄2m_AV_A2² + 1⁄2m_BV_B2²,
V_B2² = (m_A1² − m_AV_A2²) ⁄ m_B
V_B2² = [(0.500 kg)(4.00 m⁄s)² - (0.500 kg)(2.00 m⁄s)²] ⁄ (0.300 kg),
V_B2 = √20 m⁄s,
V_B2 = 4.47 m⁄s.

Conservation of the x−component of total momentum gives

m_AV_A1x = m_AV_A2x + m_BV_B2x,
(0.500 kg)(4.00 m⁄s) = (0.500 kg)(2.00 m⁄s)(cos α) + (0.300 kg)(4.47 m⁄)(cos β),

and conservation of the y−component gives

0 = m_AV_A2y + m_BV_B2y,
0 = (0.500 kg)(2.00 m⁄s)(sin α) − (0.300 kg)(4.47 m⁄s)(sin β).

There are two simultaneous equations for α and β. The simplest solution is to eliminate β as follows: We solve the first equation for cos β and the second for sin β; we then square each equation and add. Since sin² β + cos² β = 1, this eliminates β and leaves an equation that can solve for cons α and hence for α. We can then substitute this value back into either of the two equations and solve the result for β. Try to figure out the details for yourself. I'll just give you the results here

We have now concluded our tutorial on collisions. I have presented you with the theory behind collisions and examples that show that we can classify collisions according to energy considerations. A collision in which the kinetic energy is conserved is called elastic. A collision in which the total kinetic energy decreases is called inelastic. When the two bodies have a common final velocity, we say that the collision is completely inelastic. There are cases in which the final kinetic energy is greater than the initial value. The rifle recoil discussed in example 1 is an example of this.

Below is a Flash movie showing an animation of collision between two objects.

If you have any queries regarding this tutorial, do not hesitate to contact me.

Best Regards

Fidel