Elastic & Inelastic Collisions: What Is The Difference? (W/ Examples)

The term ​elastic​ probably brings to mind words like ​stretchy​ or ​flexible​, a description for something that easily bounces back. When applied to a collision in physics, this is exactly correct. Two playground balls that roll into one another and then bounce apart had what's known as an ​elastic collision​.

In contrast, when a car stopped at a red light gets rear-ended by a truck, both vehicles stick together and then move together into the intersection at the same speed – no rebounding. This is an ​inelastic collision​.

Note that inelastic collisions don't always need to show objects sticking together ​after​ the collision. For example, two train cars could start off connected, moving with one velocity, before an explosion propels them opposite ways.

Another example is this: A person on a moving boat with some initial velocity could throw a crate overboard, thereby changing the final velocities of the boat-plus-person and the crate. If this is hard to understand, consider the scenario in reverse: a crate falls onto a boat. Initially, the crate and the boat were moving with separate velocities, afterwards, their combined mass is moving with one velocity.

In contrast, an ​elastic collision​ describes the case when the objects hitting each other each start and end with their own velocities. For example, two skateboards approach each other from opposite directions, collide and then bounce back towards where they came from.

The law of conservation of momentum applies equally in either elastic or inelastic collisions in an isolated system (no net external force), so the math is the same. ​The total momentum cannot change.​ So the momentum equation shows all the masses times their respective velocities ​before the collision​ (since momentum is mass times velocity) equal to all the masses times their respective velocities ​after the collision​.

For two masses, that looks like this:

\(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\)

Where m₁ is the mass of the first object, m₂ is the mass of the second object, v_i is the corresponding mass' initial velocity and v_f is its final velocity.

​This equation works equally well for elastic and inelastic collisions.​

However, sometimes it is represented a little differently for inelastic collisions. That's because objects stick together in an inelastic collision – think of the car being rear-ended by the truck – and afterwards, they act like one large mass moving with one velocity.

So, another way to write the same law of conservation of momentum mathematically for ​inelastic collisions​ is:

\(m_1v_{1i} + m_2v_{2i} = (m_1+m_2}v_f\)

or

\((m_1+m_2}v_1 = m_1v_{1f} + m_2v_{2f}\)

In the first case, the objects stuck together ​after the collision​, so the masses are added together and move with one velocity ​after the equals sign​. The opposite is true in the second case.

An important distinction between these types of collisions is that kinetic energy is conserved in an elastic collision, but not in an inelastic collision. So for two colliding objects, the conservation of kinetic energy can be expressed as:

\(\frac {1}{2}m_1(v_{1i})^2 + \frac {1}{2}m_2(v_{2i})^2 = \frac {1}{2}m_1(v_{1f})^2 + \frac {1}{2}m_2(v_{2f})^2\)

The kinetic energy conservation is actually a direct result of the conservation of energy in general for a conservative system. When the objects collide, their kinetic energy is briefly stored as elastic potential energy before being perfectly transferred back to kinetic energy again.

That said, most collision problems in the real world are neither perfectly elastic nor inelastic. In many situations, however, the approximation of either is close enough for a physics student's purposes.

1\. A 2-kg billiard ball rolling along the ground at 3 m/s hits another 2-kg billiard ball that was initially still. After they hit, the first billiard ball is still but the second billiard ball is now moving. What is its velocity?

The given information in this problem is:

m₁ = 2 kg

m₂ = 2 kg

v_1i = 3 m/s

v_2i = 0 m/s

v_1f = 0 m/s

The only value unknown in this problem is the final velocity of the second ball, v_2f.

Plugging the rest into the equation that describes conservation of momentum gives:

\((2)(3) + (2)(0) = (2)(0) + (2)v_{2f}\)

Solving for v_2f gives v_2f = 3 m/s.

The direction of this velocity is the same as the initial velocity for the first ball.

This example shows a ​perfectly elastic collision,​ since the first ball transferred all of its kinetic energy to the second ball, effectively switching their velocities. In the real world, there are no ​perfectly​ elastic collisions because there is always some friction causing some energy the be transformed to heat during the process.

2\. Two rocks in space collide head-on with one another. The first has a mass of 6 kg and is traveling at 28 m/s; the second has a mass of 8 kg and is moving at 15 m/s. With what speeds are they moving away from each other at the end of the collision?

Because this is an elastic collision, in which momentum and kinetic energy are conserved, two final unknown velocities can be calculated with the given information. The equations for both conserved quantities can be combined to solve for the final velocities like this:

\(v_{1f}= (\frac {m_1 – m_2}{m_1 + m_2})v_{1i} + (\frac {2m_2}{m_1 + m_2})v_{2i} \newline
v_{2f}= (\frac {2m_1}{m_1 + m_2})v_{1i} + (\frac {m_2 – m_1}{m_1 + m_2})v_{2i}\)

Plugging in the given information (note that the second particle's initial velocity is negative, indicating they are traveling in opposite directions):

v_1f = -21.14m/s

v_2f = 21.86 m/s

The change in signs from initial velocity to final velocity for each object indicates that in colliding they both bounced off each other back towards the direction from with they came.

A cheerleader jumps from the shoulder of two other cheerleaders. They fall down at a rate of 3 m/s. All the cheerleaders have masses of 45 kg. How quickly is the first cheerleader moving upwards at the first moment after she jumps?

This problem has ​three masses​, but so long as the before and after parts of the equation showing conservation of momentum are written correctly, the process of solving is the same.

Before the collision, all three cheerleaders are stuck together and. But ​no one is moving​. So, the v_i for all three of these masses is 0 m/s, making the entire left side of the equation equal to zero!

After the collision, two cheerleaders are stuck together, moving with one velocity, but the third is moving the opposite way with a different velocity.

Altogether, this looks like:

\(( m_1 + m_2 + m_3)(0 ) = (m_1 + m_2)v_{1,2f} + m_3v_{3f}\)

With numbers substituted in, and setting a reference frame where ​downwards​ ​is​ ​negative​:

\((45 + 45 + 45 )(0 ) = (45 + 45 )(-3 ) + (45 )v_{3f}\)

Solving for v_3f gives v_3f = 6 m/s.