A *perfectly inelastic collision* is one in which the maximum amount of kinetic energy has been lost during a collision, making it the most extreme case of an inelastic collision. Though kinetic energy is not conserved in these collisions, momentum is conserved and the equations of momentum can be used to understand the behavior of the components in this system.

In most cases, you can tell a perfectly inelastic collision because of the objects in the collision "stick" together, sort of like a tackle in American football. The result of this sort of collision is fewer objects to deal with after the collision than you had before the collision, as demonstrated in the following equation for a perfectly inelastic collision between two objects. (Although in football, hopefully, the two objects come apart after a few seconds.)

Equation for a Perfectly Inelastic Collision:m_{1}+v_{1i}m_{2}= (v_{2i}m_{1}+m_{2})v_{f}

### Proving Kinetic Energy Loss

You can prove that when two objects stick together, there will be a loss of kinetic energy. Let's assume that the first mass, *m*_{1}, is moving at velocity ** v_{i}** and the second mass,

*m*

_{2}, is moving at velocity

**0**.

This may seem like a really contrived example, but keep in mind that you could set up your coordinate system so that it moves, with the origin fixed at *m*_{2}, so that the motion is measured relative to that position. So really any situation of two objects moving at a constant speed could be described in this way. If they were accelerating, of course, things would get much more complicated, but this simplified example is a good starting point.

m_{1}= (v_{i}m_{1}+m_{2})v_{f}

[m_{1}/ (m_{1}+m_{2})] *=v_{i}v_{f}

You can then use these equations to look at the kinetic energy at the beginning and end of the situation.K_{i}= 0.5m_{1}V_{i}^{2}K_{f}= 0.5(m_{1}+m_{2})V_{f}^{2}

Now substitute the earlier equation forV_{f}, to get:K_{f}= 0.5(m_{1}+m_{2})*[m_{1}/ (m_{1}+m_{2})]^{2}*V_{i}^{2}K_{f}= 0.5 [m_{1}^{2}/ (m_{1}+m_{2})]*V_{i}^{2}

Now set the kinetic energy up as a ratio, and the 0.5 andV_{i}^{2}cancel out, as well as one of them_{1}values, leaving you with:K_{f}/K_{i}=m_{1}/ (m_{1}+m_{2})

Some basic mathematical analysis will allow you look at the expression *m*_{1} / (*m*_{1} + *m*_{2}) and see that for any objects with mass, the denominator will be larger than the numerator. So any objects that collide in this way will reduce the total kinetic energy (and total velocity) by this ratio. We have now proven that any collision where the two objects collide together results in a loss of total kinetic energy.

### Ballistic Pendulum

Another common example of a perfectly inelastic collision is known as the "ballistic pendulum," where you suspend an object such as a wooden block from a rope to be a target. If you then shoot a bullet (or arrow or other projectile) into the target, so that it embeds itself into the object, the result is that the object swings up, performing the motion of a pendulum.

In this case, if the target is assumed to be the second object in the equation, then *v*_{2i} = 0 represents the fact that the target is initially stationary.

m_{1}+v_{1i}m_{2}= (v_{2i}m_{1}+m_{2})v_{f}m_{1}+v_{1i}m_{2}(0) = (m_{1}+m_{2})v_{f}m_{1}= (v_{1i}m_{1}+m_{2})v_{f}

Since you know that the pendulum reaches a maximum height when all of its kinetic energy turns into potential energy, you can, therefore, use that height to determine that kinetic energy, then use the kinetic energy to determine *v _{f}*, and then use that to determine

*v*

_{1i}- or the speed of the projectile right before impact.

**Also Known As: **completely inelastic collision