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### Course: Physics library>Unit 6

Lesson 2: Elastic and inelastic collisions

# What are elastic and inelastic collisions?

Collisions can be elastic or inelastic. Learn about what's conserved and not conserved during elastic and inelastic collisions.

# What is an elastic collision?

An elastic collision is a collision in which there is no net loss in kinetic energy in the system as a result of the collision. Both momentum and kinetic energy are conserved quantities in elastic collisions.
Suppose two similar trolleys are traveling toward each other with equal speed. They collide, bouncing off each other with no loss in speed. This collision is perfectly elastic because no energy has been lost.
In reality, examples of perfectly elastic collisions are not part of our everyday experience. Some collisions between atoms in gases are examples of perfectly elastic collisions. However, there are some examples of collisions in mechanics where the energy lost can be negligible. These collisions can be considered elastic, even though they are not perfectly elastic. Collisions of rigid billiard balls or the balls in a Newton's cradle are two such examples.

# Why would we ever approximate a collision as perfectly elastic?

Given that no mechanics problem we are likely to encounter involves a perfectly elastic collision, it may seem that the concept is of little practical use. However, in practice it is often very useful. This is because the requirement that kinetic energy is conserved provides an additional constraint to our equations of motion. This allows us to solve problems in which we would otherwise have too many unknowns. Often the solution will be quite adequate because the collision is 'close enough' to perfectly elastic.
Suppose a head-on elastic collision occurs between two trolleys (A and B) on a track. We want to know the final velocities (subscript f) for both the trolleys, but are only given the initial velocities ${v}_{Ai}$ and ${v}_{Bi}$. Applying conservation of momentum we can see that we have one equation with two unknowns, ${v}_{Af}$ and ${v}_{Bf}$:
${m}_{A}{v}_{Ai}+{m}_{B}{v}_{Bi}={m}_{A}{v}_{Af}+{m}_{B}{v}_{Bf}$
Because kinetic energy is also conserved, we simultaneously have another constraint:
$\frac{1}{2}{m}_{A}{v}_{Ai}^{2}+\frac{1}{2}{m}_{B}{v}_{Bi}^{2}=\frac{1}{2}{m}_{A}{v}_{Af}^{2}+\frac{1}{2}{m}_{B}{v}_{Bf}^{2}$
As we now have two equations with two unknowns, we know that we can completely solve the system using simultaneous equations to determine both velocities.
Solving these equations is somewhat tedious. For now, we simply state the result:
${v}_{Af}=\left(\frac{{m}_{A}-{m}_{B}}{{m}_{A}+{m}_{B}}\right){v}_{Ai}+\left(\frac{2{m}_{B}}{{m}_{A}+{m}_{B}}\right){v}_{Bi}$
${v}_{Bf}=\left(\frac{2{m}_{A}}{{m}_{A}+{m}_{B}}\right){v}_{Ai}+\left(\frac{{m}_{B}-{m}_{A}}{{m}_{A}+{m}_{B}}\right){v}_{Bi}$
The interesting thing about these solutions are the limiting cases that apply for different configurations of head-on collisions. These can help us gain an intuitive understanding of what happens in situations such as the elastic collisions in a Newton's cradle demonstration.
• Object A collides with an equal mass target B which is at rest:
${v}_{Af}=0$, ${v}_{Bf}={v}_{Ai}$.
The impacting object comes to a dead stop, the target gains the exact same speed as the impacting object.
This is exactly the kind of interaction we see in a Newton's cradle. When one ball is swung on one side of the cradle, one ball always comes out the other side. In principle, momentum could also be conserved if two balls were to come out, each with half the original speed. However, the collisions are (mostly) elastic. The only way to ensure conservation of both momentum and kinetic energy is if just one ball comes out.
• Object A collides with an equal mass object B. Objects have equal but oppositely directed velocity.
${v}_{Af}={v}_{Bi}$, ${v}_{Bf}={v}_{Ai}$
The two objects bounce off each other, exchanging velocity. Interestingly, this result also holds for two objects colliding with equal but opposite momentum: the objects will swap momentum. This is a very useful result which allows us to simplify otherwise complex elastic collision problems. Our article on the center of mass contains an example which makes use of this fact to simplify the calculation of an elastic collision with two moving objects.
• A heavy object collides with a much lighter target which is at rest.
The final velocity of the heavy object tends to its initial velocity. This is fairly intuitive; the light object has little effect on the heavy one.
• A light object collides with a much heavier target which is at rest.
The light object bounces off the target, maintaining the same speed but with opposite direction. The heavy target remains at rest.
Exercise 1a: A badminton player serves a shuttle. The speed of her racket is measured by high speed camera at just prior to striking the shuttle. Approximately what speed would you expect the shuttle to be traveling at after the collision?
Exercise 1b: If the racket has a mass of and the shuttle a mass of , calculate the exact speed ${v}_{s}$ assuming an elastic collision.

# What is an inelastic collision?

An inelastic collision is a collision in which there is a loss of kinetic energy. While momentum of the system is conserved in an inelastic collision, kinetic energy is not. This is because some kinetic energy had been transferred to something else. Thermal energy, sound energy, and material deformation are likely culprits.
Suppose two similar trolleys are traveling towards each other. They collide, but because the trolleys are equipped with magnetic couplers they join together in the collision and become one connected mass. This type of collision is perfectly inelastic because the maximum possible kinetic energy has been lost. This doesn't mean that the final kinetic energy is necessarily zero; momentum must still be conserved.
In the real world most collisions are somewhere in between perfectly elastic and perfectly inelastic. A ball dropped from a height $h$ above a surface typically bounces back to some height less than $h$, depending on how rigid the ball is. Such collisions are simply called inelastic collisions.

# Are there any examples of perfectly inelastic collisions?

The ballistic pendulum is a practical device in which an inelastic collision takes place. Until the advent of modern instrumentation, the ballistic pendulum was widely used to measure the speed of projectiles.
In this device, a projectile is fired into a suspended heavy wooden block. The wooden block is initially stationary. Following the collision the projectile becomes embedded in the block. Some kinetic energy gets transformed into heat, sound, and used to deform the block. However, momentum must still be conserved. Consequently, the block swings away at some speed. After the collision, the block behaves as a pendulum in which total mechanical energy is conserved. Because of this we can use the maximum height of the swing to determine the kinetic energy of the block after the collision, then using conservation of momentum we can find the initial speed of the projectile.
We know that only momentum is conserved in this collision, so the momentum of the projectile before the collision must be equal to the momentum of the projectile-block system immediately after the collision. Here we use the subscript $B$ for the block, $P$ for the projectile. ${v}_{B}$ is the velocity of the block just after the impact.
${m}_{P}{v}_{P}=\left({m}_{B}+{m}_{P}\right){v}_{B}$
after re-arranging:
${v}_{B}=\frac{{m}_{P}{v}_{P}}{{m}_{P}+{m}_{B}}$
We know that after the collision, the mechanical energy of the block-bullet system is conserved, so if the block rises up to a maximum height $h$ under a gravitational acceleration $g$ then:
$\frac{1}{2}\left({m}_{P}+{m}_{B}\right){v}_{B}^{2}=\left({m}_{P}+{m}_{B}\right)gh$
after re-arranging:
${v}_{B}^{2}=2gh$
Substituting into our previous conservation of momentum expression for the initial velocity of the block:
$\frac{{m}_{P}{v}_{P}}{{m}_{P}+{m}_{B}}=\sqrt{2gh}$
so after final rearranging:
$\overline{){v}_{P}=\frac{{m}_{P}+{m}_{B}}{{m}_{P}}\sqrt{2gh}}$
Exercise 2a: Suppose a 10 gram musket ball is fired into a 1 kg block which is part of a ballistic pendulum apparatus. It swings to a height of 0.3 m. What is the initial speed of the ball?
Exercise 2b: Suppose the musket ball in the previous exercise was replaced with a bullet of half the mass and twice the initial speed. Would it be safe to do the experiment with the same apparatus? Would you expect the same result?

# What is the coefficient of restitution?

The coefficient of restitution is a number between 0 and 1 which describes where an interaction falls on the scale between perfectly inelastic (0) and perfectly elastic (1).
For an object bouncing off of a fixed target, the coefficient of restitution is the ratio of the final ${v}_{f}$ and initial ${v}_{i}$ speed, i.e:
${C}_{R}=\frac{{v}_{f}}{{v}_{i}}$
Coefficients of restitution of common sports balls range from 0.35 for a cricket ball on a wood surface, to 0.9 for a golf ball impacting with a steel target [1]. The coefficient of restitution of a billiard ball can be up to 0.98 [2].

# Which is more damaging – a mostly elastic or mostly inelastic vehicle collision?

This depends on what you are concerned about damaging – the vehicle or the occupant!
Suppose a vehicle collides elastically with another object. The vehicle will necessarily rebound. The change in momentum as the vehicle rebounds is greater than in an equivalent inelastic collision. The force on an occupant is therefore greater and that is clearly worse for the occupant. On the other hand, because it is an elastic collision no energy will be dissipated in deforming the vehicle. Damage to the structure of the vehicle would therefore be minimized.
Modern vehicles are designed to make use of both inelastic and elastic collisions in the event of an accident. The frame of a vehicle is designed to absorb energy in a collision through deformation of crumple zones built in to the structure of the vehicle. The interior passenger compartment however is designed to be strong so that damage to the occupants is minimized.

[1] A. Haron and K. A. Ismail 2012 Coefficient of restitution of sports balls: A normal drop test in 'IOP Conference Series: Materials Science and Engineering' vol. 36 #1.
[2] Mathavan, S., Jackson, M.R. and Parkin, R.M, 2010. A theoretical analysis of billiard ball dynamics under cushion impacts. In 'Proceedings of the Institution of Mechanical Engineers, Part C: Journal of Mechanical Engineering Science', 224 (9), pp. 1863 - 1873

## Want to join the conversation?

• In Exercice 2b, solution: I do not understand how " we should recognize that the kinetic energy which must be dissipated in the block is now four times higher". Can someone explain this further?
• I believe he meant two times higher. The kinetic energy carried by the bullet in the previous problem was mv²/2, while now is (m/2).(2v)²/2 = mv². Since all the kinetic energy is transferred to the block, it dissipates twice as much energy
• >Object A collides with an equal mass object B. Objects have equal but oppositely directed velocity.

Does that mean If I use a 2-ball Newton's Cradle (sealed in vacuum, not friction of any kind, no energy loss).

If I release the two balls at the same height and at the same time, they'll bounce of each other forever?
• If there is no energy loss then yes they will bounce for ever.
• In Exercise 1b, what was the formula used to solve for the problem? I can see that the KE formula wouldn't really work, since we don't know the final velocity of the racket. So which formula was it that it used to solve for the final velocity of the shuttle?
• It was formula where the book has derived from combination of conservation of energies and momentum, which is vBf=(2mAmA+mB)vAi+(mB−mAmA+mB)vBi Because that vBi was there in the example, it simply cancels out and our equation will get in that form.
• "A light object collides with a much heavier target which is at rest.
The light object bounces off the target, maintaining the same speed but with opposite direction. The heavy target remains at rest." How can the heaver target remain at rest after having a force applied from the collision?
• Maybe the question is a little confusing?? Not sure since I dont see the whole thing.

How about if the quesiton was rephrased:

"A ball moving with velocity v strikes a wall and bounces off with velocity -v"

(This means that the collision was perfectly elastic)

Any thoughts?..
• Alright, I dont get it. In excersise 1b...
vfr = velocity, final, racket
vir = velocity, initial, racket
mr = mass, racket
ms = mass, shuttle
..

In my way:
(mr×vir²)/2+(ms×vis²)/2 = (mr×vfr²)/2+(ms×vfs²)/2
(5×0)/2+(100×20²)/2 = (5×vfr²)/2+(100×0²)/2
(100×20²)/2 = (5×vfr²)/2
100*20²=5×vfr
(100*20²)/5=*vfr²
8000=vfr²
63,24=vfr

Where is this wrong, I dont get...
• I we can't use only kinetic formula to find Vfs, because we DON'T know final velocity of racket after collision, So vfr is NOT equal to 0
• In second case of elastic collision that is:
Object A collides with an equal mass object B. Objects have equal but oppositely directed velocity.
It state that both mass of objects are equal and velocity of both object should be in opposite direction.
so if we take initial velocity of object A is positive and initial velocity of object B is negative.
so after putting those value in formula of final velocity. We get
Vaf = - Vbi and Vbf = Vai instead of Vaf = Vbi and Vbf = Vai
David Sir you done a great job as usual..
• Vaf = Vbi and Vbf = Vai are the correct expressions.
In the formula shown in the article, instead of using minus sign to denote a velocity along the negative axis, simply Vi and Vf are used to get a general relation (one that will give the correct answer according to the initial velocity directions.)

For ex: According to your assumption of B having an initial velocity along the negative axis, you will find that Vaf = - Vbi. Had I assumed B to be initially moving in a positive direction, ie., initial velocity of object B = +Vbi and A in a negative direction, - Vai, i would have found the final result to: Vaf = +Vbf.
The given expression for final velocities work for both cases.
• "Some kinetic energy gets transformed into heat, sound, and used to deform the block. However, momentum must still be conserved." Does this mean that the loss in kinetic energy is small or that the loss in kinetic energy has a small affect on the momentum? With the momentum and kinetic energy coming from the projectile and some of the kinetic energy being transferred to the surroundings after impact then 1/2mv^2 decreases?
• Its a difficult concept to work on. But you need to keep the following in mind...

Somehow: Momentum is always conserved AS momentum. (Does not convert to energy)

Kinetic energy is rarely conserved. Only in perfectly elastic collisions. BUT total energy is always conserved in whatever form.

Here is a useful thing to remember about inelastic collisions

In a perfectly inelastic collision (ie when the objects 'stick together' or coalesce, the MAXIMUM amount of KE is lost. Not necessarily ALL of it.
I like this phrase because it reminds us that the priority here is conservation of momentum. Once that has been honoured and calculated, then we can determine the resulting kinetic energy.

Hope that helps