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 vAiv_{Ai} and vBiv_{Bi}. Applying conservation of momentum we can see that we have one equation with two unknowns, vAfv_{Af} and vBfv_{Bf}:
mAvAi+mBvBi=mAvAf+mBvBfm_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:
12mAvAi2+12mBvBi2=12mAvAf2+12mBvBf2\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:
vAf=(mAmBmA+mB)vAi+(2mBmA+mB)vBi v_{Af} = \left( \frac{m_{A} - m_{B}}{m_{A}+m_{B}} \right) v_{Ai} + \left( \frac{2m_{B}}{m_{A}+m_{B}} \right) v_{Bi}
vBf=(2mAmA+mB)vAi+(mBmAmA+mB)vBi v_{Bf} = \left( \frac{2m_{A}}{m_{A}+m_{B}} \right) v_{Ai} + \left( \frac{m_{B} - m_{A}}{m_{A}+m_{B}} \right) v_{Bi}
mAvAi+mBvBi=mAvAf+mBvBfm_A v_{Ai} + m_B v_{Bi} = m_A v_{Af} + m_B v_{Bf}
12mAvAi2+12mBvBi2=12mAvAf2+12mBvBf2\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
After simplifying and rearranging of the kinetic energy equation:
mA(vAi2vAf2)=mB(vBf2vBi2)m_A ( v_{Ai}^2 - v_{Af}^2 ) = m_B ( v_{Bf}^2-v_{Bi}^2)
Using the method of factoring binomials, this can be written
Now returning to our conservation of momentum equation, this can be written in a similar form
mA(vAivAf)=mB(vBfvBi)m_A(v_{Ai}-v_{Af}) = m_B(v_{Bf}-v_{Bi})
Now dividing these two equation by each other give a simple result
vAi+vAf=vBf+vBiv_{Ai} + v_{Af} = v_{Bf} + v_{Bi}
Now we can substitute this into our original equation from conservation of momentum. We can arrange the equation to put all the VAfV_{Af} terms on the right hand side:
mAvAf=mAvAi+mBvBimB(vAi+vAfvBi)=mAvAi+mBvBimBvAimBvAf+mBvBimAvAf+mBvAf=mAvAi+mBvBimBvAi+mBvBi\begin{aligned} m_A v_{Af} &= m_A v_{Ai} + m_B v_{Bi} - m_B(v_{Ai}+v_{Af}-v_{Bi}) \\ &= m_A v_{Ai} + m_B v_{Bi} - m_B v_{Ai} - m_B v_{Af} + m_B v_{Bi}\\ m_A v_{Af} + m_B v_{Af} &= m_A v_{Ai} + m_B v_{Bi} - m_B v_{Ai} + m_B v_{Bi}\end{aligned}
Factoring out the velocities gives:
(mA+mB)vAf=(mAmB)vAi+2mBvBi(m_A + m_B) v_{Af} = (m_A-m_B)v_{Ai} + 2m_B v_{Bi}
Which can be re-arranged to give the final result for vAfv_{Af}
vAf=(mAmBmA+mB)vAi+(2mBmA+mB)vBi \boxed{v_{Af} = \left( \frac{m_{A}-m_{B}}{m_{A}+m_{B}} \right) v_{Ai} + \left( \frac{2m_{B}}{m_{A}+m_{B}} \right) v_{Bi}}
We can now substitute this back into the result of our earlier division:
vBf=vAi+vAfvBi=vAi+(mAmBmA+mB)vAi+(2mBmA+mB)vBivBi\begin{aligned} v_{Bf} &= v_{Ai}+v_{Af}-v_{Bi}\\ &= v_{Ai} + \left( \frac{m_{A}-m_{B}}{m_{A}+m_{B}} \right) v_{Ai} + \left( \frac{2m_{B}}{m_{A}+m_{B}} \right) v_{Bi}-v_{Bi} \end{aligned}
We can now group all the velocity terms and arrange over a common denominator of the sum of the masses:
vBf=(mA+mB)+(mAmB)mA+mBvAi+2mB(mA+mB)mA+mBvBiv_{Bf} = \frac{(m_A + m_B)+(m_A - m_B)}{m_A + m_B} v_{Ai} + \frac{2m_B-(m_A+m_B)}{m_A + m_B} v_{Bi}
Which can be simplified to the final form:
vBf=(2mAmA+mB)vAi+(mBmAmA+mB)vBi\boxed{v_{Bf} = \left( \frac{2m_{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:
vAf=0v_{Af} = 0, vBf=vAiv_{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.
    vAf=vBiv_{Af} = v_{Bi}, vBf=vAiv_{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.
    Here ‘tends to’ is used in the same context as is often found in calculus. Specifically, as the ratio of the masses of the heavy and light object increases towards infinity, the final velocity becomes ever closer to whatever it ‘tends to’.
  • 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 vr=20 m/sv_r=20~\mathrm{m/s} just prior to striking the shuttle. Approximately what speed would you expect the shuttle to be traveling at after the collision?
The tip of a badminton shuttle is sufficiently rigid that the collision can be approximated as an elastic collision. The mass of the racket is much larger than the shuttle. We should therefore expect the speed to be around twice the speed of the racket, or 40 m/s.
Exercise 1b: If the racket has a mass of mr=100 gramsm_r=100~\mathrm{grams} and the shuttle a mass of ms=5 gramsm_s=5~\mathrm{grams}, calculate the exact speed vsv_s assuming an elastic collision.
Substituting the given values into equation:
vs=(2mrmr+ms)vr=(2100 g(100+5) g)20 m/s=38 m/s\begin{aligned} v_s &= \left( \frac{2 m_r }{m_r + m_s} \right) v_r \\ &= \left( \frac{2\cdot 100~\mathrm{g} }{(100+5)~\mathrm{g}} \right) 20~\mathrm{m/s}\\ &= 38~\mathrm{m/s}\end{aligned}

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 hh above a surface typically bounces back to some height less than hh, 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.
Figure 1: An inelastic collision in a ballistic pendulum.
Figure 1: An inelastic collision in a ballistic pendulum.
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 BB for the block, PP for the projectile. vBv_B is the velocity of the block just after the impact.
mPvP=(mB+mP)vBm_P v_{P} = (m_B + m_P) v_{B}
after re-arranging:
vB=mPvPmP+mBv_{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 hh under a gravitational acceleration gg then:
12(mP+mB)vB2=(mP+mB)gh\frac{1}{2} (m_P + m_B) v_B^2 = (m_P + m_B) gh
after re-arranging:
vB2=2ghv_B^2 = 2gh
Substituting into our previous conservation of momentum expression for the initial velocity of the block:
mPvPmP+mB=2gh\frac{m_P v_P}{m_P + m_B} = \sqrt{2gh}
so after final rearranging:
vP=mP+mBmP2gh\boxed{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?
Substituting the numbers into the equation:
vP=0.01+1 kg0.01 kg29.81 m/s20.3 m=245 m/s\begin{aligned} v_{P} &= \frac{0.01 + 1~\mathrm{kg}}{0.01~\mathrm{kg}} \sqrt{2\cdot 9.81~\mathrm{m/s^2} \cdot 0.3~\mathrm{m}} \\ &= 245~\mathrm{m/s} \end{aligned}
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?
Because the momentum of the bullet is the same as the musket ball, we should expect the same deflection. Rearranging our previous expression to make hh the subject,
h=12g(mPvPmP+mB)2h = \frac{1}{2g}\left( \frac{m_P v_P}{m_P + m_B} \right)^2
shows that to be true, to the extent that the mass of the block is much larger than that of the bullet. However, we should recognize that the kinetic energy which must be dissipated in the block is now two times higher. This means that it is not necessarily safe to use the same block; it may explode unpredictably.

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 vfv_f and initial viv_i speed, i.e:
CR=vfviC_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