Physics & Mechanics

Coefficient of Restitution Calculator

Calculate the elasticity of a collision (COR). Determine if a collision is perfectly elastic, perfectly inelastic, or partially inelastic.

m/s
m/s
m/s
m/s
Coefficient of Restitution (e)
0.714

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Measuring the Bounce

In the real world, very few collisions are perfectly elastic (bouncing flawlessly) or perfectly inelastic (sticking together like clay). Almost all impacts fall somewhere on a sliding scale between those two extremes.

A tennis ball bounces well, but not perfectly. A basketball bounces better than a bowling ball. To quantify exactly how "bouncy" or elastic a collision is, physicists use a dimensionless number called the Coefficient of Restitution (COR), denoted by the letter $e$.

The Scale of Restitution

The Coefficient of Restitution is always a decimal value between 0 and 1:

  • $e = 1.0$: A perfectly elastic collision. The objects bounce off each other with 100% of their original relative speed.
  • $e = 0.0$: A perfectly inelastic collision. The objects hit, don't bounce at all, and stick together.
  • $e = 0.8$: A highly elastic collision (like a golf ball hitting a titanium driver).
  • $e = 0.2$: A highly inelastic collision (like dropping a heavy textbook flat on a desk).

Calculating the Coefficient

The COR is calculated by looking strictly at velocity. Specifically, we compare how fast the objects were moving toward each other before the crash (Relative Velocity of Approach) to how fast they are moving away from each other after the crash (Relative Velocity of Separation).

The Formula

e=v2fv1fv1iv2i\begin{aligned} e = \frac{|v_{2f} - v_{1f}|}{|v_{1i} - v_{2i}|} \end{aligned}

Where:
e=
Coefficient of Restitution (0 to 1)
vfv_f=
Final velocities after impact
viv_i=
Initial velocities before impact

Example Calculation

Imagine two bumper cars moving toward each other.

  • Before Crash: Car 1 moves right at $5 , \text{m/s}$. Car 2 moves left at $-3 , \text{m/s}$. Their relative velocity of approach is $5 - (-3) = \mathbf{8 , \text{m/s}}$. (They are closing the gap at $8 , \text{m/s}$).

  • After Crash: They hit and bounce. Car 1 bounces left at $-1 , \text{m/s}$. Car 2 bounces right at $3 , \text{m/s}$. Their relative velocity of separation is $3 - (-1) = \mathbf{4 , \text{m/s}}$.

  • The COR ($e$): $\frac{4}{8} = \mathbf{0.5}$.

This collision has a coefficient of 0.5. They separated at exactly half the speed at which they approached, indicating a moderate loss of kinetic energy to the rubber bumpers.

Frequently Asked Questions

Under normal passive physics, no. If e > 1, it means the objects separated faster than they approached, which violates the conservation of energy. It means they spontaneously generated energy. The only way e can be greater than 1 is if an explosion or spring releases stored chemical/mechanical energy at the exact moment of impact.

It is heavily regulated. The USGA closely monitors the COR of golf club faces to prevent a 'trampoline effect' that would allow players to hit the ball too far. Similarly, major league baseballs are strictly tested to ensure they have a consistent COR, so a 'juiced' bouncy ball doesn't result in an abnormal number of home runs.

There is a simplified formula for dropping an object onto a hard floor: $e = \sqrt{h_{bounce} / h_{drop}}$. If you drop a ball from 1 meter and it bounces back up to 0.64 meters, the COR is the square root of 0.64, which is exactly 0.8.