10.2: Surface Collisions and the Coefficient of Restitution

Last updated
Save as PDF

Page ID: 54733

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Here we will use the term surface collision to describe any instance where a body impacts and rebounds off a solid and unmoving surface. A clear example of a surface collision is a basketball bouncing off a hard floor. The basketball will have some velocity before the collision and some second velocity after the collision, with the floor exerting an impulsive force during the collision that causes this change in velocity.

To analyze this collision, we will first need to set up a normal direction (perpendicular to the surface) and a tangential direction (parallel to the surface) for our problem, and break our velocities down into components in these directions. Assuming minimal friction during the impact, we will have an impulsive impact force acting entirely in the normal direction. This fact will form the basis for our analysis.

A bouncing basketball at the instant it hits the flat, horizontal floor. The impact force exerted by the floor on the ball points straight upwards, in the normal direction; the tangential direction points along the floor to the right. The ball's initial velocity vector, before the impact with the floor, points down and to the right and makes an angle of theta_initial with the tangential axis. The ball's final velocity vector, after the impact with the floor, points up and to the right and makes an angle of theta_final with the tangential axis. — Figure \(\PageIndex{1}\): In a surface collision, the impulsive collision force will act in the normal direction. Because there is no force in the tangential direction, the velocity in the tangential direction will not change.

Because the impact force acts entirely in the normal direction, there will be no other significant force to change the momentum of the body in the tangential direction. Assuming the mass remains constant for the body, this means that the velocity must remain constant in the tangential direction because of the conservation of momentum.

\[ v_{i,t} = v_{f,t} \]

To relate the velocities in the normal direction before and after the collision, we will use something called the coefficient of restitution. The coefficient of restitution is a number between 0 and 1 that measures the "bounciness" of the body and the surface in the collision. Specifically, for a single body being bounced perpendicularly off a surface, the coefficient of restitution is defined as the speed of the body immediately after bouncing off the surface divided by the speed immediately before bouncing off the surface. If we use velocities in place of speed, we will put a negative sign in our equation because the bounce causes a change in direction for the body.

\[ \epsilon = - \frac{v_f}{v_i} \]

In instances where the body is being bounced off the surface at an angle, the impact force is entirely in the normal direction and the coefficient of restitution relationship specifically applies to the components of the velocities in the normal direction.

\[ \epsilon = - \frac{v_{f,n}}{v_{i,n}} \]

This relationship can be applied to elastic collisions (where \(\epsilon\) would be equal to 1), semi-elastic collisions (where \(\epsilon\) would be some number between 0 and 1) and inelastic collisions (where \(\epsilon\) would be equal to 0).

Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/MaY6YEWhwJc.

Example \(\PageIndex{1}\)

A basketball with an initial speed of 3 meters per second impacts a hard floor at the sixty degree angle as shown below. If the collision has a coefficient of restitution of 0.8, what is the expected speed and angle of the basketball after the impact?

A basketball at the instant it collides with a flat, horizontal floor. Its initial velocity vector points down and to the right, has a magnitude of 3 m/s, and makes a 60° angle with the floor. Its final velocity vector points up and to the right, with an unknown magnitude and making an unknown angle with the floor. — Figure \(\PageIndex{2}\): problem diagram for Example \(\PageIndex{1}\). A basketball with a known initial velocity bounces off a floor with a known coefficient of restitution.

Solution: Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\), provided by Dr. Majid Chatsaz. YouTube source: https://youtu.be/A8hl3JPUkB4.

Example \(\PageIndex{2}\)

A bounce test is used to sort ripe cranberries from unripe cranberries. In this test, cranberries are dropped vertically onto a steel plate sitting at a 45-degree angle. After the impact, a cranberry is observed to bounce off at an angle of 20 degrees below the horizontal. Based on this information, what is the coefficient of restitution for the cranberry?

Side view of a steel plate tilted at a 45° angle, slanting down and to the right. A cranberry falls straight down onto the plate, with its velocity vector after the impact pointing down and to the right at 20° below the horizontal. — Figure \(\PageIndex{3}\): problem diagram for Example \(\PageIndex{2}\). A cranberry falls straight down onto a tilted plate and bounces off at a known angle.

Solution: Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\), provided by Dr. Majid Chatsaz. YouTube source: https://youtu.be/wpPYTx52CD8.