Submitted by haku on Sat, 07/07/2018 - 23:16
Hello, I have a question on the coefficient of restitution.
A ball hits a wall with speed u and is reflected with speed v (one-dimensional motion). The coefficient of restitution of the collision is e. It (intuitively) seems clear that $$v=-eu.$$
But calculating it more formally somehow gives, using the definition of the coefficient of restitution (from https://en.wikipedia.org/wiki/Coefficient_of_restitution#Equations),
$$e=\frac{speed\ of\ wall-(-v)}{u-speed\ of\ wall}=\frac{v}{u}\ \ \ \Rightarrow \ \ \ v=eu$$ and, using conservation of momentum, $$mu=-mv \ \ \ \Rightarrow \ \ \ v=-u.$$
Equating coefficients gives e=-1.
This doesn't seem right, since e must be a number between 0 and 1.
The problem is resolved using the absolute values of the velocities, but other calculations made in the STEP II Mechanics Q4 module are made with velocities, too.
I somehow cannot understand what's wrong with the equation.
Mechanics
You can only use conservation of momentum if there are no external forces acting on the "system". So if the ball is the "system", then the wall applies an external force and so momentum is not conserved. If you consider the ball and wall as a "system" then you need to consider that the wall recoils a bit (very slightly as it is much more massive than the ball), and also there is likely to be a force acting on the wall from the ground so there is an external force.
(There are always some external forces such as air resistance, but these are usually negligible).
On the page you have linked to it does also say:
"For an object bouncing off a stationary target, $C_R$ is defined as the ratio of the object's speed after the impact to that of prior to impact:
\[
C_R = \frac{v}{u}, \text{where}
\]
$v$ is the speed of the object after impact
$u$ is the speed of the object before impact
Here, $u$ and $v$ are speeds, so you would need to insert the negative for velocity.
Newton's experimental law
The form of Newton's experimental law I prefer is:
Speed of separation = $e \times$ Speed of approach.
I think you might have a mix up with negatives. In your use of Newtons experimental law you have used "$-v$" as the velocity of the ball after the collision, and so you would expect $v$ to be positive!