# Impulse Equations

A solid sphere of mass $m$ rolls without slipping on a horizontal surface and collides with a vertical wall, elastically. The coefficient of friction between the sphere and wall is $\mu$. After the collision, the sphere follows a parabolic trajectory, with range $R$. What is the value of $\mu$ to maximize $R$?

Since the collision is elastic, we can say impulse normal impulse $J = \Delta P = 2mv$.

As frictional impulse is $\mu$ times normal impulse, $J'=2mv\mu$ (upwards). Therefore, sphere acts like projectile with horizontal velocity $v$ and vertical velocity $2v\mu$. To maximize $R = 4\mu v/g$, $\mu$ should be maximum i.e.$1$. However, this is not correct. What am I missing here?

• is the hoizontal surface smooth? Commented Mar 7, 2016 at 4:05
• Since the sphere is rolling, of course not Commented Mar 7, 2016 at 4:06
• also, friction will stop acting when slipping stops, find the relative velocity at point of contact and try something Commented Mar 7, 2016 at 4:07
• But because of the vertical frictional impulse, the body will lose contact with the surfaces Commented Mar 7, 2016 at 4:09
• it doesnt require friction for rolling. we can't solve unless coefficient of horizontal friction is given.. I think it's zero, rolling doesn't "need" friction after rolling has begun. half floor may be smooth, half may have friction Commented Mar 7, 2016 at 4:09

If the collision is elastic, then energy is conserved; however, this cannot mean that the horizontal velocity is the same on the way to the wall, and on the way back: the ball will also have a vertical velocity, and rotational kinetic energy.

This means that the impulse cannot be $2mv$ as you stated; you have to solve instead for the rotational velocity / energy, the horizontal rebound velocity / energy, and the vertical velocity / energy - and set these equal to the inbound kinetic / rotational energy of the sphere.

Note that since the sphere loses contact on the way back, the rotational velocity will be related to the vertical velocity only.

Finally - there is no reason why "maximum coefficient of friction $\mu$ = 1". It can be higher... although in this case it may end up being lower.

In the spirit of "homework like" questions, I will leave this for you to think about.

• So if I say that the horizontal velocity of the sphere is $v_1$ after the collision, the frictional impulse will be $J_f =\mu m(v_1+v)$. To find change in angular momentum, $R.J_f = I\omega_1 - I\omega$ where $\omega$ is initial angular velocity and $\omega_1$ is final angular velocity. Conserving energy, you get a lengthy quadratic in $v_1$ and $v$. Is there any other equation being missed? Commented Mar 7, 2016 at 6:20
• After the collision you have a vertical velocity which is needed for the kinetic energy and which is related to the rotational velocity after. Commented Mar 7, 2016 at 12:41
• is rotational velocity $\omega = \frac{v}{r}$? Commented Mar 24, 2016 at 17:24
• Yes that's right Commented Mar 24, 2016 at 17:26

only thing you left was the fact that friction stops acting when slipping stops. Range is maximised when upward velocity is max as horizontal is fixed

• Please consider using LaTeX for your calculations and basically any graphics program should be suitable for a sketch better than anything handrawn Commented Mar 7, 2016 at 8:49