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The handbrake of a vehicle of mass $1.5\ \mathrm{tonnes}$ completely fails while it is parked on a $30^\circ$ slope. It rolls $20\ \mathrm{m}$ down the slope before colliding with, and locking on to, a parked vehicle of mass $0.9\ \mathrm{tonnes}$.

(a) estimate the velocity of the two vehicles immediately after the collision.

(we are given that the coefficient of static friction of rubber on asphalt $= 0.9$ and the coefficient of kinetic friction is $0.7$)

For part (a):

I realize I need to find the net force down the slope, however I am unsure regarding how rolling friction works. Do I take into account static friction to be in the opposite direction as motion? Or do I ignore it? Why? (Also if I ignore friction I realize it would be easiest to just use conservation of energy.)

Once I understand this (-> how rolling friction works) I can find the velocity $20\ \mathrm{m}$ down the slope using $v = u + at$, and then using conservation of momentum find the initial velocity of the two vehicles when they collide.

Thank you so much!

(I just need help understand how rolling friction applies in this problem.)

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  • $\begingroup$ I have an answer in this question (physics.stackexchange.com/q/214544) that explains how rolling without slipping works. If there is no slipping, friction will do no work (no energy lost to friction), and you can use conservation of energy. The first thing you should probably do is determine whether or not the car slips at that slope. $\endgroup$ – tmwilson26 Dec 8 '15 at 13:02
  • $\begingroup$ Friction doesn't do work, but still, you cannot conserve energy as kinetic energy is lost during inelastic collision. So, you need to conserve energy till the moment just before collision and then conserve linear momentum. This directly gives you the answer. $\endgroup$ – ShankRam Dec 8 '15 at 14:35
  • $\begingroup$ @ShankRam The OP mentions that they understand to use conservation of momentum during the collision, the question was more about the first part of the problem involving friction on the slope. $\endgroup$ – tmwilson26 Dec 8 '15 at 15:16
  • $\begingroup$ @tmwilson26 Then it's alright. $\endgroup$ – ShankRam Dec 8 '15 at 15:21
  • $\begingroup$ Also bear in mind that when you conserve energy (if the wheels don't slip, as mentioned above to make sure of), you are deciding to ignore the kinetic energy that goes into the rotation of the wheels. That's a good assumption, especially since you know nothing about the wheels, but beware that in other problems where you do have details about a wheel, you may need to remove the rotational kinetic energy from the total. $\endgroup$ – Ken G Sep 22 '16 at 12:19
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Rolling friction is neither static nor kinetic (sliding) friction. It's caused by inelastic forces between the wheels and the road. For example, compressing the tire may take a lot of force, which is pointed up and toward the back of the vehicle. But when the tire expands (as that part of the tire is leaving the road), it may expand with less force, which is directed up and toward the front of the vehicle. Since the rearward component of the compressing regions is less than the forward component of the expanding regions, there is a net rearward force.

A rolling friction coefficient would be much less than either static or kinetic friction. Since you're not given a rolling friction coefficient or a different model for rolling friction, it looks like you can ignore it.

This page (http://www.phy.davidson.edu/fachome/dmb/PY430/Friction/rolling.html) has some decent diagrams.

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