1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinematic Car Race

  1. May 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Two cars start 10 meters up a hill. Car A has a mass of 9 grams and Car B has a mass of 2 grams. Additionally, both cars must travel the same distance to the finish line.
    a) If friction is ignored which car will win the race?
    b) If friction is not ignored, which car will win the race.


    2. Relevant equations
    u + k = u + k


    3. The attempt at a solution
    a) Both cars will finish at the same time because at any point on the track, both cars will have equal velocities.
    b) Car B will finish first because it is lighter, so it will have less of a friction force that will slow it down less. As a result, Car B will move faster and thus reach the finish line first.

    Is this right?
     
    Last edited: May 28, 2009
  2. jcsd
  3. May 28, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    What do the equations say?

    With friction or without what is the dependence on mass?
     
  4. May 28, 2009 #3
    With mgh = 1/2mv^2 the mass would cancel out, so is there so dependence on mass? This means the cars finish at the same time, right?

    Would friction play any role, or no?
     
  5. May 28, 2009 #4

    LowlyPion

    User Avatar
    Homework Helper

    Write out the equation for the case with friction.
     
  6. May 28, 2009 #5
    Is the first part correct?

    Fnet = ma
    uN = ma
    n(mg)=ma
    ng=a

    So the mass is canceled out when friction is and is not ignored?
     
  7. May 28, 2009 #6

    LowlyPion

    User Avatar
    Homework Helper

    The mass canceled out in the first one, so yes it would have to be correct wouldn't it?

    What happens in the case where you have

    Potential at the top = kinetic at the bottom + work lost to friction ?
     
  8. May 28, 2009 #7
    Friction will do more work on the heavier car, so more energy will be lost. As a result, there will be less kinetic energy to contribute to the velocity of the heavier car than the lighter car.
     
  9. May 28, 2009 #8

    LowlyPion

    User Avatar
    Homework Helper

    What does the equation say?

    m*g*h = 1/2*m*v2 - μ*m*g*cosθ*d
     
  10. May 28, 2009 #9
    Can you cancel the mass out of each term?
     
  11. May 28, 2009 #10

    LowlyPion

    User Avatar
    Homework Helper

    That would be what it looks like to me.

    And d must be the same according to the problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinematic Car Race
  1. Two cars drag racing (Replies: 3)

  2. Race Car Driving (Replies: 3)

Loading...