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Problem on Momentum and Frictionquick help please!

  1. Oct 23, 2007 #1
    Here's the problem:
    A small car (mass = 500 kg) fails to stop at the entrance to a major road until it has
    reached the middle of the carriageway. It is then struck by a sports car (mass =
    1000 kg) and the two cars lock together. Police investigators show that the skid
    marks stretch for 40 m. Assuming that the frictional force on these two vehicles
    was constant at 7.5 x 108 newtons, find

    (a) the deceleration of the two cars
    (b) the speed at which the two cars moved off after the impact
    (c) the speed of the sports car just before the impact…in miles per hour (100 mi/h = 45
    m/s approx.)

    I couldn't apply any formula of momentum here.there's even no given time and velocity.i tried to apply F=ma to get the deceleration but i think constant friction is the key here.a also have no idea how to get the other two questions.i'm kinda confused here.help please..i'd appreciate a good explanation how to solve this problem.thanks a lot.
     
  2. jcsd
  3. Oct 23, 2007 #2

    G01

    User Avatar
    Homework Helper
    Gold Member

    As you said, F=ma and the constant friction force are the key to part A. So, you know:

    The friction force and [tex]\Sigma F=ma[/tex].

    Can you use this to find anything of interest?
     
  4. Oct 23, 2007 #3
    are you sure the formula is applicable?
    i think the answer would be -500000 m/s2
    isn't that too unreal?
    how bout the constant friction?is it related to the acceleration?
    thanks.
     
  5. Oct 23, 2007 #4
    Momentum is conserved

    [tex]p_{1v1}[/tex] = momentum of vehicle 1 prior to collision
    [tex]p_{1v2}[/tex] = momentum of vehicle 2 prior to collision
    [tex]p_{2v1}[/tex] = momentum of vehicle 1 after to collision
    [tex]p_{2v2}[/tex] = momentum of vehicle 2 after to collision

    [tex]p_{1v1}+p_{1v2}=p_{2v1}+p_{2v2}[/tex]
     
  6. Oct 23, 2007 #5
    You also know how to convert between distance and acceleration:

    [tex]a(x-x_0)=\frac{1}{2}(v^2-v_0^2)[/tex]

    Using this you can find the initial velocity [tex]v_0[/tex]

    You know the friction force is μN=μmg and since m cancels out, the deceleration is μg
     
    Last edited: Oct 23, 2007
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