1. Not finding help here? Sign up for a free 30min 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!

A compact car, mass of 725 kg, is moving at 100 km.hr.

  1. Nov 3, 2003 #1
    A compact car, mass of 725 kg, is moving at 100 km.hr. What is its momentum? At what velocity is the momentum of a larger car, mass 2175 kg, equal to that of the smaller car?

    this is what i have so far:
    p=725(100 km/hr)

    but i don't kno how to change km/hr into m/sec. And i don't know how to do the second part of the question.

    Please help!
    Last edited by a moderator: Feb 7, 2013
  2. jcsd
  3. Nov 3, 2003 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hi kath102245,

    Try using a "conversion factor," like this one:

    1 kilometer
    1000 meters

    Notice that the quantities on the top and bottom are equal lengths -- they're different units, of course, but they represent exactly the same length. When you have the same thing on the top and bottom of a fraction, that fraction is equal to one. Therefore, this fraction is just one. When you multiply something by one, you do not change it, of course, so you can use such a conversion factor anywhere that it is convenient.

    Let's make use of two conversion factors:
    Code (Text):

     1 kilometer           1 hour
    -------------  and  --------------
     1000 meters         3600 seconds
    Now take your initial quantity (100 km/hr), and multiply it by these conversion factors as necessary to get the units you want. You want the result in meters per second. Start by attacking the "hours" with the second conversion factor, like this:
    Code (Text):

      100 kilometers        1 hour
     ---------------- * --------------
          1 hour         3600 seconds
    Note that there's some cool stuff happening here: since "hours" is on the top and bottom, it cancels. You're left with just kilometers per second.

    Now you can attack the "kilometers" the same way:
    Code (Text):

      100 kilometers        1 hour        1000 meters
     ---------------- * -------------- * -------------
          1 hour         3600 seconds     1 kilometer
    Now the hours cancel, and so do the kilometers. You're left with meters on top, and seconds on bottom. To get the answer, just multiply the fractions:
    Code (Text):

         100 * 1 * 1000 meters     100,000 meters
      = ----------------------- = ---------------- = about 27.77 m/s
         1 * 3600 * 1 seconds      3,600 seconds
    If you understand this, we'll go ahead and work on the next part of the problem.

    - Warren
  4. Nov 4, 2003 #3
    so then would the momentum be 20,133.25kgm/s? Then for the second part of the question...i'm thinking it would be something like this...



    This doesn't seem right...
  5. Nov 4, 2003 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Why doesn't that seem right? The small car is roughly three times less massive than the big one, and thus must have roughly three times the velocity to have the same momentum.

    I believe your answer is correct!

    - Warren
  6. Nov 4, 2003 #5
    Thanks for your help Warren!
  7. Nov 4, 2003 #6
    Warren gave you a conversion factor for the specific problem you were trying to calculate although if you find yourself in another situation like this one you have an easier way that applies to most Physics Equations (Note : this is mathematical but it does apply to physics.)

    3a ^3 4
    ----- x -----
    1 2ba

    Expand the Exponents

    3aaa 4
    ------ x -----
    1 2ba

    The "a" in the denomenator of the second fraction in the problem will cancel out one of the three "a's" in the numerator of the first fraction. Giving the Equation this apperance.

    3aa 4
    ------ x -----
    1 2b

    Now you have eliminated 2 potential confusion areas. The Equation will turn out to equal.

    3aa 4 12aa
    ------ x ----- = ------
    1 2b 2b

    You multiply the 2 top numbers and add the "a's" left, (variables hold the place of a number because everyone knows you cannot multiply a letter) coming out to 12aa. Then 1 being the identity element identifies the number being muultiplyed by it as itself, comin gout to 2b giving you the answer 12aa/2b. Now with physics if you are doing the simple equation F = m(a) you have mass = X kg and a = X m/s so you need to combine those 2 into one for the Force. The numbers are multiplyed so the units of measurment are as well. coming out like kg x m/s.

    If Warrens didn't help here was a quick lesson a little rough around the edges but that good ok hopefully it helped with your homework and if not maybe the Math part will help for Algebra.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?