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!

Momentum Head on Collision Cart Problem

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Two remote control cars with masses of 1.11 kilograms and 1.81 kilograms travel toward each other at speeds of 8.75 meters per second and 3.39 meters per second, respectively. The cars collide head-on, and the less massive car recoils with a speed of 2.99 meters per second. How much kinetic energy was lost in the collision?

    2. Relevant equations

    no idea

    3. The attempt at a solution

    I don't really understand how to start.... I was thinking we could calculate the total momentum of the system then calculate the ideal final velocity of the less massive cart then calculate the difference?

    Would you fine the total momentum of the system... thus (1.11 kg)(8.75 m)+(1.81 kg)(- 3.39m) = 3.5766 kg * m /s then divide that by 1.11 kg to find the ideal velocity of the smaller cart.... or 3.22216ish m/s.... then get the kinetic energy of that which will be 1/2(1.11)(3.2216)^2 = 5.762192 J , then find the kinetic energy of the kinetic energy of it actually, or 1/2(1.11)(2.99)^2 = 4.961755 then subtract 5.7621 - 4.9617 = .800 J ?
     
    Last edited: Oct 27, 2009
  2. jcsd
  3. Oct 27, 2009 #2
    I think you're on the right track. Use conservation of momentum to find the velocity of the other cart after the collision. The total kinetic energy before the collision can be calculated using (1/2)*m*v^2 for each of the carts. If you do the same after, the kinetic energy will be less. So you're right, subtract.
     
  4. Oct 27, 2009 #3
    Ultimately, we are trying to find [tex]\Delta[/tex]KE which is equal to KEf - KEi.

    KEf = KE1,f + KE2,f and KEi = KE1,i + KE2,i

    KEf = [tex]\frac{1}{2}[/tex]m1v1,f2 + [tex]\frac{1}{2}[/tex]m2v2,f2 and KEi = [tex]\frac{1}{2}[/tex]m1v1,i2 + [tex]\frac{1}{2}[/tex]m2v2,i2

    Therefore, [tex]\Delta[/tex]KE = [tex]\frac{1}{2}[/tex](m1v1,f2 + m2v2,f2 - m1v1,i2 - m2v2,i2)

    Of course, we don't know v2,f, so we have to use conservation of momentum to find it.

    Since the law of conservation of momentum tells us that pi = pf and we know p = mv, then it follows that m1v1,i + m2v2,i = m1v1,f + m2v2,f

    Solving for v2,f, we get v2,f = ( m1v1,i + m2v2,i - m1v1,f ) / m2

    After solving for v2,f, you can substitute that value into the [tex]\Delta[/tex]KE equation, and that should tell you the amount of KE lost.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Momentum Head on Collision Cart Problem
Loading...