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Calculating both final velocities (conservation of momentum)

  1. Sep 28, 2013 #1
    How do you calculate both final velocities in an elastic collision (assuming all momentum is conserved). For example, if a 1000kg car going at 20m/s rammed into a 800kg car going at 30m/s opposite direction, what would both their final velocities be?

    In my year 11 physics course, we only covered questions that gave you 3 of the 4 velocities, so I am interested as to how you are meant to figure this type of question out.

    Thanks for any help.
     
  2. jcsd
  3. Sep 28, 2013 #2

    Nugatory

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    Are you solving the problem in one dimension (the directions of the cars after the collision is is still in their line of travel) or two (the cars may leave the road, in opposite direction so momentum is still conserved)?

    In the easier one-dimension case conservation of kinetic energy and conservation of momentum give you two equations. You have two unknowns (the velocities after the collision) so with two equations for your two unknowns, you're just some algebra away from an answer.

    The two-dimensional case needs more information. In general, there is a different solution for each possible angle of the cars coming out of the collision.
     
  4. Sep 28, 2013 #3
    You're right. You're missing one piece of information. Using conservation of momentum only gives you a linear relationship between the two final velocities (in this case, 8*v2 = 1 + 10*v1 I believe). The problem is underconstrained.

    More than one pair of final velocities could result from this collision and still conserve momentum.

    Each possible pairing corresponds to a collision that conserves a varying amount of kinetic energy. This is where the final piece of information comes from. If you say kinetic energy is conserved (as is nearly the case in a billiard ball collision) then you also know that 0.5*mv^2 is the same before and after as well. In the case of two cars, I'd say almost none of the kinetic energy is conserved, say 10%, so you might say:

    0.5*mv^2 (before collision) = 10*[0.5*mv^2 (after collision)]

    Without knowing how much kinetic energy is conserved, you cannot say for sure what the final velocities of the two cars will be.
     
  5. Sep 28, 2013 #4
    I'm not 100% sure. But to me you could simply utilize the other vehicle's params and substitute them in with the original.

    I.e. if a car crashed into a stationary wall, what is the force, just general F=ma, with some standard velocity equation.

    In this case, car1 at 1000kg and car2 = 800kg, car1 now equals 1000+800kg. It's final velocity instead of 20m/s, can be 20+50m/s. Then the result would be the collision of car1 into a stationary object, however it is equivalent in force as the 2 opposing cars crashing to your question. Assuming speed is constant, v=u

    I just realized I didn't answer your question, haha, I gave you the resultant force produced. Either way same concept applies as mentioned above, but deal with velocity instead of force. Just play around with the idea in your mind, until you arrive at a logical equivalent. I.e. both cars crashed into a wall, which produces largest force, that will be the car with a v>0 in your question, whereas the other will have v=0, but actually calculate v.
     
  6. Sep 28, 2013 #5
    I guess the cars was a bad example. So If we replaced the cars with billiard balls of the same masses and velocities as the cars had, and all momentum was conserved, and assuming they both kept their line, whether backwards or forwards, then how would I go about finding both final velocities?

    In a collision such as this, the billiard balls should not deviate off their course as a result of the collision, should they, as it is directly in line?
     
  7. Sep 28, 2013 #6

    Nugatory

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    You have two unknowns, namely the velocity of each of two balls after the collision. Write down the equation for conservation of momentum (sum of the momenta after the collision equals the sum of the momenta before) and conservation of energy (sum of the kinetic energies after the collision equals the sum of the kinetic energies before the collision) and you have two equations. Two equations is enough to solve for your two unknowns.

    (BTW, you will actually find two solutions when you do the algebra. It's worth taking a moment to think about the physical significance of both solutions and why it makes sense that there are two solutions.... But don't do this until after you've found the solutions).
     
  8. Sep 29, 2013 #7
    Alright thanks for your help. I have the answers now after getting run through it by a friend. So I came up with the answers of -24.44 m/s and 20 m/s for the 1000kg mass, and -30 m/s and 25.56 m/s for the 800kg mass. So the 20 m/s and 30 m/s are the initial velocities and the -24.44 m/s and 25.56 m/s are the final velocities.

    Is there an easy way to do this, as it took several steps and different equations and formulae to work it out?
     
  9. Sep 30, 2013 #8

    jbriggs444

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    One standard approach would be to start by changing to a coordinate system in which the center of gravity of the system is motionless at the origin. One ball approaches from the left with a rightward velocity and the other from the right with a rightward velocity.

    Conservation of momentum implies that the velocities of the two balls (as judged against this coordinate system) are always directed oppositely to one another and that their speeds always have a fixed ratio. That ratio is the inverse of the ratio of the two masses.

    This means that the total kinetic energy of the system is a simple function of the speed of one of the balls (either ball will work).

    If the collision is dead-center on the x axis then the rebound direction is easy. All that's left is determining the rebound speed. In an elastic collision energy is conserved. Because energy is a function of the speed of the ball, its speed must be unchanged by the collision. It is simply that each ball's direction has been reversed.

    Take the reversed velocities, transform back to the original coordinate system and read off the resulting velocity.
     
  10. Sep 30, 2013 #9

    arildno

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    If you use two cars, rather than, say, two billiard balls with their center of mass at their geometric centers, you will in general need to find induced state of rotation after collision as well. That is, you'll need to use conservation of angular momentum as well, in addition to conservation of (linear) momentum
     
  11. Oct 1, 2013 #10
    @jbriggs444 Would that be something like:

    for 1000kg mass, final velocity = 1000/800 * -20?
    for 800kg mass, final velocity = 800/1000 * -30?

    For final velocity relative to initial velocity.

    @arildno So the equations would be different when having angled collisions?

    I will do some research into this. Thanks for all your help.
     
    Last edited: Oct 1, 2013
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