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Car impact at 90 degrees to each other (conservation of momentum)

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Two cars collide at right angle in the intersection of two icy roads. Car A has a mass of 1000 kg and car B 1400 kg. The cars become entangled and moved off together with a common velocity v’ in the direction indicated in figure QB4. If car A was travelling 10 m/s at the instant of impact, find
    i) v’ and [13marks]
    ii) the corresponding velocity of car B just before impact. [7 marks]

    v’ is 25 degrees to the vertical, car A on the vertical axis car b on the horizontal axis

    2. Relevant equations

    M(a)V(a) + M(b)v(b) = m(t)v'

    3. The attempt at a solution

    Ok so i have drawn a few conclutions from this, one that the impact is elastic, as i don't physically have the the values to find the coefficient of restitution and i seem to remember that as they have become entangled nothing has reflected.

    This means KE and momentum is conserved, However when canceling the Ke equations down. i'm left with effectivly the same equation just a squared in it.

    conservation of momentum
    M(a)V(a) + M(b)v(b) = m(t)v'
    10000 +1400v(b) = 2400v'

    conservation of KE
    M(a)V(a)^2 + M(b)v(b)^2 = m(t)v'^2

    100000 + 1400v(b)^2 = 2400v'^2

    substituting one into the other i get something incredibilly complicated and is unlikely to work.
  2. jcsd
  3. Jan 22, 2012 #2

    Doc Al

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    Staff: Mentor

    Since the cars become entangled, the collision is perfectly inelastic. Kinetic energy is not conserved.

    Don't forget that momentum is a vector quantity. Treat vertical and horizontal components separately.
  4. Jan 22, 2012 #3
    Oh snap.

    1000 x 10 = 2400 x cos(25) x v'

    10000 / ( 2400 x cos (25) ) = v'

    v' =4.6ms^-1

    and then reversing this for the horizontal

    1400v = 2400 x sin(25) x 4.6
    v = (2400 x sin(25) x 4.6) / 1400
    v= 3.3ms^-1
  5. Jan 22, 2012 #4

    Doc Al

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    Staff: Mentor

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