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Inelastic collision and Kinetic Energy Conservation

  1. Nov 11, 2008 #1
    A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. What is the speed of the wreckage just after the collision?

    Relevant equations: m1v1 - m2v2 = (m1+ m2) vfinal
    My attempt: This is an inelastic collision because the cars collide and stick together. So, momentum is conserved but kinetic energy is not.

    2)A 1.80-kg block slides on a rough horizontal surface. The block hits a spring with a speed
    of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient
    of kinetic friction between the block and the surface is μk = 0.560, what is the force
    constant of the spring?

    My attempt:

    1/2mv2 - Mgd - 1/2kx = 0
    mv2 -Mg
    [(1.80kg)(2m/s)(2m/s) - (0.560)(10)]/0.1meters
    k = 16 n x m
  2. jcsd
  3. Nov 11, 2008 #2
    For #1, momentum is conserved in its components.

    m1v1x + m2v2x = (m1+ m2) v'x
    m1v1y + m2v2y = (m1+ m2) v'y

    The car travelling north has no velocity in the x-direction and similarly, the car heading east has no velocity in the y-direction.

    Solve both equations to get the components on the velocity after the collision and use pythagoream theorem to solve for the final velocity and the direction
  4. Nov 12, 2008 #3
    thank you so much!!
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