#$&* course phy 201 Aug 1 4:15pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a** For an elastic collision we have m1 v1 + m2 v2 = m1 v1' + m2 v2' and v2 - v1 = -( v2' - v1'). We substitute m1, v1, m2 and v2 to obtain 450 kg * 4.5 m/s + 550 kg * 3.7 m/s = 450 kg * v1 ' + 550 kg * v2 ', or 4060 kg m/s = 450 kg * v1 ' + 550 kg * v2 ' . Dividing by 10 and by kg we have 406 m/s = 45 v1 ' + 55 v2 '. We also obtain 3.7 m/s - 4.5 m/s = -(v2 ' - v1 ' ) or v1 ' = v2 ' - .8 m/s. Substituting this into 406 m/s = 45 v1 ' + 55 v2 ' we obtain 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' . We easily solve for v2 ' obtaining v2 ' = 4.42 m/s. This gives us v1 ' = 4.42 m/s - .8 m/s = 3.62 m/s. Checking to be sure that momentum is conserved we see that the after-collision momentum is pAfter = 450 kg * 3.62 m/s + 550 kg * 4.42 m/s = 4060 m/s. The momentum change of the first car is m1 v1 ' - m1 v1 = 450 kg * 3.62 m/s - 450 kg * 4.5 m/s = - 396 kg m/s. The momentum change of the second car is m2 v2 ' - m2 v2 = 550 kg * 4.42 m/s - 550 kg * 3.7 m/s = + 396 kg m/s. Momentum changes are equal and opposite. NOTE ON SOLVING 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' FOR v2 ': Starting with 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' use the Distributive Law to get 406 m/s = 45 v2 ' - 36 m/s + 55 v2 ' then collect the v2 ' terms to get 406 m/s = -36 m/s + 100 v2 '. Add 36 m/s to both sides: 442 m/s = 100 v2 ' so that v2 ' = 442 m/s / 100 = 4.42 m/s. * STUDENT NOTE AND INSTRUCTOR RESPONSE: Ok i wasnt sure about the formulas which are:m1 + v1 + m2 + v2 = m1 + v1' + m2 + v2'and v2 -v1 = -(v2' - v1') INSTRUCTOR RESPONSE: Those are the formulas, except that you don't have any * in the first (* should replace two of your + signs; should read m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2' or just m1 v1 + m2 v2 = m1 v1' + m2 v2'). Note that the units of m1 and v1, for example, make it impossible to add m1 and v1. The meaning of the formulas is simpler and easier to remember than the formulas. If you understand the meaning you can always reconstruct the formulas: For an elastic collision: • Total momentum after = total momentum before. • Relative velocity reverses (i.e., relative velocity after collision is equal and opposite to relative velocity before collision). Recall that an elastic collision is one in which both momentum and kinetic energy are conserved. The formulas are straightfoward: m1 v1 + m2 v2 = m1 v1' + m2 v2' and1/2 m1^2+ 1/2 m2 v2^2 = 1/2 m1 v1'^2 1/2 m2 v2'^2. The second formula is more complicated than v2 - v1 = - (v2' - v1'), which combined with the first formula gives us the same results, so we use this formula instead of the second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " #$&*