query assignment 23

course phy201

‡ÿÆÜ×Ðèâ\Õœ˜ßô ˆQÌû}Îassignment #023

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Physics I

11-01-2007

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16:08:09

Query gen phy 7.27 bumper cars 450 kg at 4.5 m/s, 550 kg at 3.7 m/s, collision from back, elastic

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RESPONSE -->

I calculated the initial KE of both objects:

KEa=4556J KEb=3765J

And the intial P of both objects:

Pa=2025 Kg*m/s Pb=2035 Kg*m/s

And I know P and KE should be conserved in an elastic collision but not sure how to proceed from here.

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16:38:06

** 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. *

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RESPONSE -->

I understand and actually had worked the problem a different way but got the same answer as follows:

dv of the two objects is .8m/s * the mass of the faster object(450Kg)= a momentum ""shift"" of 360 Kg*m/s from faster to slower object. Addind this amount of momentum to the intial momentum of the slower object and dividing by the mass gives the new velocity of 4.4m/s

Subtracting this shift in momentum from the original momentum of the faster object and dividing by the mass gives the new velocity of 3.7 m/s for the faster object.

Is this incorrect?

The relative velocity of the two objects before collision is 8 m/s, with the less massive car moving faster. After collision the relative velocity will be 8 m/s, but with the more massive car moving faster. This is a general principle of elastic collisions: the relative velocity after collision is equal and opposite to the relative velocity before collision. This is the meaning of the equation v2 - v1 = -( v2' - v1').

Your reasoning doesn't produce an accurate result; the relative velocity after collision is -.7 m/s, not -.8 m/s. It's close, because the two masses are not all that different; applied to a situation in which one mass is much greater than the other it won't be that close.

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Your work looks good. See my notes. Let me know if you have any questions. &#