Query Assigment 23

course phy 201

??????????€??assignment #023

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Physics I

07-20-2007

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00:37:10

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 -->

m1= 450 kg

m2= 550 kg

v1= 4.5 m/s

v2= 3.7 m/s

450 kg (4.5 m/s) + 550 kg (3.7 m/s) = 450 kg *v1 + 550 kg * v2

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00:39:17

** 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 that diving the equation by 10 you have

406 m/s = 45 v1` + 55 v2`

and

3.7 m/s - 4.5 m/s = v2' - .8 m/s

406 m/s= 45 ( v2' - .8 m/s) + 55v2'

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Good responses. Let me know if you have questions. &#