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posted by  Maple on 8/12/2008 3:51:28 PM  |  status: Live  

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posted by zsm28 on 8/12/2008 6:04:41 PM  |  status: Live
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"Thanks a lot"
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m1 = 1.00 kg, m2 = 10.0 kg
v1 = 4.16 m/s, v2 = 2.73 m/s
find v1' and v2'
momentum conservation:
m1v1 + m2v2 = m1v1' + m2v2'
so
m1(v1 - v1') = m2(v2' - v2)
v1 - v1' = 10(v2' - v2)             (1)
elastic collision, so 
kinetic energy conservation:
m1v12/2 + m2v22/2 = m1v1'2/2 + m2v2'2/2
m1(v12 - v1'2) = m2(v2'2 - v22)
v12 - v1'2 = 10(v2'2 - v22)        (2)
(2)/(1):
v1 + v1' = v2' + v2              (3)
(1) + (3)
2v1 = 11v2' - 9v2
v2' = (2v1 + 9v2)/11 = 2.99 m/s
v1' = v2' + v2  - v1 = 1.56 m/s
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posted by Suebee on 8/12/2008 6:51:45 PM  |  status: Live
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Maple's comment:
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Two unknowns require two equations for this situation of moving particles.  You need a momentum relation and the kinetic energy relation for the particles:  Before = After
                     KEb = KEa
               1/2 m1v12 + 1/2 m2v22  = 1/2 m1v1f2  + 1/2 m2v2f2
Simplify         
                        m1 ( v12 - v1f2 ) = m2 ( v2f2 - v22 )   ...... #1
Now for momentum
                                Pi = Pf
                        m1v1 + m2v2  =  m1v1f + m2v2f
Simplify
                           m1 ( v1 - v1f )  = m2 ( v2f - v2 )  .......  #2
Divide #1 by #2; remember to factor the differnce or squares in #1; masses will cancel , results:
                              v1 + v1f  = v2f + v2
                                       v2f = v1 + v1f - v2
Substitute into #2; solve for v1f :
 
                              v1f = (m1-m2 / m1 + m2 )v1  + ( 2m2 / m1 + m2 ) v2
Likewise,                
                              v2f = ( 2m1 / m1 + m2 ) v1 - ( m1 - m2 / m1 + m2 ) v2
So sub the given:
                             v1f = 1.56 m/s   and v2f = 2.99 m/s
 
The answers make sense; the smaller mass should be moving slower after hitting a mass ten times larger, and the larger mass will get just a little boost in speed after being hit by the small mass.
 
     
 
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