Conservation of Linear Momentum

  • Thread starter mcarloni
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  • #1
mcarloni
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Two friends, Al and Jo, have a combined mass of 151 kg. At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart, because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 1.23 m/s, while Jo moves off in the opposite direction at a speed of 0.799 m/s. Assuming that friction is negligible, find Al's mass.

I know that:
m1 + m2 = 151
V(Al) = -1.23
V(Jo) = .799
m1 = 151 - m2
m1v1 = m2v2

but don't know how to separate the masses. Can anyone guide me through that process?

Greatly appreciated.
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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m1 = 151 - m2
m1v1 = m2v2
You know the two velocities, so only two unknowns m1 and m2. You have two equations relating them, so you can find both. Solve one of the equations for m1 (already done) and substitute into the other to eliminate m1. Then solve it for m2.
 
  • #3
mcarloni
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Got it!

Thank you
 

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