2-d momentum collision

  • Thread starter tamir102
  • Start date
  • #1
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Homework Statement



Two hockey pucks of equal mass undergo a collision
on a hockey rink. One puck is initially at rest, while
the other is moving with a speed of 5.4 m/s. After the
collision, the velocities of the pucks make angles of
33° and 46° relative to the original velocity of the
moving puck. Determine the speed of each puck after the
collision.

V1i=5.4m/s
m1=m2
V2i= 0
V1'=?
V2'=?

Homework Equations




P=P'
M1V1+M2V2=M1V1'+M2V2'
Eki=Ekf
1/2mv1i^2+1/2mv2i^2=1/2mv1'^2+1/2mv2'^2


The Attempt at a Solution




k so i understand we have two unknowns and thus we should have two unknown equations.

so ..

M1V1+M2V2=M1V1'+M2V2'

masses equal so they can be cancelled
and we know V2=0 so that whole part is removed

v1= v1'+v2'

5.4= v1'+v2'


5.4-v1'=v2'

^^ first unknown equation

now when i place it into 1/2mv1i^2+1/2mv2i^2=1/2mv1'^2+1/2mv2'^2
it does not give me the right answer or better yet i do not know how to continue on from this i always get

5.4-v2^2=5.4-v2^2+v2'^2


can someone please explain ...thank you in advance.
 

Answers and Replies

  • #2
kuruman
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This is a two-dimensional collision so you need to conserve momentum independently along the original direction of the puck and along a direction perpendicular to it. The problem does not mention that kinetic energy is conserved, so you may not assume that it is.
 
  • #3
8
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oh okay thank you i will try to figure it out
 

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