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Collision in 2 dimensions

  • Thread starter skonstanty
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  • #1
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Puck A has a mass of 0.025 kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with Puck B, which has a mass of 0.050 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles: Puck A 65 degrees north of east and Puck B at 37 degrees south of east.

Find the final speed of a) Puck A and b) Puck B.

Please help!!!!
 

Answers and Replies

  • #2
arildno
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Set up the equation of conservation of linear momentum; we'll take it from there..
 
  • #3
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m1vf1+m2vf2=m1v01+m2v02
 
  • #4
arildno
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That's true enough, as long as you remember that the velocities are, in general vectors (Right?).

Now, we have been given the DIRECTIONS of the resultant velocities; so what we need to determine in our problem, is the MAGNITUDES (speeds) of these velocities.

Do you agree that this is what you have to find out?
 
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yes I agree
 
  • #6
arildno
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Okay:
Can you set up the DIRECTION vectors of your two resultant velocities, in terms of their east/north COMPONENTS?
 
  • #7
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sin65 and cos 37?
 
  • #8
arildno
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That's not vectors is it?
I'll take the first one for you:
Let [tex]\vec{i}[/tex] be the unit vector eastwards along the positive x-axis.
The unit vector northwards is therefore along the positive y-axis, that is [tex]\vec{j}[/tex]
The directionvector always have unit length, and we know that the first one is 65 degrees north of east.
Hence, we have the first direction vector:
[tex]\vec{d}_{1}=\cos(65)\vec{i}+\sin(65)\vec{j}[/tex]
can you set up the other direction vector?
 
  • #9
arildno
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Just to move this thread onwards, do you understand why the second direction vector is:
[tex]\vec{d}_{2}=\cos(37)\vec{i}-\sin(37)\vec{j}[/tex]

Secondly, now that you have the direction vectors, reformulate the conservation lof linear momentum by including these!
 

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