Basic Physics question (2-d momentum)

  • #1
I can NOT figure this question out, it's the only one I havent gotten, can someone help?

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.020 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 0.040 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing.

07_32.gif


(a) Find the final speed of puck A.
______ m/s
(b) Find the final speed of puck B.
______ m/s
 

Answers and Replies

  • #2
119
0
Hi atlbraves,

usually you must propose a solution or give some hints ........but today I'll make an exception for you:

You must write the the conservation law for angular momentum:

[tex]\vec{p_{A0}}=\vec{p_A}+\vec{p_B}[/tex]

Then you have to write the above vectorial equation on Ox (the direction of the initial velocity of A) and Oy (perpendicular on it):

Ox: [tex]p_{A0}=p_A \cdot cos(\alpha_A)+p_B \cdot cos(\alpha_B)[/tex]
Oy: [tex]p_A \cdot sin(\alpha_A)-p_B \cdot sin(\alpha_B)=0[/tex]
 
  • #3
clive said:
Hi atlbraves,

usually you must propose a solution or give some hints ........but today I'll make an exception for you:

You must write the the conservation law for angular momentum:

[tex]\vec{p_{A0}}=\vec{p_A}+\vec{p_B}[/tex]

Then you have to write the above vectorial equation on Ox (the direction of the initial velocity of A) and Oy (perpendicular on it):

Ox: [tex]p_{A0}=p_A \cdot cos(\alpha_A)+p_B \cdot cos(\alpha_B)[/tex]
Oy: [tex]p_A \cdot sin(\alpha_A)-p_B \cdot sin(\alpha_B)=0[/tex]

thanks, i figured it out, i actually ended up knowing how to do it, the reason i wasnt getting the answer correct is i messed up a calculation, thanks for the help anyways
 

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