Momentum in two directions(q35)

[jack1234]

Hi, for this question
http://tinyurl.com/yvadld

This is what I have done:
in x direction
5*60=5*v1*cos30 + 2*v2*cos30 --(1)
In y direction
0=5*v1sin30 + 2*v2sin30 --(2)

From (2), v1=-(2/5)*v2 subs into (1)

And what I get is 300=-2*v2*cos30 + 2*v2*cos30

What is the problem?

 

[learningphysics]

Hi, for this question
http://tinyurl.com/yvadld

This is what I have done:
in x direction
5*60=5*v1*cos30 + 2*v2*cos30 --(1)
In y direction
0=5*v1sin30 + 2*v2sin30 --(2)

From (2), v1=-(2/5)*v2 subs into (1)

And what I get is 300=-2*v2*cos30 + 2*v2*cos30

What is the problem?

you should use:

0=5*v1sin30 - 2*v2sin30 --(2)

the two particles have opposite directions for y-components.

 

[jack1234]

Got it, answer is b:)

But does it mentioned in the question...or it just common sense?

 

[jack1234]

>>But does it mentioned in the question
What I mean is "the two particles have opposite directions for y-components."

 

[Doc Al]

You are told that each particle has a velocity that is 30 degrees from the original direction (along the x-axis). Is it possible (considering conservation of momentum) that the two particles both move to the same side of the x-axis?

Hint: What's the y-component of total momentum?

 

[jack1234]

I see...so it is base on observation:)

 

[Doc Al]

It is based on conservation of momentum.