Final Speeds After 2-Dimensional Collision

[skonstanty]

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!

 

[arildno]

Set up the equation of conservation of linear momentum; we'll take it from there..

 

[skonstanty]

m1vf1+m2vf2=m1v01+m2v02

 

[arildno]

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?

 

[skonstanty]

yes I agree

 

[arildno]

Okay:
Can you set up the DIRECTION vectors of your two resultant velocities, in terms of their east/north COMPONENTS?

 

[skonstanty]

sin65 and cos 37?

 

[arildno]

That's not vectors is it?
I'll take the first one for you:
Let \vec{i} be the unit vector eastwards along the positive x-axis.
The unit vector northwards is therefore along the positive y-axis, that is \vec{j}
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:
\vec{d}_{1}=\cos(65)\vec{i}+\sin(65)\vec{j}
can you set up the other direction vector?

 

[arildno]

Just to move this thread onwards, do you understand why the second direction vector is:
\vec{d}_{2}=\cos(37)\vec{i}-\sin(37)\vec{j}

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