# Homework Help: Collision in 2 dimensions

1. Oct 24, 2004

### 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!!!!

2. Oct 24, 2004

### arildno

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

3. Oct 24, 2004

### skonstanty

m1vf1+m2vf2=m1v01+m2v02

4. Oct 24, 2004

### 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?

5. Oct 24, 2004

### skonstanty

yes I agree

6. Oct 24, 2004

### arildno

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

7. Oct 24, 2004

### skonstanty

sin65 and cos 37?

8. Oct 24, 2004

### 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?

9. Oct 24, 2004

### 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!

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