# Simple Particle Motion (Dynamics)

1. May 17, 2012

### dunn

1. The problem statement, all variables and given/known data

A particle of mass 3.2kg is moving due west with a velocity of 6m/s. Another particle of mass 1.6kg is moving due north with a velocity of 5.0m/s. The two particles are interacting. After 2s the first particle is moving in the direction N 30° E with a velocity of 3m/s. Find the magnitude and direction of the second particle.

2. Relevant equations

Δp1 = -Δp2

3. The attempt at a solution

Okay, so I have all the information I need except that I need to decompose the direction of v1' (i.e. the velocity of the first particle after the interaction) into coordinates along the x and y axes.

Here is my data:

v1 = -6i
v1' = 3(cos(60)i+sin(60)j) = 1.5i +2.6j
v2 = 5j
v2' = ?

I write m1(v1' - v1) = -m2(v2' - v2) (i.e. Δp1 = -Δp2)

Solving for v2' I get:

(m1(v1 - v1')/m2) + v2 = v2'

I put in the relevant data and get:

-15i -0.2j = v2'

But apparently this isn't correct. I keep checking my equation against the provided answer but I can't see where my mistake is even though I know my answer is wrong. I think I made a mistake when I decomposed the vector but I don't see how.

Last edited: May 17, 2012
2. May 17, 2012

### dunn

Okay, I think the book has (YET ANOTHER!) misprint and the angle should actually be N 23° E rather than N 30° E. Could someone verify this for me? It's late, I'm tired, and maybe I made an error.

(I need a new exercise text).

Last edited: May 17, 2012
3. May 18, 2012

### BruceW

Verify it? If you write your answer using N 23° E in the calculation, then I can say if I agree with you. But since you have the provided answer, you could just check against that.

Also, your answer using N 30° E looks correct to me. And nicely set out, which is good. I'm not familiar with the notation N 30° E, but I guess it is supposed to mean 30 degrees clockwise from north, by the way you have answered the question. Maybe your exercise book meant it to be something else, which is why you get a different answer to them. Or it could just be as you say, they might have misprinted. It is annoying when that happens.

4. May 19, 2012

### dunn

Yes, N 30° E means 30 degrees clockwise from the north direction.

The problem is that the book only gives simple numerical answers so I can't check to see if I did anything wrong in particular. With this exercise there were others associated tasks (finding the change in momentum of the particles and such) which show the exact same answer as mine for N 30° E, yet the given answer to the main problem that I posted here (find the momentum of the second particle) is clearly wrong. Oh well. At least I practiced finding angles given magnitudes and direction.

5. May 19, 2012

### BruceW

I've checked it twice, and there is also all the checking you have done, so I'm pretty certain you've done the method correctly. And yes, the practice of this is the main thing. There always are at least a few misprints in textbooks. You wouldn't think it was so difficult to get rid of them all, but apparently it is.