# Basic Momentum Question

1. Mar 6, 2016

### Simoncoolstan

1. The problem statement, all variables and given/known data
A 0.30kg puck A, moving at 5m/s [W], undergoes a collision with a 0.40 puck B which is initially at rest. Pick A moves off at 4.2m/s [W 30deg N]. Find the final velocity of Puck B.

2. Relevant equations
pi=pf
mvi=mvf

3. The attempt at a solution

Let [N] and [W] be positive

Puck A
vix = 5m/s
m1= 0.30kg
vf=4.2m/s
vfx : 4.2cos30=3.6m/s
vfy : 4.2sin30=2.1 m/s

Puck B
vi=0m/s
m2=0.40kg

Solve for v2fx

m1v1ix + m2v2ix = m1v1fx + m2v2fx

(0.30)(5)+(0.40)(0)=(0.30)(3.6)+(0.40)v2fx
2.43= v2fx

Solve for v2fy

m1v1iy + m2v2iy = m1v1fy + m2v2fy

(0.30)(0)+(0.40)(0)=(0.30)(2.1)+(0.40)v2fy
-1.58 = v2fy

Find the resultant v2f

|v2f| = sqrt((2.43)2+(1.58)2)
|v2f| = 2.9 m/s

tantheta = (1.58/2.43)
theta = tan-1(1.58/2.43)
theta = 33deg

Final answer: velocity of Puck B is 2.9m/s [33deg SW].

Can someone tell me if I'm doing this correctly?:

Last edited: Mar 6, 2016
2. Mar 6, 2016

### haruspex

That step looks wrong.

By the way, you should keep some more decimal digits during the calculation or you may find the final result is rather inaccurate.

3. Mar 6, 2016

### Simoncoolstan

Corrected it, I got v2fx=3.47 m/s and theta now = 25 degrees.

Does that seem correct now?

4. Mar 6, 2016

5. Mar 6, 2016

### Simoncoolstan

Ok one second

6. Mar 6, 2016

### Simoncoolstan

Ok one second

7. Mar 6, 2016

### Simoncoolstan

(0.30)(5)+(0.40)(0)=(0.30)(3.6)+(0.40)v2fx
1.5+0=1.08+0.40v2fx
0.42=0.40v2fx
0.42/0.40=v2fx

1.05= v2fx

| v2 | = sqrt((1.05)2+(1.58)2)
= 1.9m/s

tantheta = (1.58/1.05)
theta = tan-1(1.58/1.05)
theta = 56 degrees

Answer: velocity of puck B is 1.05m/s [56deg SW]

How about now? Is my process wrong, or is it the math- this is what I get for doing my work at 6am lol.[/QUOTE]

8. Mar 6, 2016

### haruspex

That all looks ok now. "56 deg SW" is not a clear way of stating it. Better is "56 deg S of W".

Edit: You should get into the habit of doing very rough mental estimates to check your answers.
In the equation you were handling wrongly, you could see the puck A lost 1.4m/s of its x velocity, and it weighs only 3/4 of puck B. So puck B's x velocity must be 3/4 of 1.4m/s.

9. Mar 6, 2016

### Simoncoolstan

Ok thanks so much for you help!

10. Mar 6, 2016

### Simoncoolstan

Is that just a shorter way of performing the calculation?

11. Mar 6, 2016

### haruspex

whether it is shorter depends on exactly how you solved the equation. If you multiplied out all the terms first then yes it is. But more generally the numbers might have been a lot more awkward. The point is to do a mental check in terms of the approximate ratios and differences. E.g. if the masses had been 0.31 and 0.39 I would still have called it 3/4.

12. Mar 6, 2016