What is the Final Velocity of Puck B After Collision?

[Simoncoolstan]

Homework Statement

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.

Homework Equations

p_i=p_f
mv_i=mv_f

The Attempt at a Solution

Let [N] and [W] be positive

Puck A
v_ix = 5m/s
m₁= 0.30kg
v_f=4.2m/s
v_fx : 4.2cos30=3.6m/s
v_fy : 4.2sin30=2.1 m/s

Puck B
v_i=0m/s
m₂=0.40kg

Solve for v_2fx

m¹v_1ix + m₂v_2ix = m₁v_1fx + m₂v_2fx

(0.30)(5)+(0.40)(0)=(0.30)(3.6)+(0.40)v_2fx
2.43= v_2fx

Solve for v_2fy

m₁v_1iy + m₂v_2iy = m₁v_1fy + m₂v_2fy

(0.30)(0)+(0.40)(0)=(0.30)(2.1)+(0.40)v_2fy
-1.58 = v_2fy

Find the resultant v_2f

|v_2f| = sqrt((2.43)²+(1.58)²)
|v_2f| = 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?:

 

[haruspex]

(0.30)(5)+(0.40)(0)=(0.30)(3.6)+(0.40)v_2fx
2.43= v_2fx

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.

 

[Simoncoolstan]

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.

Corrected it, I got v_2fx=3.47 m/s and theta now = 25 degrees.

Does that seem correct now?

 

[haruspex]

Corrected it, I got v_2fx=3.47 m/s

Still wrong. Please post all your working.

 

[Simoncoolstan]

Ok one second

 

[Simoncoolstan]

Ok one second

 

[Simoncoolstan]

Still wrong. Please post all your working.

(0.30)(5)+(0.40)(0)=(0.30)(3.6)+(0.40)v2fx
1.5+0=1.08+0.40v_2fx
0.42=0.40v_2fx
0.42/0.40=v_2fx

1.05= v_2fx

| 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]

 

[haruspex]

(0.30)(5)+(0.40)(0)=(0.30)(3.6)+(0.40)v2fx
1.5+0=1.08+0.40v_2fx
0.42=0.40v_2fx
0.42/0.40=v_2fx

1.05= v_2fx

| 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.

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.

 

[Simoncoolstan]

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

Ok thanks so much for you help!

 

[Simoncoolstan]

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.

Is that just a shorter way of performing the calculation?

 

[haruspex]

Is that just a shorter way of performing the calculation?

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.

 

[Simoncoolstan]

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.

Ok thanks for your help!