Find speed and direction of a particle after collision

[jfnn]

SOLVED

THANK YOU

 

[CWatters]

Yes conservation of momentum is correct approach. I haven't checked your working but you seemed to jump about a bit. The following would be better order..

For the x direction: (the initial velocity of the particle is fully in the y-component, thus the initial velocity for particle one and two, which are attached, in the x-direction is 0).

(1/3m)*0 + (2/3m)*0 = (1/3m)*v1x' + (2/3m)*v2x'

0 = (1/3m)*v1x' + (2/3m)*v2x' where v1x' (the velocity of particle one in the x direction after collision is -48 m/s, negative because it is west)

0 = (1/3m)*-48 + (2/3m)*v2x'

-16m + (2/3m)*v2x' = 0 --> This is equation 1

Then keep going with..

Then I took equation 1 to find v2x'

-16m + (2/3m)*v2x' = 0 --> (2/3m)*v2x' = 16m --> Masses cancel and 16 - 2/3*v2x' --> 24 = v2x'

Then I did the same thing except for the y direction

(1/3m)*75 + (2/3m)*75 = (2/3m)v2y'

75m = (2/3m)v2y'

v2y' = 112.5 m/s

So now you have the x and y components of velocity.

then I did the pathag theorem to find v2' = 115 m/s --> Is this the correct value/way to do this?

Yes it's the correct way.

I have no clue how to find the direction

Make a drawing showing the components V2x, V2y and the resultant V2. You used Pythagoras so you know looks like a right angled triangle.

Do you know how to work out angles using Sin() , Cos() or Tan() ?

 

[CWatters]

Re...

the velocity of particle one in the x direction after collision is -48 m/s, negative because it is west

It would have been better to state at the beginning which directions you will define as positive (eg up and east).

I have no clue how to find the direction

In addition to using trig to find the angle with the horizontal... You know that one part went directly west. So what does conservation of momentum tell you about the compass direction of the other part? The V2x is positive so we know it's not got a westerly component.

 

[jfnn]

Yes conservation of momentum is correct approach. I haven't checked your working but you seemed to jump about a bit. The following would be better order..Then keep going with..

So now you have the x and y components of velocity.
Yes it's the correct way.
Make a drawing showing the components V2x, V2y and the resultant V2. You used Pythagoras so you know looks like a right angled triangle.

Do you know how to work out angles using Sin() , Cos() or Tan() ?

I know how to do the relationships I think. So it would be theta = tan^-1(v2y'/v2x')?

Therefore, I would get theta=tan^-1 (112.5/24) = 78 deg

However, how do I know what the direction is? Is it from the horizontal axis, so would be N or E or is it from another axis? How can I be sure?

Thank you so much!

 

[jfnn]

Re...
It would have been better to state at the beginning which directions you will define as positive (eg up and east).
In addition to using trig to find the angle with the horizontal... You know that one part went directly west. So what does conservation of momentum tell you about the compass direction of the other part? The V2x is positive so we know it's not got a westerly component.

OH! So since the V2x' is positive, it means that it is going east, and since the y is positive as well (it is north) I can then say it is 78 deg N of E?

 

[CWatters]

OH! So since the V2x' is positive, it means that it is going east..

Correct. If it only splits into two parts and one goes exactly west the other must have a component that is exactly east. That way momentum in the Z axis is conserved.

and since the y is positive as well (it is north) I can then say it is 78 deg N of E?

No that's not correct. Y was the vertical axis. So your 78 degrees is the angle it makes with the horizontal. I have made a diagram. Everything happens in the XY plane..

 

[CWatters]

PS I still haven't checked your calculations (just the method).

 

[jfnn]

Correct. If it only splits into two parts and one goes exactly west the other must have a component that is exactly east. That way momentum in the Z axis is conserved.
No that's not correct. Y was the vertical axis. So your 78 degrees is the angle it makes with the horizontal. I have made a diagram. Everything happens in the XY plane..

View attachment 208313

Ah okay, that makes sense now. But the angle can still be 78 degrees N of E as that implies it is north of the horizontal right?

 

[jfnn]

PS I still haven't checked your calculations (just the method).

Okay. Let me know if you do check the calculations and see any errors. I double checked it so hoping it is correct.

 

[CWatters]

No "North of east" would mean it had a component in the Z axis. It must go exactly east and upwards.

 

[jfnn]

No "North of east" would mean it had a component in the Z axis. It must go exactly east and upwards.

ahh okay.

Makes great sense now. Thank you!

 

[CWatters]

Just a heads up... Try not to delete the original post in future because it makes it less useful for other people searching the forum for help with similar problems.