Calculating Final Velocities in 2D Inelastic Collisions

[fatcats]

Hi there, so I'm not sure if this is allowed or not, but usually before I submit my work I try to check it on the internet to get full marks. I do the work as best as I can to try and get the best understanding of the subject possible. I think a) is correct but I think I messed up somewhere in b).

1. Homework Statement
a) A 0.800 kg target slides along the ice at 3.0 m/s W when it's hit by a 20g arrow moving at 260 m/s N. Find the final velocity of the target after the inelastic collision.

b) A 0.3 kg puck A moving at 5 m/s W undergoes a collision with a 0.4 kg puck B which is initially at rest. Puck A moves off at 4.2 m/s [W 30 N]. Find the final velocity of puck B.

Homework Equations

Pto = Ptf
m1v1 + m2v2 = (m1 +m2) * vfx
basic trig sohcahtoa + pythagorean theorem

The Attempt at a Solution

There is a lot of math so I have decided to just include screenshots of my work from word document, please see enclosed.

I checked b) on the internet and it seemed like the answers do not line up but maybe I did it correctly... I have diagrams made as well that I can upload if needed. To me, what does not make sense in B, is why I got a negative value for the y vector.

Please help, want to understand why/where I am wrong.

 

[Abhishek kumar]

Hi there, so I'm not sure if this is allowed or not, but usually before I submit my work I try to check it on the internet to get full marks. I do the work as best as I can to try and get the best understanding of the subject possible. I think a) is correct but I think I messed up somewhere in b).

1. Homework Statement
a) A 0.800 kg target slides along the ice at 3.0 m/s W when it's hit by a 20g arrow moving at 260 m/s N. Find the final velocity of the target after the inelastic collision.

b) A 0.3 kg puck A moving at 5 m/s W undergoes a collision with a 0.4 kg puck B which is initially at rest. Puck A moves off at 4.2 m/s [W 30 N]. Find the final velocity of puck B.

Homework Equations

Pto = Ptf
m1v1 + m2v2 = (m1 +m2) * vfx
basic trig sohcahtoa + pythagorean theorem

The Attempt at a Solution

There is a lot of math so I have decided to just include screenshots of my work from word document, please see enclosed.

I checked b) on the internet and it seemed like the answers do not line up but maybe I did it correctly... I have diagrams made as well that I can upload if needed. To me, what does not make sense in B, is why I got a negative value for the y vector.

Please help, want to understand why/where I am wrong.

Its depend how you choose your reference axis if you choose west as positive x-axis and north as positive y-axis then east and south will be negative axis.if you got negative value of y component that means velocity is opposite to your reference axis.

 

[fatcats]

I understand that part of it, I just don't understand why a puck moving west hitting a stationary puck would then go north and why the other one goes south
And I don't know whether or not I am correct in that because to me it seems weird that the pucks would move in that way

 

[haruspex]

why a puck moving west hitting a stationary puck would then go north and why the other one goes south

Nothing strange about that. It is not a head on collision, so one will go to the right and one to the left, but both will go somewhat Westerly.

 

[Abhishek kumar]

I understand that part of it, I just don't understand why a puck moving west hitting a stationary puck would then go north and why the other one goes south
And I don't know whether or not I am correct in that because to me it seems weird that the pucks would move in that way

You mentioned here after collision puck A moves 4.2m\s W 30 N resovling it to x and y component and and seperately applying conservation of momentum with puck B you will get both component of puck B may be positive or negative.

 

[haruspex]

A few issues in part a.
You need to keep another digit of precision until the end.
Given the two velocity components after impact, does it seem right that the movement is more West than North?
In the working, you have a couple of extraneous "kg" units.

 

[haruspex]

Similar issues in part b. (I am assuming EW is the x axis.)
Check your sin and cos usage.
Keep an extra digit.
Check whether the final velocity is more to the S or more to the W.

 

[fatcats]

alright, it's very late here, so I'm going to take a go at fixing the math and post my new results tomorrow
thanks for letting me know what was wrong

 

[fatcats]

North/West designated positive for both questions

a) Vfx = 2.93
Vfy = 6.34
Vf = 6.94
I got the same answers as before, but added the digits you recommended.

For the angle I got 24.7 (dividing vfx by vfy). If I divide vfy by vfx I get 65.2 which sounds more correct, no?
I have the triangle drawn out but to be honest this is something that in general confuses me in physics. In my drawing, where do I put the given angle after drawing vfx and vfy pointing in the correct directions. I put mine with vfx as the opposite that I am solving for. Is this wrong?

b) Is this better: Vy1 = 2.1, Vx1 = 3.64
Was all I did mess up which was which there?

V2fx = 3.27
V2fy = -1.58
V = 3.63
Angle = Vfx/Vfy: -66.48

Vfy/fx -25.79 degrees

Same issue with angle, I think I do v2fy/v2fx. I don't know why I'm having such an issue with all of a sudden because I was pretty good at trig...

 

[haruspex]

I put mine with vfx as the opposite that I am solving for. Is this wrong?

It depends on the form you intend to quote in the answer. You gave an answer as W(angle)N. That would mean it is the angle to the x axis, measured clockwise from that axis.

Was all I did mess up which was which there?

my guess was that you got sin and cos crossed over, but it comes to the same.

 

[fatcats]

With the angle, I want to say W (angle) N. So for a: W 65.2 N?
I am visualizing it on the x-axis of the cartesian plane so I use Vfy as the opposite.
And for b: S 26 W

Does my work look correct now?

 

[haruspex]

With the angle, I want to say W (angle) N. So for a: W 65.2 N?
I am visualizing it on the x-axis of the cartesian plane so I use Vfy as the opposite.
And for b: S 26 W

Does my work look correct now?

You still have b wrong.
You took (finally) the angle with tangent vy/vx. That is the angle between the trajectory and the x axis, so it would be W(angle)S.

 

[fatcats]

Is that my only issue with B?
My bad, that was silly and I miswrote that >.< I always write it W/E (angle) S/N.
Thanks for all your help, appreciate it!

 

[haruspex]

Is that my only issue with B?
My bad, that was silly and I miswrote that >.< I always write it W/E (angle) S/N.
Thanks for all your help, appreciate it!

I think it is ok now, but I have not rechecked all the arithmetic after your corrections.

 

[fatcats]

Alright I'm triple checking the arithmetic and submitting it, thanks again!