Calculating Final Velocities in 2D Inelastic Collisions

In summary, the target slides along the ice at a speed of 3.0 m/s W when hit by a 20g arrow, but afterwards the target moves off at a velocity of 4.2 m/s W 30 N.
  • #1
fatcats
34
1
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.
 

Attachments

  • work for a) (target and arrow).png
    work for a) (target and arrow).png
    7.5 KB · Views: 543
  • work for b) (two pucks).png
    work for b) (two pucks).png
    8.6 KB · Views: 559
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  • #2
fatcats said:
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.
 
  • #3
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
 
  • #4
fatcats said:
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.
 
  • #5
fatcats said:
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.
 
  • #6
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.
 
  • #7
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.
 
  • #8
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
 
  • #9
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...
 
  • #10
fatcats said:
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.
fatcats said:
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.
 
  • #11
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?
 
  • #12
fatcats said:
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.
 
  • #13
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!
 
  • #14
fatcats said:
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.
 
  • #15
Alright I'm triple checking the arithmetic and submitting it, thanks again!
 

Related to Calculating Final Velocities in 2D Inelastic Collisions

What is the difference between impulse and momentum?

Momentum is a measure of an object's tendency to continue moving in the same direction with the same speed. It is calculated by multiplying an object's mass by its velocity. Impulse, on the other hand, is the change in an object's momentum over time. It is calculated by multiplying the force applied to an object by the time it is applied.

What is the conservation of momentum?

The conservation of momentum states that in a closed system, the total momentum before an event is equal to the total momentum after the event. This means that in any interaction between two objects, the total momentum of the system remains constant.

How is momentum conserved in 2D collisions?

In 2D collisions, the total momentum in the x-direction and the y-direction must both be conserved. This means that the sum of the momentums in each direction before the collision must be equal to the sum of the momentums in each direction after the collision.

What is the difference between elastic and inelastic collisions?

In elastic collisions, both momentum and kinetic energy are conserved. This means that after the collision, the objects involved will have the same total momentum and same total kinetic energy as before the collision. In inelastic collisions, only momentum is conserved. This means that after the collision, the objects involved will have the same total momentum, but the total kinetic energy may be different due to energy being lost to other forms, such as heat or sound.

How does the angle of impact affect the momentum of a collision?

The angle of impact can affect the direction and magnitude of the momentum transfer between two objects during a collision. If the objects collide head-on, the momentum transfer will be in the same direction as the initial momentum of the objects. If the objects collide at an angle, the momentum transfer will be in a direction that is a combination of the initial momentum and the angle of impact. The magnitude of the transferred momentum will also be affected by the angle of impact, with a larger angle resulting in a smaller magnitude of transferred momentum.

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