Vector subtraction with momentum

[Lolagoeslala]

Homework Statement

solve this change in velocity when the V2` = 19.04 m/s [S59W]
and V2=16.32 m/s [S36W]

Homework Equations

The Attempt at a Solution

I solved this using the components method first, for which i got the answer 7.536 m/s [W 26.78 N]

And now for the vector subtraction piece:

I did this diagram:
http://s1176.beta.photobucket.com/user/LolaGoesLala/media/dffdfd.jpg.html

So this is what i did
I used cosine law like this:
Δv^2 = (19.04)^2 + (16.32)^2 -2(19.04)(16.32)cos23
Δv= 7.536

then i used sine law like this:

7.536/sin23 = 19.04/sinx
for x = 81°

and then when i place it into the diagram and subtract 36 from it, i get 45°. So obviously it would be [W 45 N] but by using the components method i get a totally different angle why?

 

[TSny]

7.536/sin23 = 19.04/sinx
for x = 81°

There's more than one solution for x.

 

[Lolagoeslala]

There's more than one solution for x.

Umm what do you mean by one solution..
according to the diagram i made... this is the x we were looking for.,..

 

[TSny]

For example, the inverse sine of 0.5 has more than one answer, not just 30 degrees.

 

[Lolagoeslala]

For example, the inverse sine of 0.5 has more than one answer, not just 30 degrees.

I am confused.. where are you getting the 0.5 and 30 degress from?

 

[TSny]

I just made up an example to illustrate an important point. What is the value of sin^-1(0.5)? Your calculator will probably give you 30^o. But that's not the only answer.

 

[Lolagoeslala]

I just made up an example to illustrate an important point. What is the value of sin^-1(0.5)? Your calculator will probably give you 30^o. But that's not the only answer.

so what's the other answer then?

 

[TSny]

Are you familiar with the unit circle representation of the sine function? See http://maine.edc.org/file.php/1/tools/TrigUnitCircle2.html for example and click the "y = sinθ" box and the "show coordinates" box. Then click and drag the green dot on the horizontal axis of the graph. Can you find another angle besides 30^o where the sine function has a value of 0.5?

 

[Lolagoeslala]

Are you familiar with the unit circle representation of the sine function? See http://maine.edc.org/file.php/1/tools/TrigUnitCircle2.html for example and click the "y = sinθ" box and the "show coordinates" box. Then click and drag the green dot on the horizontal axis of the graph. Can you find another angle besides 30^o where the sine function has a value of 0.5?

so 149 degrees?

 

[TSny]

Yes, actually 150 degrees. So, both 30 degrees and 150 degrees would be solutions to sin x = 0.5. Note that if you add 30 and 150 you get 180 degrees. The sine function has the property that if two angles add to 180 degrees, then the sine of the angles will be equal. So, what would be another angle that would have the same sine as 81 degrees?

 

[Lolagoeslala]

Yes, actually 150 degrees. So, both 30 degrees and 150 degrees would be solutions to sin x = 0.5. Note that if you add 30 and 150 you get 180 degrees. The sine function has the property that if two angles add to 180 degrees, then the sine of the angles will be equal. So, what would be another angle that would have the same sine as 81 degrees?

I am guessing. 99°

 

[TSny]

No need to guess. Grab the calculator and see if sin 99^o = sin 81^o.

From your diagram, does the angle "x" look like its greater than or less than 90 degrees?

 

[Lolagoeslala]

No need to guess. Grab the calculator and see if sin 99^o = sin 81^o.

From your diagram, does the angle "x" look like its greater than or less than 90 degrees?

It looks like its greater

 

[Lolagoeslala]

No need to guess. Grab the calculator and see if sin 99^o = sin 81^o.

From your diagram, does the angle "x" look like its greater than or less than 90 degrees?

Well these velocities are used for finding the change in velocity for the second ball. the first ball had a velocity change of 18.02 m/s [E 32 S] and had a mass of 0.165 kg

I need to find the mass for the second ball.
And this equation was given.. m2 = (v1`-v1 / v2 - v2`) m1
So i changed the diagram so that it would be v2 - v2` and i get 27 as the angle and the angle i get is [E 27 E] and the velocity of 7.536 m/s. I used that information to cancel out the [E32S] and [E27E] and divided the velocityand multiplied by the mass of the first ball. I got the final mass to be 0.39 kg. Was my process correct?

 

[TSny]

Well these velocities are used for finding the change in velocity for the second ball. the first ball had a velocity change of 18.02 m/s [E 32 S] and had a mass of 0.165 kg

I need to find the mass for the second ball.
And this equation was given.. m2 = (v1`-v1 / v2 - v2`) m1
So i changed the diagram so that it would be v2 - v2` and i get 27 as the angle and the angle i get is [E 27 E] and the velocity of 7.536 m/s.

Did you mean to type [E 27 S]?

I used that information to cancel out the [E32S] and [E27E] and divided the velocityand multiplied by the mass of the first ball. I got the final mass to be 0.39 kg. Was my process correct?

I don't understand how you are cancelling angles.

But something's wrong. Do you see why the direction of the change in velocity of m2 should be exactly in the opposite direction of the change in velocity of m1? But that's not what you're getting. I don't see any error in how you got the change in velocity of m2. So, either v2 or v2' individually is wrong, or the change in v1 is wrong.

 

[Lolagoeslala]

Did you mean to type [E 27 S]?

I don't understand how you are cancelling angles.

But something's wrong. Do you see why the direction of the change in velocity of m2 should be exactly in the opposite direction of the change in velocity of m1? But that's not what you're getting. I don't see any error in how you got the change in velocity of m2. So, either v2 or v2' individually is wrong, or the change in v1 is wrong.

Ok well let's continue on with finding the angle for Change in velocity for the second ball so it is 99 degrees I am i correct? but how would i represent that .. because in my personal notes i showed the whole process of finding the angle "x" to be 81 degrees. should i just write beside the answer for x that ... x can have two angle and then continue on with that?

So i do use 99 degrees as for my x...
I would have a diagram like this... http://s1176.beta.photobucket.com/user/LolaGoesLala/media/dffdfd-1.jpg.html

So it would be [W 27 N] so i think now i can cancel out the [ E 32 S] yes the angles are a bit off but i think that would the mistake in the experiment data collected..

 

[TSny]

Ah, these are experimental values. Good.
So, if you chalk up the difference between 27 degrees and 32 degrees as experimental error, I believe your calculation for m2 (.39 kg) is correct.

 

[Lolagoeslala]

Ah, these are experimental values. Good.
So, if you chalk up the difference between 27 degrees and 32 degrees as experimental error, I believe your calculation for m2 (.39 kg) is correct.

Oh so i guess the angles have to be completely opposite like we just found to cancel out correct?? BUT HOW sould i explain to my teacher that the 81 is same as 99... i mean what should i right in my notes?

 

[TSny]

Well, 81 is not the same as 99. Only 99^o is the correct answer for your problem. It's just that the sine of 81^o is the same as the sine of 99^o. You're going to run into that problem when you use the law of sines to find an angle larger than 90 degrees. I think the component method is the best way to subtract vectors. You got the correct answer that way.

Alternately, you could use the law of sines to find the other unknown angle in the triangle, and then find the angle x from the fact that the three angles must add up to 180 degrees.

 

[Lolagoeslala]

Well, 81 is not the same as 99. Only 99^o is the correct answer for your problem. It's just that the sine of 81^o is the same as the sine of 99^o. You're going to run into that problem when you use the law of sines to find an angle larger than 90 degrees. I think the component method is the best way to subtract vectors. You got the correct answer that way.

Alternately, you could use the law of sines to find the other unknown angle in the triangle, and then find the angle x from the fact that the three angles must add up to 180 degrees.

On my sheet with this question they give the equation

m2 = (v1`-v1 / v2 - v2`)m1

So does that mean the change in velocity should be with v2-v2`
not v2`-v2 ? that how we did it

 

[PeterO]

Oh so i guess the angles have to be completely opposite like we just found to cancel out correct?? BUT HOW sould i explain to my teacher that the 81 is same as 99... i mean what should i right in my notes?

Having used the cosine rule to find the size of the resultant vector you are after, you know all 3 sides of this triangle.
If you re-apply the Cosine rule to find the next angle, you will not get a dual answer situation - you would just get the 91^o.
The sine rule is certainly simpler (arithmetically) but has the ambiguous answer issue to deal with on many occasions.

 

[Lolagoeslala]

Having used the cosine rule to find the size of the resultant vector you are after, you know all 3 sides of this triangle.
If you re-apply the Cosine rule to find the next angle, you will not get a dual answer situation - you would just get the 91^o.
The sine rule is certainly simpler (arithmetically) but has the ambiguous answer issue to deal with on many occasions.

oh okayyy...

On my sheet with this question they give the equation

m2 = (v1`-v1 / v2 - v2`)m1

So does that mean the change in velocity should be with v2-v2`
not v2`-v2 ? that how we did it

 

[PeterO]

oh okayyy...

On my sheet with this question they give the equation

m2 = (v1`-v1 / v2 - v2`)m1

So does that mean the change in velocity should be with v2-v2`
not v2`-v2 ? that how we did it

Multiply that out and then interpret:

m2 = (v1`-v1 / v2 - v2`)m1

m2(v2 - v2`) = (v1`-v1)m1

m2(v2 - v2`) = -(v1-v1`)m1

what does that tell you.

 

[Lolagoeslala]

Multiply that out and then interpret:

m2 = (v1`-v1 / v2 - v2`)m1

m2(v2 - v2`) = (v1`-v1)m1

m2(v2 - v2`) = -(v1-v1`)m1

what does that tell you.

Umm I am not sure what you are trying to say,,,
that the intial velocity for the second ball is subtracted from final velocity of the second ball?

 

[PeterO]

Umm I am not sure what you are trying to say,,,
that the intial velocity for the second ball is subtracted from final velocity of the second ball?

what does m(v'-v) represent?

 

[Lolagoeslala]

what does m(v'-v) represent?

the momentum...

 

[PeterO]

the momentum...

the CHANGE in momentum!

so the earlier post merely says

Δp₂ = - Δp₁

which is nothing new.

 

[Lolagoeslala]

the CHANGE in momentum!

so the earlier post merely says

Δp₂ = - Δp₁

which is nothing new.

but see it says..
v2 - v2` specifically in the equation..
and according to our vector subtraction graph...

should the head be towards the intial velocity rather then the final velocity in ym diagram?

 

[PeterO]

but see it says..
v2 - v2` specifically in the equation..
and according to our vector subtraction graph...

should the head be towards the intial velocity rather then the final velocity in ym diagram?

OK
m2 = (v1`-v1 / v2 - v2`)m1

m2(v2 - v2`) = (v1`-v1)m1

m2(v2 - v2`) = -(v1-v1`)m1

multiply through by -1 we get

m2(v2` - v2) = - (v1` - v1) m1

Do you prefer to see it in that order?

Doesn't change the concept.

 

[Lolagoeslala]

OK
m2 = (v1`-v1 / v2 - v2`)m1

m2(v2 - v2`) = (v1`-v1)m1

m2(v2 - v2`) = -(v1-v1`)m1

multiply through by -1 we get

m2(v2` - v2) = - (v1` - v1) m1

Do you prefer to see it in that order?

Doesn't change the concept.

So the vector subtraction we did is correct..
i think its correct since it doesn;t change much correct?