Vector subtraction with momentum

In summary, the equation for finding the change in velocity for the second ball is: v2 - v2` = 27and the angle is [E 27 E]
  • #1
Lolagoeslala
217
0

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?
 
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  • #2
Lolagoeslala said:
7.536/sin23 = 19.04/sinx
for x = 81°

There's more than one solution for x.
 
  • #3
TSny said:
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.,..
 
  • #4
For example, the inverse sine of 0.5 has more than one answer, not just 30 degrees.
 
  • #5
TSny said:
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?
 
  • #6
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 30o. But that's not the only answer.
 
  • #7
TSny said:
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 30o. But that's not the only answer.

so what's the other answer then?
 
  • #8
Are you familiar with the unit circle representation of the sine function? See http://maine.edc.org/file.php/1/tools/TrigUnitCircle2.html [Broken] 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 30o where the sine function has a value of 0.5?
 
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  • #9
TSny said:
Are you familiar with the unit circle representation of the sine function? See http://maine.edc.org/file.php/1/tools/TrigUnitCircle2.html [Broken] 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 30o where the sine function has a value of 0.5?

so 149 degrees?
 
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  • #10
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?
 
  • #11
TSny said:
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°
 
  • #12
No need to guess. :smile: Grab the calculator and see if sin 99o = sin 81o.

From your diagram, does the angle "x" look like its greater than or less than 90 degrees?
 
  • #13
TSny said:
No need to guess. :smile: Grab the calculator and see if sin 99o = sin 81o.

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

It looks like its greater
 
  • #14
TSny said:
No need to guess. :smile: Grab the calculator and see if sin 99o = sin 81o.

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?
 
  • #15
Lolagoeslala said:
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.
 
  • #16
TSny said:
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..
 
  • #17
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.
 
  • #18
TSny said:
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?
 
  • #19
Well, 81 is not the same as 99. Only 99o is the correct answer for your problem. It's just that the sine of 81o is the same as the sine of 99o. 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.
 
  • #20
TSny said:
Well, 81 is not the same as 99. Only 99o is the correct answer for your problem. It's just that the sine of 81o is the same as the sine of 99o. 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
 
  • #21
Lolagoeslala said:
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 91o.
The sine rule is certainly simpler (arithmetically) but has the ambiguous answer issue to deal with on many occasions.
 
  • #22
PeterO said:
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 91o.
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
 
  • #23
Lolagoeslala said:
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.
 
  • #24
PeterO said:
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?
 
  • #25
Lolagoeslala said:
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?
 
  • #26
PeterO said:
what does m(v'-v) represent?

the momentum...
 
  • #27
Lolagoeslala said:
the momentum...

the CHANGE in momentum!

so the earlier post merely says

Δp2 = - Δp1

which is nothing new.
 
  • #28
PeterO said:
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?
 
  • #29
Lolagoeslala said:
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.
 
  • #30
PeterO said:
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?
 
  • #31
Lolagoeslala said:
So the vector subtraction we did is correct..
i think its correct since it doesn;t change much correct?

It was the correct subtraction

Many people - in order to ensure they do subtractions in the "correct" order do the following

Vector A - Vector B = Vector A + (-Vector B)

In other words you reverse the (appropriate) vector and do the standard head to tail resultant.
That avoids which particular tail to tail vector you are after.
since in this problem you wanted the ~19 - ~16, you would reverse the 16 vector and add.

That confirms that the direction you drew it was correct.
 
  • #32
Lolagoeslala said:
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

The equation should read m2 = |v1'-v1 |/ |v2 - v2'| m1 where the numerator and denominator of the fraction are magnitudes of the velocity changes. So, it won't matter whether you write |v2 - v2'| or |v2' - v2|. They are equal.
 
  • #33
TSny said:
The equation should read m2 = |v1'-v1 |/ |v2 - v2'| m1 where the numerator and denominator of the fraction are magnitudes of the velocity changes. So, it won't matter whether you write |v2 - v2'| or |v2' - v2|. They are equal.

oh okay..
but see i found the experimental error...

so i did 27 - 32 / 32

and i get 15.6 % ...
but my teacher said u can have a maximum of 8% what is this? :(
 
  • #34
TSny said:
The equation should read m2 = |v1'-v1 |/ |v2 - v2'| m1 where the numerator and denominator of the fraction are magnitudes of the velocity changes. So, it won't matter whether you write |v2 - v2'| or |v2' - v2|. They are equal.

disagree. Absolute value not needed as one of the subtractions is written as final - initial while the other is written initial - final. This reinforces the idea that the gain in momentum of one object matches the loss in momentum of the other ie Δp1 = -Δp2, or if you prefer Δp1 + Δp2 = 0
It doesn't matter if you do include absolute value, but it is not necessary.
 
  • #35
I don't think it's very meaningful to calculate a % error for an angle. For example, suppose you had an angle that was supposed to be 2 degrees but your experimental value was 3 degrees. That's pretty good - only one degree off. That's just as good as an experimental value of 101 degrees for an angle that should be 100 degrees. But the % error in the first case would be calculated as 100%*(3-2)/2 = 50 %. In the second case, 100%*(101-100)/100 = 1%.

Are you sure that you weren't suppose to find the % error in the mass m1?
 
Last edited:
<h2>1. What is vector subtraction with momentum?</h2><p>Vector subtraction with momentum is a mathematical operation that involves subtracting the momentum vector of one object from another. It is commonly used in physics and engineering to calculate the net momentum of a system.</p><h2>2. How is vector subtraction with momentum different from regular vector subtraction?</h2><p>The main difference between vector subtraction with momentum and regular vector subtraction is that momentum is a vector quantity that includes both magnitude and direction, whereas regular vectors only have magnitude. This means that when subtracting momentums, both the magnitude and direction must be taken into account.</p><h2>3. What is the formula for vector subtraction with momentum?</h2><p>The formula for vector subtraction with momentum is: <br> Resultant Momentum = Initial Momentum - Final Momentum</p><h2>4. What are some real-life applications of vector subtraction with momentum?</h2><p>Vector subtraction with momentum is used in various fields such as physics, engineering, and robotics. Some examples of its applications include calculating the net momentum of a rocket, determining the velocity of a moving object after a collision, and predicting the trajectory of a projectile.</p><h2>5. What are some common mistakes when performing vector subtraction with momentum?</h2><p>One common mistake is forgetting to take into account the direction of the momentum vectors. Another mistake is incorrectly identifying the initial and final momentums, which can lead to incorrect results. It is important to carefully label and double-check all vectors before performing the subtraction.</p>

1. What is vector subtraction with momentum?

Vector subtraction with momentum is a mathematical operation that involves subtracting the momentum vector of one object from another. It is commonly used in physics and engineering to calculate the net momentum of a system.

2. How is vector subtraction with momentum different from regular vector subtraction?

The main difference between vector subtraction with momentum and regular vector subtraction is that momentum is a vector quantity that includes both magnitude and direction, whereas regular vectors only have magnitude. This means that when subtracting momentums, both the magnitude and direction must be taken into account.

3. What is the formula for vector subtraction with momentum?

The formula for vector subtraction with momentum is:
Resultant Momentum = Initial Momentum - Final Momentum

4. What are some real-life applications of vector subtraction with momentum?

Vector subtraction with momentum is used in various fields such as physics, engineering, and robotics. Some examples of its applications include calculating the net momentum of a rocket, determining the velocity of a moving object after a collision, and predicting the trajectory of a projectile.

5. What are some common mistakes when performing vector subtraction with momentum?

One common mistake is forgetting to take into account the direction of the momentum vectors. Another mistake is incorrectly identifying the initial and final momentums, which can lead to incorrect results. It is important to carefully label and double-check all vectors before performing the subtraction.

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