Simple Canon Question (Conservation Of Momentum): Frame of Reference

In summary: Ah so if its the other equation, I was wrong initiallyso at first it would be:Vbc = Vbg - Vcgmeaning Vbg = Vbc + VcgThen the rest is the same:0 = MbVbg + McVcg0 = Mb(Vbc + Vcg) +McVcg0 = MbVbc + MbVcg +McVcgVcg = -MbVbc/(Mb+Mc) = -72*55/(72+1300) = -2.9m/sAnd THATS correctSo it was just a dumb mistake with the
  • #1
Kermit_the_Phrog
22
3
Homework Statement
In Fig below, a mobile cannon of mass M=1300 Kg fires a ball of mass m=72 kg in the positive direction of an x axis, at velocity Vbc= 55 m/s relative to cannon. Because of the firing, the cannon recoils (moves) with velocity Vcg relative to the ground. What are the magnitude and direction of Vcg?
Relevant Equations
Pi=Pf
Canon, Dumb.png

Since Pi = Pf,

0 = MbVbg + McVcg

I just need to express Vbg in terms of Vbc and Vcg (that is, I need to express the velocity of the ball relative to the ground in terms that I know/want to solve for):

by reference frames:

Vbc = Vbg + Vcg
so Vbg = Vbc -Vcg

Now I can sub in and solve

0 = MbVbg + McVcg
0 = Mb(Vbc -Vcg) + McVcg
0 = MbVbc - MbVcg +McVcg
MbVcg - McVcg = MbVbc
Vcg(Mb - Mc) = MbVbc

therefore,
Vcg = 72*55/(72-1300)
= -3.2 m/s
which would be a.

HOWEVER the teacher's answer key states that the answer is b, -2.9 m/s.
this answer is obtained if the denominator was " (- mb - mc) ".

Thus, did I just mess up somewhere with my signs? I can't find mistake, and any help deducing this would be appreciated, as I was pretty confident about my answer until seeing that it was wrong.
 
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  • #2
Kermit_the_Phrog said:
Thus, did I just mess up somewhere with my signs?

Yes. The signs are inconsistent between the two equations

0 = MbVbg + McVcg
Vbg = Vbc -Vcg

Give it some more thought.
 
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  • #3
TSny said:
Yes. The signs are inconsistent in the two equations

0 = MbVbg + McVcg
Vbg = Vbc -Vcg

Give it some more thought.

Is the momentum equation supposed to be0 = MbVbg - McVcg

Because Vcg is pointed in the opposite direction to Vbg?
 
  • #4
TSny said:
Give it some more thought.

If that's true, you get

Vcg = MbVbc/(Mb+Mc)
= 72*55/(72+1300)
= 2.9

which is then negative through intuition (since its moving in the opposite direction?)

Though i suppose this works, I feel like there's a better / more accurate way to show the signs in my algebra.

Thoughts?
 
  • #5
Kermit_the_Phrog said:
Is the momentum equation supposed to be0 = MbVbg - McVcg

Because Vcg is pointed in the opposite direction to Vbg?
Assuming you are taking right as positive in all cases, your original momentum equation was correct.
In writing relative motion equations, the safest approach is always to write them in the form
Motion of A relative to B = Motion of A relative to common frame - Motion of B relative to common frame.
 
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  • #6
haruspex said:
Assuming you are taking right as positive in all cases, your original momentum equation was correct.
In writing relative motion equations, the safest approach is always to write them in the form
Motion of A relative to B = Motion of A relative to common frame - Motion of B relative to common frame.

Ah so if its the other equation, I was wrong initially

so at first it would be:
Vbc = Vbg - Vcg

meaning
Vbg = Vbc + Vcg

Then the rest is the same:

0 = MbVbg + McVcg
0 = Mb(Vbc + Vcg) +McVcg
0 = MbVbc + MbVcg +McVcg
Vcg = -MbVbc/(Mb+Mc)
= -72*55/(72+1300)
= -2.9m/s

And THATS correct

So it was just a dumb mistake with the first reference frame equation

Thank you for your help!
 
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What is the conservation of momentum?

The conservation of momentum is a fundamental law in physics that states that the total momentum of a closed system remains constant over time, unless an external force acts upon it. This means that the total momentum of all objects involved in a collision or interaction will be the same before and after the interaction.

What is a frame of reference?

A frame of reference is a set of coordinates that are used to describe the position, motion, and interactions of objects. It can be thought of as a point of view or perspective from which an observer makes measurements and observations.

How does the conservation of momentum apply to different frames of reference?

The conservation of momentum applies to all frames of reference, as it is a fundamental law of physics. However, the observed momentum of objects may differ depending on the frame of reference. This is because momentum is a vector quantity, meaning it has both magnitude and direction, and different frames of reference may have different orientations or axes.

Why is the conservation of momentum important?

The conservation of momentum is important because it is a fundamental law of nature that helps us understand and predict the behavior of objects in motion. It is used in a wide range of fields, including engineering, physics, and astronomy, to analyze and design systems and to make accurate predictions about the outcomes of collisions and other interactions.

Can the conservation of momentum be violated?

No, the conservation of momentum is a fundamental law of physics and cannot be violated. However, in some cases, it may appear to be violated if all external forces acting on a system are not taken into account. This is why it is important to carefully define the system and consider all external forces when applying the law of conservation of momentum.

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