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r.maliak
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#1
say for example we took the forward motion as positive
and after collision truck A continues going forward but truck B goes back
would the impulse force for truck A be m(v+u) its plus because its still going in the same direction and for truck B its m(v-u)
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and after collision truck A continues going forward but truck B goes back
would the impulse force for truck A be m(v+u) its plus because its still going in the same direction and for truck B its m(v-u)
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username4407536
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#2
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#2
You define -> as the positive direction
so
You assume that direction of motion of A is unchanged, and direction of motion ofB is reversed
Let's represent these velocities by
and 
Impulse on A =![m_A [V_A - U_A]](https://www.thestudentroom.co.uk/latexrender/pictures/47/475921191747d31d2af919977d56067a.png "m_A [V_A - U_A]")
Impulse on B =![m_B [V_B - U_B]](https://www.thestudentroom.co.uk/latexrender/pictures/af/aff601a2f0eddd6141661e548eba6b2b.png "m_B [V_B - U_B]")
I hope this helps
so
You assume that direction of motion of A is unchanged, and direction of motion ofB is reversed
Let's represent these velocities by
and Impulse on A =
Impulse on B =
I hope this helps
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r.maliak
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#3
(Original post by golgiapparatus31)
You define -> as the positive direction
so
You assume that direction of motion of A is unchanged, and direction of motion ofB is reversed
Let's represent these velocities by and
Impulse on A =
Impulse on B =
I hope this
You define -> as the positive direction
so
You assume that direction of motion of A is unchanged, and direction of motion ofB is reversed
Let's represent these velocities by
and Impulse on A =
Impulse on B =
I hope this
is the original equation m(v-u) but becuase its -3.1 it becomes v+3.1
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username4407536
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#4
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#4
(Original post by r.maliak)
are you sure for B its v-u or its meant to be v+u
is the original equation m(v-u) but becuase its -3.1 it becomes v+3.1
are you sure for B its v-u or its meant to be v+u
is the original equation m(v-u) but becuase its -3.1 it becomes v+3.1
Impulse for B = mb(vb - (-3.1)) = mb(vb + 3.1)
also you said it's an "impulse force"
this is wrong, an impulse is not a force. it is a change in momentum
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r.maliak
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#5
(Original post by golgiapparatus31)
Impulse for A = ma(va-2.8)
Impulse for B = mb(vb - (-3.1)) = mb(vb + 3.1)
also you said it's an "impulse force"
this is wrong, an impulse is not a force. it is a change in momentum
Impulse for A = ma(va-2.8)
Impulse for B = mb(vb - (-3.1)) = mb(vb + 3.1)
also you said it's an "impulse force"
this is wrong, an impulse is not a force. it is a change in momentum
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Eimmanuel
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#6
(Original post by golgiapparatus31)
....
an impulse is not a force. it (An impulse) is a change in momentum
....
an impulse is not a force. it (An impulse) is a change in momentum
I don't know what you have learnt about an impulse or how your teacher defines impulse.
According to your statement "it is a change in momentum", you seem to define impulse as the change in momentum.
I highly recommend that you check with your A level physics specification whether you can define impulse as the change in momentum.
You can see the actual definition of impulse in physics in the following links and none of them define impulse as the change in momentum.
https://phys.libretexts.org/Bookshel...isions_(Part_1)
https://www.scribd.com/document/390742215/Chapter-11
Impulse is related to change in momentum via impulse-momentum theorem: impulse is equal to the change in momentum.
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username4407536
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#7
(Original post by Eimmanuel)
I edited your reply to explain what you seem to imply.
I don't know what you have learnt about an impulse or how your teacher defines impulse.
According to your statement "it is a change in momentum", you seem to define impulse as the change in momentum.
I highly recommend that you check with your A level physics specification whether you can define impulse as the change in momentum.
No thanks.
You can see the actual definition of impulse in physics in the following links and none of them define impulse as the change in momentum.
https://phys.libretexts.org/Bookshel...isions_(Part_1)
https://www.scribd.com/document/390742215/Chapter-11
Impulse is related to change in momentum via impulse-momentum theorem: impulse is equal to the change in momentum.
I edited your reply to explain what you seem to imply.
I don't know what you have learnt about an impulse or how your teacher defines impulse.
According to your statement "it is a change in momentum", you seem to define impulse as the change in momentum.
I highly recommend that you check with your A level physics specification whether you can define impulse as the change in momentum.
No thanks.
You can see the actual definition of impulse in physics in the following links and none of them define impulse as the change in momentum.
https://phys.libretexts.org/Bookshel...isions_(Part_1)
https://www.scribd.com/document/390742215/Chapter-11
Impulse is related to change in momentum via impulse-momentum theorem: impulse is equal to the change in momentum.
I didn't say impulse is defined as change in momentum. Nowhere did I imply it is a definition.
"Impulse is change in momentum" is equivalent to saying "impulse is equal to change in momentum" because of how English grammar works. That was all I said, I didn't say it was a definition.
Also note that integrals are not a syllabus item for "my A level physics specification".
There is no need to be condescending about it.
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Eimmanuel
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#8
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#8
(Original post by golgiapparatus31)
Um... your first link doesn't work.
Um... your first link doesn't work.
(Original post by golgiapparatus31)
… I didn't say impulse is defined as change in momentum. Nowhere did I imply it is a definition.
… I didn't say impulse is defined as change in momentum. Nowhere did I imply it is a definition.
(Original post by golgiapparatus31)
… Also note that integrals are not a syllabus item for "my A level physics specification".
… Also note that integrals are not a syllabus item for "my A level physics specification".
(Original post by golgiapparatus31)
…"Impulse is change in momentum" is equivalent to saying "impulse is equal to change in momentum" because of how English grammar works. That was all I said, I didn't say it was a definition.
…"Impulse is change in momentum" is equivalent to saying "impulse is equal to change in momentum" because of how English grammar works. That was all I said, I didn't say it was a definition.
In physics, the “=” sign that appears has different meanings. In terms of definition of physical quantities, the “=” sign is replaced by an “is” while in terms of cause-and-effect equation like N2L or Impulse-momentum theorem, the “=” sign is replaced with “is equal to”.
Here is an example of N2L from a standard University Physics:
Newton’s second law of motion: If a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force. The mass of the body times the acceleration of the body equals the net force vector.
N2L based on linear momentum:
In situations where the mass is changing with time, one must use an alternative statement of Newton’s second law: The time rate of change of momentum of a particle is equal to the net force acting on the particle, or
Another example based on Impulse-momentum theorem from an A level equivalent Physics text.
The quantity FnetΔt is called the impulse of the force, and is usually denoted by J. It is the product of the average force times the time for which the force acts. The impulse is also equal to the change in momentum.
For the same text, you can try to scroll up a few pages to see how the text defines linear momentum based on p = mv and states N2L on page 98.
An example on the definition of work done:
The work W done on a system by an agent exerting a constant force on the system is the product of the magnitude F of the force, the magnitude Δr of the displacement of the point of application of the force, and cos θ, where θ is the angle between the force and displacement vectors:
Another example of impulse-momentum.
This snippet is from this book. You can scroll down to read about the authors of this physics text.
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