A Level Further Maths Impulse Question

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liamlarner2020
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How would you solve this am I correct in thinking, to begin with drawing a diagram of the scenario and using the conservation of momentum:
m1u1+m2u2=m1v1+m2v2

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mqb2766
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Certainly draw a diagram, the constraint that no further impacts means that there is a constraint on the velocity of B after impact.

The coefficient of restitution means that energy is lost. It's based on the relative speed ratio. Have you covered this? That, together with momentum, should give part a).
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liamlarner2020
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(Original post by mqb2766)
Certainly draw a diagram, the constraint that no further impacts means that there is a constraint on the velocity of B after impact.

The coefficient of restitution means that energy is lost. It's based on the relative speed ratio. Have you covered this? That, together with momentum, should give part a).
do you mean newtons law of restitution i have covered that yes so shall i use the conservation of linear momentum first and then solve for VB to get the first few marks
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mqb2766
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(Original post by liamlarner2020)
do you mean newtons law of restitution i have covered that yes so shall i use the conservation of linear momentum first and then solve for VB to get the first few marks
Yes.
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liamlarner2020
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(Original post by mqb2766)
Yes.
here is my working out so far
(4a) utilizing conservation of linear momentum:
m1u2+m2u2=m1v1+m2v2
4mu+km0=4mVb+KmVc
im stuck with how to continue do i just do:
4mu=4mVb and now solve for Vb where does newtons law of restitution come into it
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Qxi.xli
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(Original post by liamlarner2020)
How would you solve this am I correct in thinking, to begin with drawing a diagram of the scenario and using the conservation of momentum:
m1u1+m2u2=m1v1+m2v2

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Hey just wondering which website did you get that question from xx
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liamlarner2020
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(Original post by Qxi.xli)
Hey just wondering which website did you get that question from xx
its in the additional assessment materials made public by edexcel
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mqb2766
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(Original post by liamlarner2020)
here is my working out so far
(4a) utilizing conservation of linear momentum:
m1u2+m2u2=m1v1+m2v2
4mu+km0=4mVb+KmVc
im stuck with how to continue do i just do:
4mu=4mVb and now solve for Vb where does newtons law of restitution come into it
Do you know what you're trying to show?
Where did the Vc go, can you simplify that equation, ...
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liamlarner2020
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(Original post by mqb2766)
Do you know what you're trying to show?
Where did the Vc go, can you simplify that equation, ...
4mu=4mVb+kmVc is that it could you subtract 4m from both sides to give
u=Vb+KmVc or not
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mqb2766
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(Original post by liamlarner2020)
4mu=4mVb+kmVc is that it could you subtract 4m from both sides to give
u=Vb+KmVc or not
I agree with the bold.
Subtract isnt correct though, I hope you can see why.

Do you know what you're trying to prove, do you know what the restitution gives?
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liamlarner2020
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(Original post by mqb2766)
I agree with the bold.
Subtract isnt correct though, I hope you can see why.

Do you know what you're trying to prove, do you know what the restitution gives?
restitution gives 1/4 not sure how to continue
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mqb2766
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(Original post by liamlarner2020)
restitution gives 1/4 not sure how to continue
Look in your book/online to find out what the restitution equation is.
Did you draw the diagram, how will no further impacts occur ... I have no idea whether you've simplified the momentum equation correctly.
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liamlarner2020
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(Original post by mqb2766)
Look in your book/online to find out what the restitution equation is.
Did you draw the diagram, how will no further impacts occur ...
Newton's Law of restitution states that:
for a direct collision on a smooth plane
Speed of rebound/speed of approach = e
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mqb2766
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(Original post by liamlarner2020)
Newton's Law of restitution states that:
for a direct collision on a smooth plane
Speed of rebound/speed of approach = e
In terms of this problem the equation is ...
Last time,
* have you done the diagram
* Have you simplified the momentum equation
* Do you know, mathematically, what you're trying to show ...
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liamlarner2020
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(Original post by mqb2766)
In terms of this problem the equation is ...
Last time,
* have you done the diagram
* Have you simplified the momentum equation
* Do you know, mathematically, what you're trying to show ...
I have drawn the diagram and got the equation down to 4mu=4mVb+KmVc should I solve for Vb
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mqb2766
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(Original post by liamlarner2020)
I have drawn the diagram and got the equation down to 4mu=4mVb+KmVc should I solve for Vb
The question doesn't ask to to solve for Vb.

Honest advice is to try some simpler skill builder questions first, before coming back to this. The amount of help you seem to need means I'll be basically doing the question for you.
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liamlarner2020
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(Original post by mqb2766)
The question doesn't ask to to solve for Vb.

Honest advice is to try some simpler skill builder questions first, before coming back to this. The amount of help you seem to need means I'll be basically doing the question for you.
could you show me this one then I can try question 6 which is another one like this independently just because my exam is tomorrow and need to do as much as possible in case collisions like this come up
this is the question 6 i want to try alone if you can show me this one
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mqb2766
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There are lots of worked examples here
https://madasmaths.com/archive/maths...collisions.pdf
Or in your textbook or ...
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JaJa18796
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so we can use e, since it equals to( speed of separation of particles)/(speed of approach of particles), to get :
(Vc-Vb)/U=1/4
we can then substitute this into the momentum equation to get rid of Vc. This helps us so we can solve for Vb and then saying Vb=bu where b is a constant we can cancel out you. with b alone we can say b is great or equal to 0 for no further collision, since if it was negative it would travel back and collide with A. then solve for k and hay presto.

not sure if that makes sense, can send how I did it, not sure if correct tho, do you have the markscheme? This is M2 right?
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