# A Level Further Maths Impulse Question

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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

m1u1+m2u2=m1v1+m2v2

Last edited by liamlarner2020; 2 weeks ago

0

reply

Report

#2

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).

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).

Last edited by mqb2766; 2 weeks ago

1

reply

(Original post by

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).

**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).

Last edited by liamlarner2020; 2 weeks ago

0

reply

Report

#4

(Original post by

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

**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

0

reply

(Original post by

Yes.

**mqb2766**)Yes.

(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

0

reply

Report

#6

(Original post by

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

**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

0

reply

(Original post by

Hey just wondering which website did you get that question from xx

**Qxi.xli**)Hey just wondering which website did you get that question from xx

1

reply

Report

#8

(Original post by

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

**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

Where did the Vc go, can you simplify that equation, ...

0

reply

(Original post by

Do you know what you're trying to show?

Where did the Vc go, can you simplify that equation, ...

**mqb2766**)Do you know what you're trying to show?

Where did the Vc go, can you simplify that equation, ...

u=Vb+KmVc or not

0

reply

Report

#10

(Original post by

u=Vb+KmVc or not

**liamlarner2020**)**4mu=4mVb+kmVc**is that it could you subtract 4m from both sides to giveu=Vb+KmVc or not

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?

Last edited by mqb2766; 2 weeks ago

0

reply

(Original post by

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?

**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?

0

reply

Report

#12

(Original post by

restitution gives 1/4 not sure how to continue

**liamlarner2020**)restitution gives 1/4 not sure how to continue

Did you draw the diagram, how will no further impacts occur ... I have no idea whether you've simplified the momentum equation correctly.

Last edited by mqb2766; 2 weeks ago

0

reply

(Original post by

Look in your book/online to find out what the restitution equation is.

Did you draw the diagram, how will no further impacts occur ...

**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 ...

for a direct collision on a smooth plane

Speed of rebound/speed of approach = e

0

reply

Report

#14

(Original post by

Newton's Law of restitution states that:

for a direct collision on a smooth plane

Speed of rebound/speed of approach = e

**liamlarner2020**)Newton's Law of restitution states that:

for a direct collision on a smooth plane

Speed of rebound/speed of approach = e

Last time,

* have you done the diagram

* Have you simplified the momentum equation

* Do you know, mathematically, what you're trying to show ...

0

reply

(Original post by

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 ...

**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 ...

0

reply

Report

#16

(Original post by

I have drawn the diagram and got the equation down to 4mu=4mVb+KmVc should I solve for Vb

**liamlarner2020**)I have drawn the diagram and got the equation down to 4mu=4mVb+KmVc should I 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.

0

reply

(Original post by

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.

**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.

this is the question 6 i want to try alone if you can show me this one

0

reply

Report

#18

There are lots of worked examples here

https://madasmaths.com/archive/maths...collisions.pdf

Or in your textbook or ...

https://madasmaths.com/archive/maths...collisions.pdf

Or in your textbook or ...

Last edited by mqb2766; 2 weeks ago

0

reply

Report

#19

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?

(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?

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top