M2 Question on collisons

Heinemann maths book (orange)

pg 106, question 5.... restitution nd stuff

no idea on how to do it.... any help appreciated!

cheers

Reply 1

MdSalih

With these questions.. all you need to remember to do is:

Do conservation of momentum- then apply Newtons Law's of Restitution...

You then have two equations... solve them for the two velocities for after the collision.. they will be in terms of e and u normally.

After that.. think about the situation... in this question... the first particle is going in backwards direction... so the final vely of that will be < 0
the other one is moving forwards so vely will be > 0

and there you have the inequalities.

I'll do the working in a second and post it up.

MdSalih

Reply 2

ThugzMansion7

9

KingAS

Heinemann maths book (orange)

pg 106, question 5.... restitution nd stuff

no idea on how to do it.... any help appreciated!

cheers

use

e = Sep / App

so what is the seperation speed of the particles
and with what speed do they approach each other
Hence your eq. for e

Reply 3

KingAS

OP

yeah i did get two equations but i had three unknowns in this case... e, u and v... thanks for the help so far, waiting for the sols... thanks ...rep

btw do you mean conservation of momentum? or mass?? :confused:

Reply 4

ThugzMansion7

9

just working it out. won't be a min

Reply 5

MdSalih

By Conservation of Momentum:

3(8) + 12(0) = 3(V1) + 12(V2)
24 = 3V1 + 12V2
8 = V1 + 4V2 ...................[1]

By Newton's Law of Restitution:

e = (V2 - V1)/8
8e = V2 - V1 ...................[2]
------------------------------
For V2 -> [1] + [2] :

8e + 8 = 5V2
so...
V2 = (8/5)(e+1)

For V1 -> [1] - 4[2] :
8 - 32e = 5V1
so...
V1 = (8/5)(1-4e)
------------------------------
Now, given Particle A is going in the opposite direction.. V1 < 0

Therefore,

(8/5)(1-4e) < 0
-4e < -1
e > 1/4

Also, particle B is going in +ve direction.. so V2 > 0

Therefore,

(8/5)(e+1) > 0
e > -1
------------------------------

So, e > 1/4

MdSalih

Reply 6

MdSalih

KingAS

yeah i did get two equations but i had three unknowns in this case... e, u and v... thanks for the help so far, waiting for the sols... thanks ...rep

btw do you mean conservation of momentum? or mass?? :confused:

Sorry.. yes I ment Conservation of Momentum

hmm.. not sure how you've got 'u' in there... you are given the initial speed of particle A.. so you know the approach speed as particle B is at standstill.

Although, if they do give you a u in there as appose to the initial speed - they normally cancel out at some point, i.e. when you do the equalities.

MdSalih

Reply 7

KingAS

OP

wow thanks a lot!!! nice diagram as well! hey about restitution, will the speed of separation be (v1 + V2) instead of (v2-v1)?
anyways, we get the same answer!

and for the conserv of mmtm, should it be - 3v1?.... on the momentum after bit?

thanks

M2 Question on collisons

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