You are Here: Home

# Newton's third law Tweet

Physics and electronics discussion, revision, exam and homework help.

Announcements Posted on
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. Newton's third law
Why isnt the recoil velocity equal to 350ms-1 by newton's third law,since they are both action reaction forces?(contact forces)
Attached Thumbnails

2. Re: Newton's third law
You sorta answered your own question, 350 is the velocity not the force.

A better way to think of it is that the momentum is conserved!!...total momentum before = total momentum after
3. Re: Newton's third law
Newton's 3rd Law does not state that the velocities should be equal. It states that the forces are equal and opposite and leads to the fact that, in this case, the momentum of the bullet in one direction is equal and opposite to the momentum of the rifle in the other. This is a result of the total momentum before firing being zero. After firing the total is still zero but now made up of the sum of a plus value in one direction and a minus in the other (opposite) direction.

The velocities would only be equal if the mass of the bullet was equal to the mass of the rifle. It clearly isn't. The velocities of bullet and rifle will depend on their mass because momentum is mass x velocity.
Last edited by Stonebridge; 27-07-2012 at 14:01.
4. Oh...i mixed them up! Thanks for pointing that out! I have another question here, the answers show these step for finding combined recoil velocity, with the recoil velocity of the rifle found to be 3.5ms-1 in the previous part of the same question.

I dont get why it is (Mm+Mr) Vm+r when it is supposed to be MmVm+MrVr for total final momentum.

Doesnt seem like a correct factorisation to me,but couldnt find another way to add the two velocities :/
Attached Thumbnails

5. Re: Newton's third law
They are asking for "the combined recoil velocity of the man and rifle". (vm+r)
So on the one side you have the bullet with its mass and velocity mbvb, and on the other side you have the combined mass of the man and rifle, mm + mr with whatever velocity they have, vm+r
The combined momentum of the man plus rifle in the one direction is mmvm+r + mrvm+r
this can also be written (mm+mr)vm+r

The m+r is just a subscript meaning man and rifle combined.
vm+r means the combined single velocity of man and rifle together.
6. I recognise that they are asking for the combined velocity of man and rifle, but shouldn't the final momentum (when considering the man and rifle only) ,be MmVm + MrVr ? That is the final momentum so how can they change the Vm or Vr to Vm+r just because the question wants to find Vm+r? Vm isnt equal to Vm+r ,neither is Vr

Hmm even if everything is considered combined momentum, then why isnt the initial momentum in the picture Mm+Mr(Um+Ur)?
The initial momentum in the picture (which is the answer) is MmUm+MrUr which is not equal to Mm+Mr(Um+Ur) the supposed initial "combined momentum"
Last edited by Vadevalor; 28-07-2012 at 10:26.
7. Re: Newton's third law
(Original post by Vadevalor)
I recognise that they are asking for the combined velocity of man and rifle, but shouldn't the final momentum (when considering the man and rifle only) ,be MmVm + MrVr ? That is the final momentum so how can they change the Vm or Vr to Vm+r just because the question wants to find Vm+r? Vm isnt equal to Vm+r ,neither is Vr

Hmm even if everything is considered combined momentum, then why isnt the initial momentum in the picture Mm+Mr(Um+Ur)?
The question is not well written.
The man is holding the rifle so it is assumed they have the same speed of recoil.
You don't need vm and vr separately. The are equal.
The answer calls that vm+r
This is the value you have to find.
The combined mass of the man and rifle is mm + mr and they have a velocity vm+r
So their combined momentum is (mm+mr)vm+r

The initial momentum in the picture (which is the answer) is MmUm+MrUr which is not equal to Mm+Mr(Um+Ur) the supposed initial "combined momentum"
The initial momentum of the system is zero. This is the same as in the first part of the question.
You then know that the momentum of the man and rifle combined in the one direction is equal to the momentum of the bullet in the other direction.
This is then the same as the first part of the question.

The answer in the book should show
72 vr+m = 0.02 x 350

That is
combined mass of man and rifle times combined speed of man and rifle = momentum of bullet.

These two momenta are equal and opposite because the initial total momentum was zero.
I explained this in my 1st post.
Ignore the book. It's confusing.
Last edited by Stonebridge; 28-07-2012 at 12:15.
8. Wow i understand you more than the book :P you're amazing !!

This was posted from The Student Room's iPhone/iPad A