Further Mechanics, Unit 4 question

Hello, currenbtly stuck on thi

A shell of mass 12 kg moving horizontally at 320 ms^ -1 explodes into three fragments A,B and C, which continue to move in a common vertical plane.

Fragment A: mass 2 kg continues at 450ms^-1,45 degree angle above the horizontal.

Fragment B: mass 6kg continues at 400ms^-1 horizontally.

Fragment C: mass m continues at speed v, in a direction theta to the horizontal.

1) Calculate the size of the momentums of the shell before it explodes and of fragments A and B afterwards

2) Draw a vector diagram showing these momentums and hence deduce the value of m, v and theta?

I only got the momentum of the shell correct. My other answers were wrong, so starting with 1) the answers to momentums of A and B were 1800 Ns and 2520 Ns respectively.

Thanks!

(edited 10 years ago)

Reply 1

Stonebridge

Original post by Jaydude

Hello, currenbtly stuck on thi

A shell of mass 12 kg moving horizontally at 320 ms^ -1 explodes into three fragments A,B and C, which continue to move in a common vertical plane.

Fragment A: mass 2 kg continues at 450ms^-1,45 degree angle above the horizontal.

Fragment B: mass 6kg continues at 400ms^-1 horizontally.

Fragment C: mass m continues at speed v, in a direction theta to the horizontal.

1) Calculate the size of the momentums of the shell before it explodes and of fragments A and B

2) what is the value of momentums of A and B afterwards? what is the value of m, v and theta?

I only got the momentum of the shell correct. My other answers were wrong, so starting with 1) the answers to momentums of A and B were 1800 Ns and 2520 Ns respectively.

Thanks!

We need to know how you did this to find out where you've gone wrong.

The priciples you apply are

The sum of the forward horizontal components of momentum of the 3 particles after they fragment is equal to the forward momentum of the single particle before.

The two fragments that move upwards and downwards after fragmentation have equal and opposite momentum components in the vertical direction.

Can you show us your working, please?

Reply 2

Jaydude

Sure;

I calculated the momentum of A to be (2x 450) = 900kgms^-1

And momentum of B : (6x400) = 2400kgms^-1

These are wrong though, and I thought the question is simply asking the momentum. Do i take trig into account? Why? I am just calculating the mometum of these fragments respective to their directions... Do i Calculate the horizontal component of the momentum instead? If so, then why is B wrong, as its direction is horizontal...

Thanks!

Reply 3

Stonebridge

Unless someone has changed the laws of physics since I last looked your two answers are correct.
Are you sure the question is correctly posted? Where are the answers from?
There is something not right here. Is the question complete?
Assuming it is... (I may regret this!)
The next step is to find the horizontal components of the three momenta. You have two of them, A and B. The three add to give the initial momentum of the original fragment, which you also have.
This gives you a value for the horizontal momentum component of C, mv cos theta.
Can you get to that point?.

Reply 4

Jaydude

The question is from textbook called hodders physics a2. I don't wwant to post a link to the answers as it requires a login, But I've copied the question exactly right! And the answers they have given.
The second part of the question required a vector drawing, they have drawn a trapezium shape and went on to calculate the mass, theta using a protractor and using accurate measurements and ratio went onto calculate the momentum. (of fragment C) I understand this to an extent, just the first part of the question has tripped me over!

I think the best thing for me to do is ask my physics teacher as I can show him the actual solutions, but it's half term so that will be quite a wait!

Thanks though for replying!

Just to make sure, ive added in the exact wording of the question, but its still the same.

Posted from TSR Mobile

(edited 10 years ago)

Reply 5

Stonebridge

Original post by Jaydude

The correct version of the question makes much more sense.
It's done by drawing and not by calculation. Ignore what I said as I assumed it was asking for a calculation.
Even so your answers for those 1st two (A and B) are correct.

(edited 10 years ago)

Reply 6

Jaydude

Original post by Stonebridge

The correct version of the question makes much more sense.
It's done by drawing and not by calculation. Ignore what I said as I assumed it was asking for a calculation.
Even so you answers for those 1st two (A and B) are correct.

Just to make sure, can you answer part 2? :biggrin:

i.e the vector drawing, i just want to see the rough shape of the vector diagram you do, compare it to the answer, and see where I have gone wrong.

Reply 7

Stonebridge

Original post by Jaydude

Just to make sure, can you answer part 2? :biggrin:

i.e the vector drawing, i just want to see the rough shape of the vector diagram you do, compare it to the answer, and see where I have gone wrong.

Well no I can't because it doesn't work that way. You do the diagram and we help you correct it if it's wrong.

Reply 8

angelwattings

from which examiner board is this exercise?

Original post by Jaydude

Reply 9

neometalero

The momentum before the collision must be equal to the momentum after the collision so:
12 x 320 = 6x400 + square root {[900 sin (45)]^2 + [900 cos(45)]^2} + momentum (C).
That gives you a 540Kg.m/s for C.

Finding m, v and the angle is not so easy. You gotta do the same but decomposing for the three objects their momentum in the horizontal and vertical axes. For fragment B is easy cause it's moving only horizontally. For fragment A we have:
X: mvcos(45)
Y: mvsin(45)
Both positive cause we can consider that is moving to the rigth and upwards.

For fragment C we have:
X: -mvcos(angle)
Y: -mvsin(angle)
It does not say anything but it makes sense to assume that fragment C is going downwards. The tricky question is, is it going backwards also? We could think so cause fragments A and B are moving forward.

Now you gotta join all the equations correspondent to the X axes and also on the Y axis and solve them. You only have 2 equations but 3 unknowns so you need a third equation, which is the one for momentum of fragment C: mv = 540 kg.m/s.