<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

Further Mechanics...Help!

rdfzmax

The question:
A particle A of mass 2kg lies on the edge of a table of height 1m. It's connected by a light inelastic string 0.65m long to particle B of 3kg which is lying on the table 0.25m from the edge(AB is perpendicular to the edge)
If A is pushed gently over the edge, find the velocity with which B begins to move.
Find also the impulsive tension in the string.

I think particle B will be pulled after A travels 0.4m, so I find the time for this period first(=sqrt0.08). Then consider the impulse, which is the weight of A times the time it acts(sqrt0.08). When the string becomes taut, A,B will move with same speed, if it's v, then change in momentum should be v(3+2)-0. The answer however, is not correct. Can anyone point out my mistake in reasoning?

Thanks a lot!

Reply 1

rnd

I confess I don't know what impulsive tension is but I think your t value is wrong. Also did you need t? You can find v directly. If you wouldn't mind defining impulsive tension for me I'll have a go at the rest of the question.

Reply 2

rdfzmax

rnd

Thanks alot...Impulsive tension is the impulse in a jerking string. Examples in my book finds impulsive tension by considering the impulse on the particle of one end only.
The answer is 1.12m/s and 3.36Ns
Thanks...

Reply 3

rnd

The more I think about this the less I think I understand. I will sit back and await enlightenment.

Reply 4

steve2005

rdfzmax

I think particle B will be pulled after A travels 0.4m, so I find the time for this period first(=sqrt0.08). Then consider the impulse, which is the weight of A times the time it acts(sqrt0.08). When the string becomes taut, A,B will move with same speed, if it's v, then change in momentum should be v(3+2)-0. The answer however, is not correct. Can anyone point out my mistake in reasoning?

Thanks a lot!

I have not looked at the question in detail but I think impulse involves mass and not weight.

Reply 5

rdfzmax

steve2005

I have not looked at the question in detail but I think impulse involves mass and not weight.

Impulse is calculated by F*t?

Reply 6

waterinabottle

I thought it was F*v?

Reply 7

rdfzmax

waterinabottle

I thought it was F*v?

No, Impulse is force acting for certain time, which is equal to change in momentum=mv-mu

Reply 8

waterinabottle

rdfzmax

No, Impulse is force acting for certain time, which is equal to change in momentum=mv-mu

My mistake, thanks for the correction.

Reply 9

steve2005

I am confused by the title of this tread.

Is Further Mechanics a subject title or are you simply asking for further help with mechanics? If a subject what is it's reference number?

Reply 10

rdfzmax

steve2005

I am confused by the title of this tread.

Is Further Mechanics a subject title or are you simply asking for further help with mechanics? If a subject what is it's reference number?

Further Maths Mechanics...not sure about reference number

Reply 11

ghostwalker

Work out the velocity of A at the time the string becomes taut ( use v^2=u^2+2as)

Then conservation of momentum will give you speed of the two masses just after the string has tightened (note they will now have the same velocity).

Then impulse is change in momentum. Just apply that to either mass.

Reply 12

rnd

Thanks ghostwalker. I should have known that.

Reply 13

rdfzmax

ghostwalker

Thanks ghostwalker...is conservation of momentum only applicable to the moment the string tightened? I think this moment it's still under external force, doesn't Conservation law fail here?

Reply 14

ghostwalker

rdfzmax

Thanks ghostwalker...is conservation of momentum only applicable to the moment the string tightened?

In this question yes.

I think this moment it's still under external force, doesn't Conservation law fail here?

Conservation of momentum doesn't fail, as the collision (which is effectively what it is), takes place in an "instance of time". The effect of any external force in zero (for an instance).