# Suspended Particle Questions

1. Homework Statement
Particle A and B are attached by an inextensible string with masses 0.5 and 0.4 respectively

10N
^
|
{a}
|
|
{b}

take a to be 9.8 downwards

When the 10N is applied the particles move upwards. Ginf the tension in the lower string and the acceleration of the system.

2. Homework Equations
Kinematic Ones
Ignore air resistance

3. The Attempt at a Solution

A:
10 - T = 0.5a

B:T + 10 = 0.4a

BUT i think straight away this is wrong as what about the efffect of gravity???

Thanks

Last edited:

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tiny-tim
Homework Helper
Hi thomas49th! Hint: work out the acceleration of the whole system (the tensions won't matter, since they're internal).

Then work out the tensions! yeh i got it in the end.

but now i have a harder question :(

Two particles P and Q of masses 3kg and 6kg respectively are attached to the ends of a light inextensible string. The string passes over a smooth pulley. The system is released from rest wit both masses a distance of 2m above the horizontal floor. Find how long it takes for particle Q to hit the floor. Assuming particle P does not reach the pulley find its greatest height above the floor.

The time was easy to find (1.11s)
but i dont know how to find the maximum height particle P reaches. I would of though 4 metres (2+2) but it must be travelling at such a speed that it continues after particle Q has hit the floor.

Can someone please so me how to do the second part please.
Thanks :)

tiny-tim
Homework Helper
Hi thomas49th! Hint: speed of P at 4m must be same as speed of Q at 0m.

From then on, P is just decelerating under gravity.

So … ? errr Q at 0m must be

V= 0 + (1/3)(g)(1.1)
= 3.629....
and then use
s= ut + 0.5at²

where u = 3.629..

a = (1/3)(9.8) and t = 1.11111..
but that gives me 6. point somthing when the answer is 4.67 is the back off the book :(

where is my error(s)?

Cheerz :)

tiny-tim
Homework Helper
… conservation of energy …

Hi thomas49th! Not sure what you've done …

But shouldn't a be the full 9.8 for P on its own?

And why not use conservation of energy? ill try this step by step:

when Q hits the ground it's v becomes P intial velocity (u)
the accleration of the system, is a 1/3 of g because i worked it out using f = ma OR does it change back to the orginal 9.8??? Can you explain.

Thanks

tiny-tim
Homework Helper
… new ball-game … !

when Q hits the ground it's v becomes P intial velocity (u)

Yes! the accleration of the system, is a 1/3 of g because i worked it out using f = ma OR does it change back to the orginal 9.8??? Can you explain.
Forget the original acceleration … this is a completely new ball-game!

There's nothing moving except P … and the string is completely slack (no tension, so no force).

So it's the same as if P was in free-fall (which it is!).

So you would definitely use g.

But if you use conservation of energy, the acceleration won't come into it, will it? how can i apply the conservation of energy

energy before = energy after

in class i dont think we've used the conservation of energy apart from calculations involving elastic collsions between particles.

Thanks :)

tiny-tim
Homework Helper
in class i dont think we've used the conservation of energy apart from calculations involving elastic collsions between particles.
ah … so you haven't done PE + KE = constant?

(potential energy + kinetic energy = constant.)

we've dont avbout PE and KE being transfered, but that is in the m2 silbeius
this is the m1 silibus, thus easier.

I tried using a = 9.8 but my answer was wrong:

a = 9.8
u = 3.629
t = 1.11
the plug them into s = ut + 0.5at²

and i get 10.069...
so then i tried swapping the sign around for a and i got -2.005
:S Im stuck
Where next?

Thanks

tiny-tim
Homework Helper
we've dont avbout PE and KE being transfered, but that is in the m2 silbeius
this is the m1 silibus, thus easier.
ok, we'll do it the way in the m1 syllabus! t = 1.11
erm … where did that come from?? well the first part of the question was to find the time for particle Q to hit the floor.

Tht was
Q:
6g - T = 6a

P: T + 3g = 3a

so combine and rearrange giving a to be 1/3(9.8) - can you see why i was using that

then stick it in s = ut + 0.5at² giving t = 1.11s
:)

tiny-tim
Homework Helper
… then stick it in s = ut + 0.5at² giving t = 1.11s
:)
Yes I know! … but what's it doing here?

1.11 is the time for Q to hit the ground. What has that to do with the time for P to reach the top?

(There is a relationship, obviously … but are they the same?)

well do i need to find the time using v = u+at
t = u/a EDIT(v = 0)
then put this into s = ut + 0.5at²
giving me 2.6786.. then add this onto 2 metres giving me 4.69 which is close, but i have rounded when using u = 3.629 coudl this of made that 0.2 of a difference

Thank you ;)

well do i need to find the time using v = u+at
t = u/a EDIT(v = 0)
then put this into s = ut + 0.5at²
giving me 2.6786.. then add this onto 2 metres giving me 4.69 which is close, but i have rounded when using u = 3.629 coudl this of made that 0.2 of a difference

Thank you ;)
I wouldn't get into the habit of using different decimal places for each part of any equation, that'll likely make your answer off, sometimes significantly. If the answer is to 2dp, I'd calculate everything to 3dp and then round up to give the answer to 2dp's. Try it for yourself, see if you can get the answer by being consistent with rounding.

Noooooooooooooooooooooo i get 1.9999 when not rounding and that + 2 = 3.9999. WHAT? I dont know how to solve this question. Can you walk me through please?
Thanks :)

tiny-tim
Homework Helper
Noooooooooooooooooooooo i get 1.9999 when not rounding and that + 2 = 3.9999.
Hi thomas49th! Looks bad. Show us your full working (starting from t = 1.106, u = 3.614), so we can see what went wrong. so t = $$\sqrt{\frac{12}{\frac{9.8}{}}}$$

i found v of Q when it hits the ground using
v = u + at
= $$0 +\frac{9.8}{3}$$

then v of Q would be u of P

so u = 3.6147...
then i need to find t using v = u + at
t = v - u / a NOTE that v will be 0 at greaest height
= u / a
= -3.6147 / -9.8 (are the signs right?)
= 0.36884...
s = ut + 0.5at²
i get
s = 3.6147...(0.36884...) + 0.5(-9.8)(0.36884...)²

and i get the wrong answer :( Where have i gone wrong!

tiny-tim
Homework Helper
oh thomas49th … why didn't you put the last line in?

It's 1.3333 - .6666 = .6666.

For some reason, you added, despite your own minus-sign! Incidentally, you could also have "run the film backwards", and just used u = 0, which would have given the same result! (Now why should that be? )

sorry :( im lost. what last line are we talking about? where has 1.33333 come from?

Thanks

tiny-tim
Homework Helper
3.6147...(0.36884...) = 1.33333 … what did you have?

so was my calculations all right. I dont see how i can put the last line in. Am i missing out something really simple?

s = 3.6147...(0.36884...) + 0.5(-9.8)(0.36884...)²
this gives me 0.66666 as my answer but it should be 4.37m in the back of the book. Im lost :(

Can you give me the rules of taking stuff to be positive and negative. I think i might be getting muddled up

Thanks ;)

tiny-tim
Homework Helper
… can anyone else help … ?

Hi thomas! I'm sorry I've taken so long to reply, but I can't work out how they got that answer.
Can you give me the rules of taking stuff to be positive and negative. I think i might be getting muddled up
No … you're ok … it's just that you put the minus sign in at the end of post #19, and then forgot that it was there! I'll reproduce the question here, for convenience:
Two particles P and Q of masses 3kg and 6kg respectively are attached to the ends of a light inextensible string. The string passes over a smooth pulley. The system is released from rest wit both masses a distance of 2m above the horizontal floor. Find how long it takes for particle Q to hit the floor. Assuming particle P does not reach the pulley find its greatest height above the floor.