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Stuck on a mechanics 1 question.

I'm stuck on the following question:
Particles A and B, each of mass 0.6kg, are joined by a light inextensible string. The string passes over a smooth pulley at the edge of a smooth horizontal platform. A is held at rest on the platform. B hangs vertically below the pulley at a height h m above the floor. A is released, with the string taut, and the particles start to move. There is no air resistance.

i) Find the tension in the string and the acceleration of A

ii)Hence find the speed of A after it has travelled a distance of 2m

iii) When A has moved a distance of 2m it becomes detached from the string. From this instant B takes a further 0.2seconds to reach the floor. Find the value of h.

iv) Find also the total time for which B is in motion before it reaches the floor.

I have absolutely no idea how to do these. I tried resolving with F=ma, but that's getting me nowhere. Please help!
Reply 1
Avatar for Al_YG
Al_YG
5
Which question is this from the book? Give me the page number ill try to help!
Original post by Al_YG
Which question is this from the book? Give me the page number ill try to help!


Page 52 from OCR mechanics 1 (advanced maths essentials)
Reply 3
Avatar for Al_YG
Al_YG
5
I don't do OCR. However what you first have to do is see the force exerted by the mass (of particle B) to the floor. To find this out you do 9.8 multiplied by 0.6kg
Reply 4
Avatar for H.b9
H.b9
5
Original post by naturolocer123
I'm stuck on the following question:
Particles A and B, each of mass 0.6kg, are joined by a light inextensible string. The string passes over a smooth pulley at the edge of a smooth horizontal platform. A is held at rest on the platform. B hangs vertically below the pulley at a height h m above the floor. A is released, with the string taut, and the particles start to move. There is no air resistance.

i) Find the tension in the string and the acceleration of A

ii)Hence find the speed of A after it has travelled a distance of 2m

iii) When A has moved a distance of 2m it becomes detached from the string. From this instant B takes a further 0.2seconds to reach the floor. Find the value of h.

iv) Find also the total time for which B is in motion before it reaches the floor.

I have absolutely no idea how to do these. I tried resolving with F=ma, but that's getting me nowhere. Please help!


Do you have the correction for the first question?I tried to calculate it and it gave me a tension of 3N and 5m/s^2 for acceleration
Reply 5
Avatar for Al_YG
Al_YG
5
Because particle A is on a smooth surface, there is no friction acting against it therefore the only force in the whole entire world that can cause particle A to move is the tension force. Using F=MA we know that the tension force is equal to the mass (which is 0.6) time by the acceleration of the particle, which is clearly just 'a' as we don't know the acceleration. Therefore T=0.6a
Original post by Al_YG
I don't do OCR. However what you first have to do is see the force exerted by the mass (of particle B) to the floor. To find this out you do 9.8 multiplied by 0.6kg


I have no idea where that would get me. What would that give? F?

Original post by H.b9
Do you have the correction for the first question?I tried to calculate it and it gave me a tension of 3N and 5m/s^2 for acceleration


Those are the right answers.

Original post by Al_YG
Because particle A is on a smooth surface, there is no friction acting against it therefore the only force in the whole entire world that can cause particle A to move is the tension force. Using F=MA we know that the tension force is equal to the mass (which is 0.6) time by the acceleration of the particle, which is clearly just 'a' as we don't know the acceleration. Therefore T=0.6a


I got that far. No idea what to do next though. I must have done something wrong with the 0.6kg particle (I'm getting 0.6g-T=0.6a, which has got to be wrong)
(edited 8 years ago)
Reply 7
Avatar for Al_YG
Al_YG
5
If you had figured out the force being exerted by particle B to the floor, you would get simply that the force is equal to the mass multiplied by the acceleration. The mass of particle B is 0.6 and multiply by the acceleration (which is 9.8 for ALL objects). Therefore the force exerted by particle B is 5.88N.

However , However the tension force in the string opposes the force of the particle going downwards so 5.88N-TN is equal to the mass of the object multiplied by the acceleration. Therefore 5.88N-TN= 0.6a
Reply 8
Avatar for H.b9
H.b9
5
And you do the same for particle b
Fb=mb*a
Fb=(mb*10)-T
Therefore
(mb*10)-T=mb*a
Then substitute T with ma*a
You will have then an equation of
(mb*10)-ma*a=mb*a
(=) mb*10=(mb*a)+(ma*a)
(=) mb*10=a(mb+ma)


Legends:
T- tension
mb-mass of b
ma-mass of a
a- acceleration
Fb- total force on b

I used 10m/s^2 instead of 9.8 m/s^2
Reply 9
Avatar for Al_YG
Al_YG
5
Now you have two equations ,
First one is : T=0.6(a)
Second one is: 5.88-T=0.6(a)

Now solve simultaneously
Reply 10
Avatar for Al_YG
Al_YG
5
If you just look at my posts you will get the same answers, there is no other way of explaining it better. Instead of me using 9.8 use your 10 if that's for your exam board, the answer will be the same anyway.
Original post by Al_YG
If you had figured out the force being exerted by particle B to the floor, you would get simply that the force is equal to the mass multiplied by the acceleration. The mass of particle B is 0.6 and multiply by the acceleration (which is 9.8 for ALL objects). Therefore the force exerted by particle B is 5.88N.

However , However the tension force in the string opposes the force of the particle going downwards so 5.88N-TN is equal to the mass of the object multiplied by the acceleration. Therefore 5.88N-TN= 0.6a


Yup, I got that thanks. For some reason I was taking off 0.6a from 0.6a which was confusing me. The rest should be fine.
Reply 12
Avatar for Al_YG
Al_YG
5
If you still don't get it just send me a message and ill try to explain there.
Reply 13
Avatar for H.b9
H.b9
5
Did I get it right?
I just applied what I learned from the other day on my 11th year class
I'm from Portugal
Resitting the class at the moment
Reply 14
Avatar for Al_YG
Al_YG
5
OK then no problem I'm glad to have helped :smile:
Reply 15
Avatar for H.b9
H.b9
5
Original post by Al_YG
If you just look at my posts you will get the same answers, there is no other way of explaining it better. Instead of me using 9.8 use your 10 if that's for your exam board, the answer will be the same anyway.


Yeah it will be almost be the exact answer
Original post by H.b9
Did I get it right?
I just applied what I learned from the other day on my 11th year class
I'm from Portugal
Resitting the class at the moment


Yeah, you got it right. The rest of the questions were a breeze. One silly mistake got me stuck for so long... Thanks again!
Reply 17
Avatar for H.b9
H.b9
5
No problem
😁
Can you share how you solved the second question ?
Kinda confuse on it😅


Posted from TSR Mobile
Original post by H.b9
No problem
😁
Can you share how you solved the second question ?
Kinda confuse on it😅


Posted from TSR Mobile


Use suvat so

s = 2
u = 0
v = ?
a = 4.9
t = not needed

use v^2=u+2as and substitute in the values. You should get 4.32ms-1
Reply 19
Avatar for H.b9
H.b9
5
I always forget that formula
Thanks 👍🏼

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