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Confusing pulley question m1

Screenshot_2016-03-09-20-06-13.pngHi for part b for this question, I am stuck on it. So I have already answered a but for b how do I know when the string breaks. How do you find v because the string can break at any moment right? I know that s is h and u is 0 and a is 49/150 ms-2. Next I am not sure. Thanks in advance.
Original post by coconut64
Screenshot_2016-03-09-20-06-13.pngHi for part b for this question, I am stuck on it. So I have already answered a but for b how do I know when the string breaks. How do you find v because the string can break at any moment right? I know that s is h and u is 0 and a is 49/150 ms-2. Next I am not sure. Thanks in advance.


the acceleration cannot be positive
Reply 2
Avatar for Notnek
Notnek
21
Original post by coconut64
Screenshot_2016-03-09-20-06-13.pngHi for part b for this question, I am stuck on it. So I have already answered a but for b how do I know when the string breaks. How do you find v because the string can break at any moment right? I know that s is h and u is 0 and a is 49/150 ms-2. Next I am not sure. Thanks in advance.

You're meant to find the distance in terms of h.

So use s = h and do SUVAT to find v in the same way as you would if you had s equal to some number.

Try this and post your working if you get stuck.
Reply 3
Avatar for Notnek
Notnek
21
Original post by drandy76
the acceleration cannot be positive

The OP first has to find the velocity when the string breaks.

Taking left as positive, the acceleration will be positive before the string breaks.
(edited 8 years ago)
Original post by notnek
The OP first has to find the velocity when the string breaks.

Taking left as positive, the acceleration will be positive.


ahh thought they were focusing soley on the distance after the spring breaks
Reply 5
Avatar for B_9710
B_9710
18
Need to use the SUVAT equations to finding velocity at the point the string breaks. Then use SUVAT again to find the distance moved from that point. The only force once the string breaks acting on A in the horizontal direction is the frictional force.
Reply 6
Avatar for coconut64
coconut64
OP
16
Original post by drandy76
the acceleration cannot be positive

If you sub in 0.2 into the equation you fpund in a it wil. The markscheme says so as well.
Reply 7
Avatar for coconut64
coconut64
OP
16
Original post by B_9710
Need to use the SUVAT equations to finding velocity at the point the string breaks. Then use SUVAT again to find the distance moved from that point. The only force once the string breaks acting on A in the horizontal direction is the frictional force.


How do I find u ? I know that v would be 0 and value of a and s. Thanks
Reply 8
Avatar for coconut64
coconut64
OP
16
Original post by notnek
The OP first has to find the velocity when the string breaks.

Taking left as positive, the acceleration will be positive before the string breaks.


But i have no idea when the string breaks though. I domt know what u is . thanks
What about this way...what do you think?
Can you upload the corresponding mark scheme?
Thanks
in a hurry.pdf663.3KB
Reply 10
Avatar for Notnek
Notnek
21
Original post by depymak
What about this way...what do you think?
Can you upload the corresponding mark scheme?
Thanks

This is beyond M1.
Reply 11
Avatar for Notnek
Notnek
21
Original post by coconut64
But i have no idea when the string breaks though. I domt know what u is . thanks

You said u = 0 in your first post which is correct.

You're starting from t = 0 when the particle is at rest. And you need to find the velocity of the particle just as the string breaks, which is when s = h.
yes, I should know. Thanks.
Original post by notnek
This is beyond M1.

thus,
v2=u2-2|a|s
has to be used twice.
At first with the predefined acceleration and zeroth initial velocity, where s=h,
then we redefine acceleration of A, reuse the above equation where final velocity is.. and distance is unknown.
I like question c, though.
(edited 8 years ago)
Original post by coconut64
If you sub in 0.2 into the equation you fpund in a it wil. The markscheme says so as well.


I was talking about when the string breaks, thought you had already solved the first part


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