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M1 help
Hi.
Could someone please help me with the following M1 level question? I would really appreciate any help as I am completely stuck at the moment.
Two particles P and Q have mass 0.5 kg and m kg respectively, where m < 0.5. The particles are connected by a light inextensible string which passes over a smooth, fixed pulley. Initially P is 3.15 m above horizontal ground. The particles are released from rest with the string taut and the hanging parts of the string vertical. After P has been descending for 1.5 s, it strikes the ground. Particle P reaches the ground before Q has reached the pulley.
(a) Show that the acceleration of P as it descends is 2.8 m/s^2.
I got this part correct.
(b) Find the tension in the string as P descends.
I worked out the correct answer of 3.5 N.
(c) Show that m = 5/18.
I got the answer right here.
When P strikes the ground, P does not rebound and the string becomes slack. Particle Q then moves freely under gravity, without reaching the pulley, until the string becomes taut again. Find the time between the instant when P strikes the ground and the instant when the string becomes taut again.
I used v = u + at to work out the speed (v) of Q before the string becomes slack but I am not sure what to do next. Could someone please help.
Thank you.
Cathy
Could someone please help me with the following M1 level question? I would really appreciate any help as I am completely stuck at the moment.
Two particles P and Q have mass 0.5 kg and m kg respectively, where m < 0.5. The particles are connected by a light inextensible string which passes over a smooth, fixed pulley. Initially P is 3.15 m above horizontal ground. The particles are released from rest with the string taut and the hanging parts of the string vertical. After P has been descending for 1.5 s, it strikes the ground. Particle P reaches the ground before Q has reached the pulley.
(a) Show that the acceleration of P as it descends is 2.8 m/s^2.
I got this part correct.
(b) Find the tension in the string as P descends.
I worked out the correct answer of 3.5 N.
(c) Show that m = 5/18.
I got the answer right here.
When P strikes the ground, P does not rebound and the string becomes slack. Particle Q then moves freely under gravity, without reaching the pulley, until the string becomes taut again. Find the time between the instant when P strikes the ground and the instant when the string becomes taut again.
I used v = u + at to work out the speed (v) of Q before the string becomes slack but I am not sure what to do next. Could someone please help.
Thank you.
Cathy
Reply 1
Well, there isn't any tension acting on it anymore, only gravity. So work out how long it takes for Q to return to the height it was when P hit the ground.
Reply 2
CathyLou
OP
2
Totally Tom
Well, there isn't any tension acting on it anymore, only gravity. So work out how long it takes for Q to return to the height it was when P hit the ground.
Thank you for your help.
How do I know how high Q was when P hit the ground?
Cathy
Reply 3
I don't think you do, you only know it in terms of how high P is. Say its distance when P hits the ground is h, it carries on at a velocity v, until gravity over comes it, and then it goes back down to h. You can model h as = to 0.
So we have s=ut+1/2at² so 0=ut-4.9t² you know what u is so just work it out for t.
So we have s=ut+1/2at² so 0=ut-4.9t² you know what u is so just work it out for t.
Reply 4
CathyLou
OP
2
Totally Tom
I don't think you do, you only know it in terms of how high P is. Say its distance when P hits the ground is h, it carries on at a velocity v, until gravity over comes it, and then it goes back down to h. You can model h as = to 0.
So we have s=ut+1/2at² so 0=ut-4.9t² you know what u is so just work it out for t.
So we have s=ut+1/2at² so 0=ut-4.9t² you know what u is so just work it out for t.
Thank you so much for your help.
I have the correct answer now.
Cathy
Reply 5
hi, how do you work out part c) m = 5/18
Reply 6
also does anyone know what paper this is from?
thanks
thanks
Reply 7
Well you have ma=F
And F=Tension-mg
so you have ma=Tension-mg so m=tension/(a+g)
And F=Tension-mg
so you have ma=Tension-mg so m=tension/(a+g)
Reply 8
Originally Posted by Totally Tom
I don't think you do, you only know it in terms of how high P is. Say its distance when P hits the ground is h, it carries on at a velocity v, until gravity over comes it, and then it goes back down to h. You can model h as = to 0.
...i know i'm being really stupid abt this, bt could someone please explain to me why h=0, coz i seem to be under the impression that Q would have moved up the ramp by a distance equivalent to the distance between P and the ground, since the string is inextensible? or does the string never become taut when P is on the ground, so Q moving down the ramp will pull P up to its position before the system was released from rest?
thank youuuu =]
I don't think you do, you only know it in terms of how high P is. Say its distance when P hits the ground is h, it carries on at a velocity v, until gravity over comes it, and then it goes back down to h. You can model h as = to 0.
...i know i'm being really stupid abt this, bt could someone please explain to me why h=0, coz i seem to be under the impression that Q would have moved up the ramp by a distance equivalent to the distance between P and the ground, since the string is inextensible? or does the string never become taut when P is on the ground, so Q moving down the ramp will pull P up to its position before the system was released from rest?
thank youuuu =]
Reply 9
Hi,
firstly, it's not a ramp, but more of a pulley system. Following that,
h = 0 is because when P hits the ground (and does not rebound), there is no more tension in the string.
This means that now the only forces acting on Q is gravitational force, ie g.
When Q goes up at this moment, maybe x m off the ground, the string becomes loose, but when it comes down, and is less than x m off the ground, the string will become taut again.
So h = 0, for the string to become taut again, ie, Q is at the same height from the ground, but this time going downwards.
Using s = ut + 1/2 at^2 , taking upwards accel as positive
0 = t (u -1/2at)
hence t = 0 , or t = 2u/a
The two solutions simply correspond to the time taken. t = 0 is at the start, h = 0, as Q starts to go upwards with a loose string.
t = 2u / a is at the end point, when h = 0 as Q goes downwards, string becomes taut.
Hope this helps, and isn't too confusing.
firstly, it's not a ramp, but more of a pulley system. Following that,
h = 0 is because when P hits the ground (and does not rebound), there is no more tension in the string.
This means that now the only forces acting on Q is gravitational force, ie g.
When Q goes up at this moment, maybe x m off the ground, the string becomes loose, but when it comes down, and is less than x m off the ground, the string will become taut again.
So h = 0, for the string to become taut again, ie, Q is at the same height from the ground, but this time going downwards.
Using s = ut + 1/2 at^2 , taking upwards accel as positive
0 = t (u -1/2at)
hence t = 0 , or t = 2u/a
The two solutions simply correspond to the time taken. t = 0 is at the start, h = 0, as Q starts to go upwards with a loose string.
t = 2u / a is at the end point, when h = 0 as Q goes downwards, string becomes taut.
Hope this helps, and isn't too confusing.
Reply 10
thank you very much for your explanations...
i have sort of just resorted to recognising this type of questions, and sub in s=0
what bothers me is that why Q does not move a distance of 3.15m towards the pulley, since the string is inextensible and P doesn't rebound...
so now i just think that everything goes back to their original positions...
incidentally, i done the paper this question is from yesterday - edexcel m1 june 2007
i have sort of just resorted to recognising this type of questions, and sub in s=0
what bothers me is that why Q does not move a distance of 3.15m towards the pulley, since the string is inextensible and P doesn't rebound...
so now i just think that everything goes back to their original positions...
incidentally, i done the paper this question is from yesterday - edexcel m1 june 2007
Reply 11
Hi,
just to clarify, Q does move 3.15 m up towards the pulley as P falls towards the ground.
However, when P hits the ground, Q still has some velocity (upwards), so it will still travel some distance upwards, before its velocity reaches zero, and then goes down again.
just to clarify, Q does move 3.15 m up towards the pulley as P falls towards the ground.
However, when P hits the ground, Q still has some velocity (upwards), so it will still travel some distance upwards, before its velocity reaches zero, and then goes down again.
Reply 12
THANKYOU!!!!!!!!!!!!!!!!!!!! it's so annoying, i couldn't find any examples of those in the textbook. exams tomorrow =/
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