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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
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.
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.
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
hi, how do you work out part c) m = 5/18
also does anyone know what paper this is from?
thanks
thanks
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)
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 =]
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.
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
THANKYOU!!!!!!!!!!!!!!!!!!!! it's so annoying, i couldn't find any examples of those in the textbook. exams tomorrow =/
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