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    I'm really stuck on part c of this question, when does it become taut, why and what does it mean?:


    Two particles, A and B are connected by a light inextensible string which passes over a smooth, fixed pulley. A has a mass of 7kg and B has a mass of 3kg. The particles are released from rest with the string taut, and A falls until it strikes the ground at a speed of 5.9ms-1. Assume that A does not rebound after hitting the ground.

    a) Find the time taken for A to hit the ground.

    b) How far will B have travelled when A hits the ground?

    c) Find the time (in s) from when A hits the ground until the string becomes taut again.



    If you can help I'd be very greatful. Thanks.
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    (Original post by BrainJuice)
    I'm really stuck on part c of this question, when does it become taut, why and what does it mean?:


    Two particles, A and B are connected by a light inextensible string which passes over a smooth, fixed pulley. A has a mass of 7kg and B has a mass of 3kg. The particles are released from rest with the string taut, and A falls until it strikes the ground at a speed of 5.9ms-1. Assume that A does not rebound after hitting the ground.

    a) Find the time taken for A to hit the ground.

    b) How far will B have travelled when A hits the ground?

    c) Find the time (in s) from when A hits the ground until the string becomes taut again.


    If you can help I'd be very greatful. Thanks.
    So for a) I got 1.505s and b) I got 2.95m
    For c) the string becomes slack when A hits the ground (meaning the string has no tension). This means particle B is travelling freely under gravity when partice A hits the ground.
    You can use suvat to work out the time it takes for B to reach it's maximum height by using v=u+at. Where v=0 and u=5.9 and a=-g I got t at this point to be 0.602 seconds. Add it on to your previous answer at a) and you should get 2.107 seconds.
    So that is the time it takes to reach max height. Now the particle falls back down and the string becomes taut again (meaning it has tension). Assuming air resistance is negligible. It should take 0.602 seconds again to the point of the string being taut because the particle is still moving freely under gravity, so 0.602*2+1.505=2.71 seconds
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    (Original post by Funnycatvideos)
    So for a) I got 1.505s and b) I got 2.95m
    For c) the string becomes slack when A hits the ground (meaning the string has no tension). This means particle B is travelling freely under gravity when partice A hits the ground.
    You can use suvat to work out the time it takes for B to reach it's maximum height by using v=u+at. Where v=0 and u=5.9 and a=-g I got t at this point to be 0.602 seconds. Add it on to your previous answer at a) and you should get 2.107 seconds.
    So that is the time it takes to reach max height. Now the particle falls back down and the string becomes taut again (meaning it has tension). Assuming air resistance is negligible. It should take 0.602 seconds again to the point of the string being taut because the particle is still moving freely under gravity, so 0.602*2+1.505=2.71 seconds
    Thanks! but the mark scheme says that b is 4.44m and c is 1.20s.
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    (Original post by BrainJuice)
    Thanks! but the mark scheme says that b is 4.44m and c is 1.20s.
    String is taut whilst B is descending. When B hits the ground, both A and B are travelling at 5.9ms-1. B comes instantaneously to rest, but A carries on now moving under gravity and the string is slack. A travels up to a maximum height and then back down until it gets to the point where it first went slack. See below.

    Funnycatvideos's working was more or less correct for c), in that it is 2 x 0.602 = 1.2...

    There's no requirement to add the time from part a), as during that motion, the string was taut, and we're only interested in when the string was slack.
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    (Original post by ghostwalker)
    String is taut whilst B is descending. When B hits the ground, both A and B are travelling at 5.9ms-1. B comes instantaneously to rest, but A carries on now moving under gravity and the string is slack. A travels up to a maximum height and then back down until it gets to the point where it first went slack.

    Funnycatvideos's working was more or less correct for c), in that it is 2 x 0.602 = 1.2...

    There's no requirement to add the time from part a), as during that motion, the string was taut, and we're only interested in when the string was slack.
    Thanks but I'm still having great trouble in understand why the string becomes tense when B is at it's starting position but on the other side. Surely it would be falling at that point and the string would be slack?
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    (Original post by BrainJuice)
    Thanks but I'm still having great trouble in understand why the string becomes tense when B is at it's starting position but on the other side. Surely it would be falling at that point and the string would be slack?
    Read it wrong. Whoops. Got 1.2 in the end anyway
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    (Original post by BrainJuice)
    Thanks but I'm still having great trouble in understand why the string becomes tense when B is at it's starting position but on the other side. Surely it would be falling at that point and the string would be slack?
    Sorry, I got the A and B the wrong way round. Post should read:

    String is taut whilst A is descending. When A hits the ground, both A and B are travelling at 5.9ms-1. A comes instantaneously to rest (and remains there on the ground), but B carries on moving upwards, now moving under gravity, and the string is slack. B travels up to a maximum height and then back down until it gets to the point where it first went slack - which is the point it was at when A reached the ground.

    Edit: Another thread on this question came up a few days ago and I did a diagram at the time, unfortunately I haven't kept it, and the search facility is proving useless.
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    (Original post by ghostwalker)
    Sorry, I got the A and B the wrong way round. Post should read:

    String is taut whilst A is descending. When A hits the ground, both A and B are travelling at 5.9ms-1. A comes instantaneously to rest (and remains there on the ground), but B carries on moving upwards, now moving under gravity, and the string is slack. B travels up to a maximum height and then back down until it gets to the point where it first went slack - which is the point it was at when A reached the ground.

    Edit: Another thread on this question came up a few days ago and I did a diagram at the time, unfortunately I haven't kept it, and the search facility is proving useless.
    Thanks, it's a bit hard to understand but I think I got it and you guys definitely helped.
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    (Original post by ghostwalker)
    Edit: Another thread on this question came up a few days ago and I did a diagram at the time, unfortunately I haven't kept it, and the search facility is proving useless.
    Was this:



    (Original post by BrainJuice)
    Thanks, it's a bit hard to understand but I think I got it and you guys definitely helped.
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    (Original post by Zacken)
    Was this:
    That's the one - well found.
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    (Original post by Funnycatvideos)
    So for a) I got 1.505s and b) I got 2.95m
    For c) the string becomes slack when A hits the ground (meaning the string has no tension). This means particle B is travelling freely under gravity when partice A hits the ground.
    You can use suvat to work out the time it takes for B to reach it's maximum height by using v=u+at. Where v=0 and u=5.9 and a=-g I got t at this point to be 0.602 seconds. Add it on to your previous answer at a) and you should get 2.107 seconds.
    So that is the time it takes to reach max height. Now the particle falls back down and the string becomes taut again (meaning it has tension). Assuming air resistance is negligible. It should take 0.602 seconds again to the point of the string being taut because the particle is still moving freely under gravity, so 0.602*2+1.505=2.71 seconds
    Why is u= 5.9? Is it because B only moves upwards once A reaches the ground.. and as they both have the same acceleration u would be A's final velocity???
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    (Original post by Starmock99)
    Why is u= 5.9? Is it because B only moves upwards once A reaches the ground.. and as they both have the same acceleration u would be A's final velocity???
    Because A hits the ground at 5.9ms^-1 while the string was taut. The particles move at the same acceleration when the strings are taut and so they travel at the same velocity too becausre they start from rest. So the very instant the sting becomes slack B is travelling at an initial velocity of 5.9
 
 
 
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