# Thread: Mechanics 1 - pulley question.

1. ## Mechanics 1 - pulley question.

Hi, i don't really get question (e), is there anyone explain question (e)?

Here is the question.

6. 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 3.15m 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.

acceleration is 2.8 ms^-2
Tension is 3.5 N
and m = 5/18

(e)
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.

hence, find the time between the instant when P strikes the ground and the instant when the string becomes taut again.

----------------------------------------------------------------------

Mark scheme says

v = u + at => 4.2 = -4.2 + 9.8t
therefore, t = 6/7

but i dont really get this.. could anyone explain why v is -4.2 too??

Thank you.

2. I haven't carried out the calculation but I assume that the speed of the particle Q at the point P hits the ground is 4.2 $ms^{-1}$ which is in the opposite direction to gravity i.e. upwards. The string will be taut again when the particle reaches this point again and by conservation of energy this means that it will have the same speed but in the opposite direction. Hence:

final velocity, $v = 4.2 ms^{-1}$ (downwards so positive)
initial velocity, $u = -4.2 ms^{-1}$ (upwards so negative).

Does it make sense?

If not let me know.

but i am not sure what this mean...

The string will be taut again when the particle reaches this point again and by conservation of energy this means that it will have the same speed but in the opposite direction

can you elaborate or explain a bit more please..?

sorry... and thank you very much.

4. Sure:

So before P hits the ground the string is taught and so the mass of P is pulling Q upwards.

When P hits the ground it no longer exerts a pull on the string and hence there is no longer any force pulling Q upwards. (There's no tension in the string).

However, at this time Q has speed (and kinetic energy) upwards but the only force acting on it is gravity downwards which is slowing it down.

So from this time on the particle Q is doing much the same as you would expect from a ball that you threw straight up (vertically) in the air. It would travel upwards until it reached its highest point (at which point its speed is 0) then fall back down.

What I meant by conservation of energy is that when P hits the ground Q has a certain amount of kinetic energy which is converted to potential energy as it travels upwards and then back into kinetic energy on the way down. So it must end up with the same kinetic energy at any particular height whether it's going up or down. (The sum of potential and kinetic energy remains a constant and potential energy depends on the height of the particle above ground).

Finally when the string is taut it means it is stretched out and when it is not it is shorter than it's normal full length i.e. it's slack. So we know that after P hit the ground there was a period of time when the string was slack. However, when it got back to the same height above the ground (that it was at when Q hit the ground) then the string would be taut as it is at full length again.

OK, I hope that was more helpful. If that was a longer/ simpler explanation than was required then I'm sorry as I was just trying to be completely clear.