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# m1 pulley help watch

1. for part b), i found the v which was 2.42 and then used it to find the distance using v2 = u2 x 2as , and got 0.3 m...... but in the ms it did total distance = 0.3*2 = 0.6 m. could someone explain why it multiplied by two
thanks
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2. You've calculated the the distance A travels upwards - it will then travel the same distance back down in order for the string to become taught again.
3. (Original post by JLegion)
You've calculated the the distance A travels upwards - it will then travel the same distance back down in order for the string to become taught again.
wait why does it have to travel back the same distance to become taught again???!
4. (Original post by pondsteps)
wait why does it have to travel back the same distance to become taught again???!
When it is still moving upwards (or back down) he line is still slack to some degree. For the line to be taut, the block mst be at its starting height (the height when the other one hits the floor)
5. (Original post by an_atheist)
When it is still moving upwards (or back down) he line is still slack to some degree. For the line to be taut, the block mst be at its starting height (the height when the other one hits the floor)
but the other already hit the ground so how will it be taught again :|
6. (Original post by pondsteps)
but the other already hit the ground so how will it be taut again :|
FTFY

Because the string on the other side of the pulley, the side that isnt on the floor, is not taut;its mass is still in motion. The string will be taut when neither block is in motion
7. (Original post by pondsteps)
but the other already hit the ground so how will it be taught again :|
See attached:
8. (Original post by pondsteps)
but the other already hit the ground so how will it be taught again :|
If you think about it, ball 2 is accelerating upwards immediately before Ball 1 hits the ground. Thus when ball 1 hits the ground the tension in the rope disappears. Ball 2 is "catapulted" upwards in the sense that it continues moving in the same direction only it acts as a free falling object since the only force acting upon it is weight. Only when it returns to the position it was in when Tension is removed (i.e. at the moment when ball 1 hits the ground and tension disappears), does the string become taut again.
9. (Original post by pondsteps)
wait why does it have to travel back the same distance to become taught again???!
It's been explained in previous posts, but I'll try to explain it anyway.

As soon as B hits the ground, the string is no longer taught and becomes slack. Let's call this point P.
Since A has acceleration it continues travelling upwards until it's velocity reaches zero - this is what you have calculated; the distance travelled by A after B hits the ground to A's highest point.

At this point, the string is still slack, and hence A will travel back down due to gravity being the only force acting on A - as the string is slack, there is no tension holding / pulling A back upwards.
A will fall back down until the string is taught again - it will return to being taught when A reaches Point P again.

So, you've calculated the distance from point P to A's maximum height - you must double it because A will fall back down again to the point P before the string is taught again.

(Original post by ghostwalker)
See attached:
This explains everything better than any words can :P

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