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1. Particles p and q of masses 2m kg and m kg respectively are attached to the ends of a large and inextensible string which passes over a smooth fixed pulley. They both hang at a distance of 2 meters above horizontal ground. The system is released from rest.
Given that the particle Q does not reach the pulley:
Find the greatest height that Q reaches above the ground
2. (Original post by allegrosquir)
Particles p and q of masses 2m kg and m kg respectively are attached to the ends of a large and inextensible string which passes over a smooth fixed pulley. They both hang at a distance of 2 meters above horizontal ground. The system is released from rest.
Given that the particle Q does not reach the pulley:
Find the greatest height that Q reaches above the ground
Great.

You tried anything, or do you want us to do this Q for you?
3. I found that a=3.3 and v=3.6 (speed of p when it hits the ground) . I checked my solutions ant they are right.
I tried to do the question. But my answer is wrong. I think that u=0 as they are released form rest and v=3.6 and a=3.3. So the answer should be 3.96( 1.96+2). But it is wrong.
4. (Original post by RDKGames)
Great.

You tried anything, or do you want us to do this Q for you?
5. (Original post by allegrosquir)
6. (Original post by RDKGames)
Here
Attached Images

7. (Original post by RDKGames)
Is there anything wrong with the working?
8. (Original post by allegrosquir)
Is there anything wrong with the working?
When P hits the floor, Q becomes a projectile upwards. If the final speed of P is 3.6 then the initial speed of the projectile Q is 3.6 upwards. You must treat Q as a projectile now and find the highest it goes.
9. (Original post by RDKGames)
When P hits the floor, Q becomes a projectile upwards. If the final speed of P is 3.6 then the initial speed of the projectile Q is 3.6 upwards. You must treat Q as a projectile now and find the highest it goes.
Does the final speed of equal to 0?
10. (Original post by allegrosquir)
Does the final speed of equal to 0?
Yes.
11. (Original post by RDKGames)
Yes.
But then the distance will be negative?
12. (Original post by allegrosquir)
But then the distance will be negative?
Well it shouldn't be. You need to pick and assume upwards as the +ve direction and you'll be good.
13. (Original post by RDKGames)
Well it shouldn't be. You need to pick and assume upwards as the +ve direction and you'll be good.
Okay, thank you!
14. (Original post by allegrosquir)
Okay, thank you!
I am a bit confused now. Do I use this: 0=3.6^2 -9.8*2*s ?
15. (Original post by allegrosquir)
I am a bit confused now. Do I use this: 0=3.6^2 -9.8*2*s ?
Yes.

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