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Q.13--- why u/cos@ and not 2ucos@?

Q.13--- why u/cos@ and not 2ucos@?
 

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Grade:11

9 Answers

Neeti
571 Points
8 years ago
Assuming that speed of the mass is v  and that of the ends P and Q is given as u.
Since M has velocity v the tiny length of thread which is attached to the mass directly will also have velocity v (vertically upwards) .
Now the component of this velocity along each string = v cos \Theta therefore the velocity of the two strings is  v cos \Theta .
 
But since each string is inectinsible and uniform, the entire string should have the same velocity and it is given that the velocity of each string is u ( velocity of ends P and Q = velocity of each string = u ) 
 
So you have v cos \Theta = u 
or v = u / cos \Theta
 
Please feel free to ask any follow up question and approve the answer if you think it’s useful :) 
Raisinten
12 Points
8 years ago
Well, I liked your answer but why isn’t it ucos@?
Neeti
571 Points
8 years ago
I know why you would think that. Surprisingly i had once got stuck in the same question :P 
Well for the simple reason, you cannot just take a component of u in the vertically upward direction and assume that it’s the velocity of the mass. Because the Mass here is a different object, it’s not a part of the string but is connected to the string that too through another small piece of thread. Getting it? 
Raisinten
12 Points
8 years ago
Even if it is not connected by a different string, the velocity still remains u/cos@ isn’t it? Why is that?
Neeti
571 Points
8 years ago
No i mean to say, it’s not a part of the string , it is connected to a string. like think about it this way, a car is moving and a body is sitting on top of it ( then the are considered to be a single system having the same velocity) 
and a man is attached to a car and the car is taking a turn, the man’s velocity won’t be a component of the velocity of the car.
Did this help? 
 
Raisinten
12 Points
8 years ago
Yes but it would be better if there is further explanation and what shall I consider if I have to face a variation of this kind of numerical?
 
Neeti
571 Points
8 years ago
I’m really sorry i don’t know how else i should explain it. You have to visualize it yourself. 
Ok try this, i’m sure you’ve watched harry potter. in the second movie when Ron drives the car in air and there’s one moment when Harry is about to fall.
Now the car suppose has a velocity at a particular angle to the horizontal . Now since Ron is sitting IN the car, (part of the system ) the component of the velocity in the horizontal direction will also be the velocity of ron in the horizontal direction but that’s not the case for Harry who’s hanging down the car. He will have a different velocity in that direction. 
Now please tell me you understood this? Try to visualize it, once you understand the difference, you’ll know when a body is directly in the system and when it is attached to the system.
and then you’ll understand the approach. (when directly in the system, the logic of direct component shall be applicable) and when not directly a part of the moving body, you apply the concept which i have used. 
 
Is it clear now? 
Raisinten
12 Points
8 years ago
Yes it is clear now. Thank you. Good example :)
Neeti
571 Points
8 years ago
Haha You are most welcome, i’m glag you understood it :) 

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