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AS Physics P1 MCQs Preparation Thread.

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Initial KE = 2 * final KE ( because the ball rebounds to half its original height)
So, 1/2*m*u^2 = 2*1/2*m*v^2
1/2*m*u^2 = m*v^2
Rearrange in the form v/u and you get 1/(root) 2
can u explain a bit? y did u multiply by 2 when it was half hieght? :S
 
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haha easy. spending time and short tempering will not help. its the game of concept :D

see in 13th mcq. you should recall that torque of a couple is one of the force into perpendicular distance between the two same forces. means F into 1.20 = 900 into .20 , this will give you force 150 N i.e the very dear and sweet B option!
for the minimum force to be applied we need to get same torque below and on the spindle in order to create rotation effect so that the poor weight can be lifted

15th: height decreases by 2 in both containers. pressure reduces by 4 times altogether...no convincing answer available at the moment.
thanks for ur help ^^
 
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initial momentum of the ball is 2mv before collision... if collision was perfectly elastic one, all the momentum would have been gained by the ball... and this would have made the change in the momentum equal to 4mv.. 2mv-(-2mv)... now it is given that collision is inelastic, it wil losse some of its momentum to the wall.. so now change in momentum will be less than 4mv but cant be less than the initial momentum... so the answer is C.. i think you cannot verify the answer by the calculations...
oh okay thanks a lot :)
 
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