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AS Physics P1 MCQs Preparation Thread.

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A trolley runs from P to QAt P, the kinetic energy of the trolley is 5kJ. Between P and Q, the work the trolley does against
friction is 10kJ.
What is the kinetic energy of the trolley at Q?
A 35kJ B 45kJ C 55kJ D 65kJ
plzzz help guys!
 
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A trolley runs from P to QAt P, the kinetic energy of the trolley is 5kJ. Between P and Q, the work the trolley does against
friction is 10kJ.
What is the kinetic energy of the trolley at Q?
A 35kJ B 45kJ C 55kJ D 65kJ
plzzz help guys!
go to page 79 of this thread and ull find me explaining the whole NOv 2011 P1 variant 1 and 2..and ull find that i explained this question with details that will help u alot
 
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no its B π/2rad
oh sry i mis-calculated..yes i got it π/2rad..see first u get the wavelength which is V/F= lamba..so 300/540 = 0.68 so if the length is 0.17..so 0.68 lambda = 0.17m so lambda = 0.17 / 0.68 = 0.25 is the difference between the wavelength..so if lambda/2 = π..then lambda/4 = π/2 so B...hope i helped
 

N.M

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yeah, how do we do this??
see as you throw the ball downwards the G.P.E converts into K.E
so mgh=1/2mu^2
make u the subject
(2mgh/m)^-0.5=u
u=(2mgh)^-0.5

then take out separately an expression for v
so u said that the ball rises half the height than b4
so 1/2mv^2=mgh/2
make v the subject
you will get this after simplification
v= (gh)^-0.5

now substitute the value of v and u in to
v/u
and u will get the answer

hope its clear now
 
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PLEASE SOME HELP!! I really need to find out whether I am wrong or not!!!! In this the answer is D, and i get it how they got the answer(.03 +.02), but isnt the uncertainty (.03+.02)/(16.24-12.78)uncer.jpg
 

Tkp

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A trolley runs from P to QAt P, the kinetic energy of the trolley is 5kJ. Between P and Q, the work the trolley does against
friction is 10kJ.
What is the kinetic energy of the trolley at Q?
A 35kJ B 45kJ C 55kJ D 65kJ
plzzz help guys!
ans is b.becz the total energy will be 55 as u missed 1 inf that the ep is 45.so 55-10=45
 
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