IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: halosh92 on March 07, 2010, 03:15:17 pm

Title: physics AS HELPPP
Post by: halosh92 on March 07, 2010, 03:15:17 pm: physics cie AS
2006 jun paper 1
Q 1
Q9
Q12
Q15
Q17
Q18
Q24
Q33

thx ;D ;D ;D ;D ;D ;D ;D ;D ;D
Title: Re: physics AS HELPPP
Post by: astarmathsandphysics on March 07, 2010, 03:58:28 pm: Will have to do it when I get home
Title: Re: physics AS HELPPP
Post by: tmisterr on March 07, 2010, 05:09:39 pm: Q1 vector quantities have both magnitude(size) and direction. answer is A (displacement and Acceleration)
Q12 from law of conservation of momentum, moment before=momentum after. Call speed after momentum x and their combined mass after collision will be m+3m=4m. take one direction as positive and the other negative
so m*-2v+3m*v=3m*x
-2mv+3mv=4mx
mv=4mx
x=v/4 which is A
Q15 length of bar is 2.4m long. Half the distance is 1.2 m long so distance from 300N to pivot is 0.4.
Clockwise moment is 300*0.4=120. Anticlockwise moment is 200*0.8=160. for it to be in equilibrium CM=ACM so you need to add a torque of 40Nm in the clockwise direction for it to equal 160. Answer is A.
Q17 v²=u²+2as. from the infromation 0²=10²+2*10*a
a=-5ms^-2
now u is 30 and s is asked
0²=30²-2*-5*s
s=90 Answer is D
Q18 potential energy=mass*gravity*height
mg=4 so PE= 4*30=120J answer is A
Q24 intensity is directly proportional to Amplitude²*frequecy² knowing this will help you solve. you can just replace values from the first graph into the equation, find and find k, and use it to find the amplitude and frequency of the second graph. Its a little hard to explain via writing but answer should be B
Q33 Answer is B. Use the equation Vout=Vin*R1/(R1+R2). Vin to the top two resistors is 12 and to the bottom two is also 12 since they are connected in parallel. Voltage across 500 ohm resistor (voltage at X) will be 4 and across 2000 ohm resistor (voltage at Y)is 8. the difference is 4.
Title: Re: physics AS HELPPP
Post by: halosh92 on March 07, 2010, 05:48:34 pm: Quote from: tmisterr on March 07, 2010, 05:09:39 pm
Q1 vector quantities have both magnitude(size) and direction. answer is A (displacement and Acceleration)
Q12 from law of conservation of momentum, moment before=momentum after. Call speed after momentum x and their combined mass after collision will be m+3m=4m. take one direction as positive and the other negative
so m*-2v+3m*v=3m*x
-2mv+3mv=4mx
mv=4mx
x=v/4 which is A
Q15 length of bar is 2.4m long. Half the distance is 1.2 m long so distance from 300N to pivot is 0.4.
Clockwise moment is 300*0.4=120. Anticlockwise moment is 200*0.8=160. for it to be in equilibrium CM=ACM so you need to add a torque of 40Nm in the clockwise direction for it to equal 160. Answer is A.
Q17 v²=u²+2as. from the infromation 0²=10²+2*10*a
a=-5ms^-2
now u is 30 and s is asked
0²=30²-2*-5*s
s=90 Answer is D
Q18 potential energy=mass*gravity*height
mg=4 so PE= 4*30=120J answer is A
Q24 intensity is directly proportional to Amplitude²*frequecy² knowing this will help you solve. you can just replace values from the first graph into the equation, find and find k, and use it to find the amplitude and frequency of the second graph. Its a little hard to explain via writing but answer should be B
Q33 Answer is B. Use the equation Vout=Vin*R1/(R1+R2). Vin to the top two resistors is 12 and to the bottom two is also 12 since they are connected in parallel. Voltage across 500 ohm resistor (voltage at X) will be 4 and across 2000 ohm resistor (voltage at Y)is 8. the difference is 4.

thankyouuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu soooooooooooooooooo muchhhhhhhhh! that wasvery gd! ;) ;D ;D
Title: Re: physics AS HELPPP
Post by: nid404 on March 08, 2010, 04:33:20 pm: thanks tmisterr
Title: Re: physics AS HELPPP
Post by: halosh92 on March 08, 2010, 06:33:44 pm: 2005 nov paper 1:
Q6
Q7
Q8
Q12
Q21
Q31
Title: Re: physics AS HELPPP
Post by: astarmathsandphysics on March 09, 2010, 07:52:54 am: 6)B acceleration of gravity is constant 9,8
7)C use s=ut+1/2at^2 twice
$s_1=1/2at_1^2$
$s_2=1/2at_2^2$
subtract
$s_2 -s_1 =1/2at_2^2 -1/2at_1^2$
$h=1/2a(t_2^2 -t_1^2)$
$a=2h/(t_2^2 -t_1^2)$
8)A zero speed at start then constant force means constant acceleration hence gradient
Title: Re: physics AS HELPPP
Post by: astarmathsandphysics on March 09, 2010, 08:02:09 am: 12)moments about P
10*0.6+100*0.1=20x+0.4*20
16=20x+8 so x=0.4
D
21)Y=Fl/Ae
e=Fl/AY
half the diameter means 1/4 of area so 1/4/1/4=1 and extension is not changed
C
31)F=Eq=V/d*q=200/5*10-3 *1.6*10^-19 =6.4*10-15
A
Title: Re: physics AS HELPPP
Post by: tmisterr on March 09, 2010, 02:54:15 pm: ur welcome :)