Phiscis Paper1 M/J 2007 and O/N 2007

[1992]

Can someone help on M/J 2007: Q3, 11, 12, 23 and 40 please
Also O/N 2007: Q 16, 17, 18 ,23 and 31. I always get these types of questions wrong :s

 

[intel1993]

M/j 2007:
Q3:
u just have to look in which equation left side = right side...
v=(g lemda)^1/2
ms^-1= {ms^-2 . m)^1/2
= (m^2 / s^-2)^1/2
m/s = m/s (hence proved so A ans...)

 

[intel1993]

Q11:
total momentum = m1 v1 +m2v2
=20000(20) +(900.-30) (use minus sign with 30 coz in opposite direction of lorry)
= 373 kN...

 

[intel1993]

Q12:
m1v1+ m2v2= (m1+m2)v
4.2 + 4.1=(4+2)v
v= 2

k.e=1/2 mv^2
K.e= 1/2 . 6 .2^2
k.e= 12j

 

[intel1993]

Q40:
the paricle with less charge density will have lowest speed......

charge density= proton no/ neutron no


u will get the ans to be lithium..............

 

[henglin]

hey hows it going?
im willing to help in physics past papers,

just add me on facebook : heng lin from budapest

 

[1992]

ty well explained . would u explain also the one of Q23, i dont know how it works the graph
Also if u post the ones O/N 2007 would be very grateful

 

[mzzzz]

well fr q23, wen acceleration is max, velocity is zero
at S, the gradient of tangent will b zero tht is velocity will be zero
which means max acceleration

its a rule of simple harmonic motion

 

[1992]

well fr q23, wen acceleration is max, velocity is zero
at S, the gradient of tangent will b zero tht is velocity will be zero
which means max acceleration

its a rule of simple harmonic motion

ok, i understood thanks
could u do the ones of O/N 2007.: Q16, 17 and 18, these 3 only

 

[PlanetMaster]

ok, i understood thanks
could u do the ones of O/N 2007.: Q16, 17 and 18, these 3 only

May/June 2007 Paper 1:
Question 16:
At position 3, the rope is stretched at maximum so there's no movement therefore Kinetic energy will be minimum.
The elastic being stretched out at maximum makes elastic potential energy maximum.
So its 'C'.

Question 17:
Density of P > Density of Q (Given)
Density = Mass/Volume
Since np and nq are atoms per unit volume,
MPnP > MQnQ

Question 18:
P = hpg
h1 = 100000/(1030x9.81)
h2 = 450000/(1030x9.81)
Depth below surface = h2 - h1
Depth below surface = 44.535 - 9.897
Depth below surface = 34.6 (correct to 3 sig)

 

[1992]

ty