physics

[nid404]

got it Thanks!

more questions!
Nov 02 pp1
9, 11, 12, 14, 17, 28, 36

Q12) 60m-30m= 2mv
        30m=2mv
           v=15cm/s
Q14) 50x +100X10= 20X60
             50x=200
                x=4cm

+4cm from the pivot=44cm

Q17) no change in k.e because velocity is constant

Q28 and 36 I've also explained this elsewhere...will look





        

        

[halosh92]

got it Thanks!

more questions!
Nov 02 pp1
9, 11, 12, 14, 17, 28, 36

14) 50g is on the other side
therefore clockwise = anticlockwise     let z be the unknown distance so:
(20g * 60/100) + ( 100g * 10/100) = (50g * z)
g=9.8 and convert cm to m
so z=44 cm

17) constant speed , kinetic energy= 1/2mv^2
no change in velocity so no chang ein kiinetic energy therfore change=0

28) lamda = ax/D
increase lambda , increase x

36) find current first so:
first loop and 2nd loop total resistance (15*15/15+15) =7.5
so total resistance = 7.5
total Voltage = 2V
R=V/I
I=2/7.5 =0.2666666
THATS TOTAL CURRENT.......current in one loop divide by 2 = 0.133333
so IR=V
0.1333 * 5 =0.6666
and  check the choices 2/3= 0.666666
so its A

[halosh92]

Q12) 60m-30m= 2mv
        30m=2mv
           v=15cm/s
Q14) 50x +100X10= 20X60
             50x=200
                x=4cm

+4cm from the pivot=44cm

Q17) no change in k.e because velocity is constant

Q28 and 36 I've also explained this elsewhere...will look



i already did nid

        

        

[nid404]

oh gr8 thanks 

[Phosu]

11) speed of approach=speed of separation
uA - uB = vA - vB
different directions lets take this direction -----------> as (+)
so
uA-(-uB)=vA + vB

could you plz explain the difference between speed of approach and speed of seperation. how do we know that they are same?
wen and how can we use this formula? and why cant we just use m1u1+m2u2=m1v1+m2v2

plus wen u solve the ans u've given, it gives uA+uB=vA+vB but the mark scheme says its uA+uB=vA-vB

36) find current first so:
first loop and 2nd loop total resistance (15*15/15+15) =7.5
so total resistance = 7.5
total Voltage = 2V
R=V/I
I=2/7.5 =0.2666666
THATS TOTAL CURRENT.......current in one loop divide by 2 = 0.133333
so IR=V
0.1333 * 5 =0.6666
and  check the choices 2/3= 0.666666
so its A

i got till the total current part, but why did u take the resistance as 5 in the next step?

Thanks alot for solving

[halosh92]

could you plz explain the difference between speed of approach and speed of seperation. how do we know that they are same?
wen and how can we use this formula? and why cant we just use m1u1+m2u2=m1v1+m2v2

plus wen u solve the ans u've given, it gives uA+uB=vA+vB but the mark scheme says its uA+uB=vA-vB

i got till the total current part, but why did u take the resistance as 5 in the next step?

Thanks alot for solving

because the resistor is between the terminal look at the second loo[, and view the resistor as a single component in a battery..........nid will explain q11 better than me
np POSH 

[nid404]

Q11) this is gonna be long...

In an elastic collision...momentum as well as kinetic energy is conserved,,
conservation of momentum
total momentum before collision=total momentum after collision.
mu1+mu2=mv1+mv2

conservation of k.e
1/2mu1²+ 1/2mu2²= 1/2mv1²+1/2mv2²

cancel all the halfs first, then cancel all the 'm' terms.

in steps...here's how u do it

m(u1-v1)=m(v2-u2) 1)
m(u1²-v1²)= m( v2²-u2²)  2)

since (u1²-v1²)= (u1-v1)(u1+v1)
     &  ( v2²-u2²)=(v2-u2)(v2+u2)   

divide the 2nd eqn by the first
m(u1²-v1²)      m( v2²-u2²)
---------------------------------= ------------------------------------
m(u1-v1)                                                   m(v2-u2)

m [(u1-v1)(u1+v1)]=[m(v2-u2)(v2+u2)]
-------------------  ------------------
m(u1-v1)                  m(v2-u2)

and then cancelling gives
(u1+v1)=(v2+u2)
or
u1-u2=v2-v1

now coming back to the question.
u1-u2 in this case will be positive because u for the other body is in the opp direction
u1-(-u2)=u1+u2
v2-v1 will be as it...both move in the same direction after collision
so u1+u2=v2-v1

that was tiring....i really hope you got it

[Phosu]

thank you soo much for that,
it was really helpful!

[astarmathsandphysics]

Thanks too. Really gratelul cos I had to recover a hard disk during exam season - forced to be absent for 2 days

[nid404]

pleasure is all mine 

[Phosu]

Jun 03 pp1
5, 35 and 36

[halosh92]

Jun 03 pp1
5, 35 and 36

35) C
because electric field strength is the same at any point between the plates, so its constant

36) D
E=V/d
E=700/(5/1000)
mm to m
therefore its D and direction is always from (+) to (-)

[neno]

can sum1 solve these questions:
i have the answers but i dont understand them

#Q9-mayjune 2004
#Q25,Q20,Q11-octnov 2004
#Q7-mayjune 2005
#Q34,Q33-octnov 2005

[nid404]

June 04
Q9) s= v2 sin 2theta / g    tan theta=1.25/10
 10= v2 sin (2X7.125)/ 9.8
  v2= 10/ 0.025=398.125
       v = root of 398.125= 20

Nov 04

Q25) wavelength= 2 times distance between corresponding antinodes. Hence C
Q11) mv-(-mv)=2mv to the left. opp direction
Q20) A density of P is half that of Q...It rises  twice that of Q

[nid404]

Q7 june 05
---> I have no idea 

Nov 05

Q33) D  What do you not understand here?

Q34) R=v/I
       the gradient at point C is least steep. indicates V/I is least...resistance is least