URGENT!!!! PHYSICS MCQ QUESTIONS

[superduper2009]

Can sum1 please help me with the following questions -->
oct/nov 2009 --> q 13,14,15,29
Thanks in advance
superduper2009

[astarmathsandphysics]

is that 9701/11 or 9702/12

[superduper2009]

it's 9702/11

[ilmarien]

For number 15...
Find first the velocity of both trolley by the conservation of momentum.
m1 x v1 = m2 x v2
2 x 2 = 1 x v2
v2 = 4 m/s

Then, find the kinetic energy of both trolley after moving (because the kinetic energy is generated by the potential energy of the spring).
Spring energy = KE1 + KE2 = (0.5x1x4x4) + (0.5x2x2x2)
So... 8+4 = 12 and the answer is D

And for number 29...
Remember the formula F = Eq?
F is force, E is electric field and Q is the charge.
The charge of an electron is 1.6x10^-19 (I think you can find this in the data booklet on the front pages of the question paper)
So, you just need to times the electric field with the charge...
F = 3x10^7 x 1.6x10^-19
The answer is 4.8x10^-12, which is A.

Sorry I can't help you on the other two.
Good luck for the exam!

[superduper2009]

thanks anywayz...
astarmathsandphysics cn u help me out in the other 2 plz

[ifraha18]

plz plzz help me out wid nov 2009 var 11 q3???

[superduper2009]

read both currents at 0 voltage...
u get 2.8 and 0.6
2.8/0.6 = 4.7
answer is B...
can u help me with q 13,14

[ifraha18]

nd q 6 nd 9 nov 2009

[superduper2009]

for q 6... when the ball is dropped... the displacement increases... when it bounces it comes nearer to the point where u left it... displacement decreases...then increases then decreases and so on and so forth...
the other properties do not change in this way... therfore answer is B

fr 1 9... just add up the momentum before collision...... 2mu - mu = mu
in the answers ... only A and B have a total momentum of mu...
but in D the spheres stick together...therfore it is inelastic collision...
The answer is A...

do u know q 13 and 14 please???

[ifraha18]

thnnkuu soo mchhhhh


i m sori i dont understnd dese..

[Tammet]

Anybody there??

[Tammet]

Anybody there??

[~Mimi~]

Nov 2009
 
#14


the easiest method seemed to be substitution here.
assume m=2, Initial v= 4,
so E(initial)= 1/2 *mv^2= 16 joules

now at the top, the vertical component of velocity is 0. There is no acceleration in the horizontal direction, so the horizontal component of velocity (the only velocity at the top) is the same as the beginning.
This is  v cos 45, again using v=4, horizontal component of velocity at the top= 4 cos 45
now find E at the top
E(new)= 1/2 * 2 * (4 cos 45)^2 = 8 joules
now compare the new E to the initial E.  8/16=1/2=====> 0.5E (The answer is A)....

[astarmathsandphysics]

13. Tension=torque at Q/r=3/0.1=30
Torque at P=tension*distance=30*0.15
14. horizontal component of velocity=Vcos45
at top only horizontal component is only one since vertical velocity is zero so KE is 1/2m(vcos45)^2 =1/4mv^2 =1/2of inital KE
15. From conservation of momentum 2*2=4*1 so velocity of 1kg mass is 4
KE=1/2mv^2 for both trolleys 1/2*2*2^2+1/2*1*4^2=12
29 A use F=EQ

[astarmathsandphysics]

nd q 6 nd 9 nov 2009
6 B the ball is bouncing. They have it upside down
9 A separation speed=approach speed=2u