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Author Topic: ALL CIE PHYSICS DOUBTS HERE !!!  (Read 176857 times)

Offline Master_Key

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #495 on: April 11, 2011, 12:36:52 pm »
Quote from: ashish on April 10, 2011, 06:21:59 am
hmm one factor is air resistance, as the speed increases (acceleration) the net force acting on the body is affected therefore there is a change in acceleration.

Net force = Force by engine - Retarding Force(Air resistance, Friction, etc.)

As the speed increase the Retarding Force(Air resistance, Friction, etc.) increases. You have learn't this. Falling bodies have a maximum terminal velocity. Their velocity will not increase after this. So at this time NF = 0.

F=MA
0=MA
A=0.

As F decrease F is directly proportional to A.

Acceleration decreases. The acceleration is not constant.

Hope it helps.
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Offline mdwael

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #496 on: April 11, 2011, 10:53:43 pm »
A ball of mass "m" makes a perfectly elastic head on collision with a second ball, of mass "M", initially at rest. The second  ball moves off with half the original speed of the first ball.
(a) Express "M" interms of "m".
(b) Determine the fraction of the original kinetic energy retained by the ball of mass "m" after the collision.

Pls urgently answer this question from my text book its so confusing and I dont get the correct answer in the back of the text book :(
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Offline EMO123

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #497 on: April 12, 2011, 09:53:39 am »
Quote from: Master_Key on April 12, 2011, 07:54:40 am
Is this question complete? I am trying to answer it, i think it must have something else too.

(a)If both the balls move with half the velocity of ball "m" then

By using momentum

m1v1 = m1v1+M2v2

Take v = 1

m = 0.5m + 0.5M

m = M

If "m" stays stationary and only ball "M" moves with half the velocity then

m1v1 = M2v2

v2 = 2v1

M = 2m.

(b) Fraction of Ek

Ek = 1/2mv2.

v=0.5u

Ek before collision

Ek = 1/2m(u)2.

Ek after collision

Ek = 1/2m(.5u)2.


(1/2m(.5u)2)/(1/2m(u)2)

=.25u/u

=1/4th  Ek
something is missing in this question
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Offline HUSH1994

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #498 on: April 12, 2011, 02:46:56 pm »
MJ 2006 P1 Q15, I get the answer 40N anticlockwise as the moment at the 300N force is 120 while at the 200N force 160,so the difference is 40N and should be anticlockwise,can anyone explain it to me?and question 31 on the same paper please?
« Last Edit: April 12, 2011, 03:08:57 pm by ~~!$!HUSH!$!~~ »
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Offline mdwael

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #499 on: April 12, 2011, 03:19:37 pm »
Quote from: Master_Key on April 12, 2011, 07:54:40 am
Is this question complete? I am trying to answer it, i think it must have something else too.

(a)If both the balls move with half the velocity of ball "m" then

By using momentum

m1v1 = m1v1+M2v2

Take v = 1

m = 0.5m + 0.5M

m = M

If "m" stays stationary and only ball "M" moves with half the velocity then

m1v1 = M2v2

v2 = 2v1

M = 2m.

(b) Fraction of Ek

Ek = 1/2mv2.

v=0.5u

Ek before collision

Ek = 1/2m(u)2.

Ek after collision

Ek = 1/2m(.5u)2.


(1/2m(.5u)2)/(1/2m(u)2)

=.25u/u

=1/4th  Ek


That is actually wrong because you didnt use the law (velocities of approach= velocities of seperation)  M=3m
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Offline mdwael

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #500 on: April 12, 2011, 03:23:30 pm »
Can someone solve this? Tha answer in my textbook says its 15%

" In ordere to strengthen her legs, an athlete steps on a box and then down again 30 times per minute. The girl has mass 50kg and the box is 35cm high. The exercise lasts 4.0 minutes and as a result of the exercise, her leg muscles generate 120kJ of heat energy. Calcualate the efficiency of the leg muscles (g= 10m/s^2) "
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Offline tmisterr

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #501 on: April 12, 2011, 08:07:58 pm »
work done while stepping up = energy = F*d where F is the athletes weight (her legs do work by lifting this weight upwards) and d is the distance.
work done for one step up is (50*10)*0.35=175J. while stepping down, her weight does the work so don't worry about that. ok so she does this 30 times in one minute for four minutes so the total work done (energy from legs) is equal to 175*30*4=21,000J. Her legs dissipate 120,000 joules in the form of heat energy. so the total energy from the legs is 21,000+120,000=141,000J
efficiency is equal to (work out/work in) * 100. Work input by the legs was 141,000J. but 120,000J was dissipated as heat energy so the useful work (work out, which is the energy used during the exercise) is 21,000J. replacing these values into the equation above

efficiency=(21,000/141,000)*100=14.8936......% which is approximately 15% efficiency
« Last Edit: April 12, 2011, 08:09:35 pm by tmisterr »
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Offline Master_Key

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #502 on: April 13, 2011, 02:54:05 pm »
Quote from: mdwael on April 12, 2011, 03:19:37 pm

That is actually wrong because you didnt use the law (velocities of approach= velocities of seperation)  M=3m

Does the question specify that the BALLS MOVE IN OPPOSITE DIRECTION?

Sorry for that but in this case :-

m1v1 + M1v1 = M2v2 - m1v1

Ball "M" is stationary.

m1v1 = M2v2 - m1v2

m1v1 + m1v2 = M2v2

m1+.5m1 = .5M2

1.5m1 = .5M2

M2 = 2*1.5m1

M2 or M = 3m

So did you get your answer. I thought something was missing and it was the diagram. I forgot to add case 3 in my first reply.

Case 3 :- ball "m" moves in opposite direction with .5 velocity

ball "M" moves in direction of impact with .5 velocity.
« Last Edit: April 13, 2011, 03:09:35 pm by Master_Key »
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Offline Master_Key

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #503 on: April 13, 2011, 02:58:28 pm »
(b) Fraction of Ek

Ek = 1/2mv2.

v=0.5u

Ek before collision

Ek = 1/2m(u)2.

Ek after collision

Ek = 1/2m(.5u)2.


(1/2m(.5u)2)/(1/2m(u)2)

=   .25u
------------
       u

= 1
  -- th  Ek
   4
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Offline Master_Key

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #504 on: April 13, 2011, 03:07:34 pm »
Quote from: ~~!$!HUSH!$!~~ on April 12, 2011, 02:46:56 pm
MJ 2006 P1 Q15, I get the answer 40N anticlockwise as the moment at the 300N force is 120 while at the 200N force 160,so the difference is 40N and should be anticlockwise,can anyone explain it to me?and question 31 on the same paper please?

Q15 - Should be clockwise.

as 160N acting anti-clockwise

to get in equilibruim

160 - 160 = 0
 

-160N is 160 N in clockwise direction.

And you correctly got 40 N so i am not doing it.


Q31

charge = 1.6 * 10-19

4.8A = 4.8C

Rate of Flow =       4.8
                   ------------
                    1.6 * 10-19

Rate of Flow = 3 * 1019 s-1
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Offline mdwael

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #505 on: April 13, 2011, 04:04:41 pm »
Quote from: Master_Key on April 13, 2011, 02:54:05 pm
Does the question specify that the BALLS MOVE IN OPPOSITE DIRECTION?

Sorry for that but in this case :-

m1v1 + M1v1 = M2v2 - m1v1

Ball "M" is stationary.

m1v1 = M2v2 - m1v2

m1v1 + m1v2 = M2v2

m1+.5m1 = .5M2

1.5m1 = .5M2

M2 = 2*1.5m1

M2 or M = 3m

So did you get your answer. I thought something was missing and it was the diagram. I forgot to add case 3 in my first reply.

Case 3 :- ball "m" moves in opposite direction with .5 velocity

ball "M" moves in direction of impact with .5 velocity.

Yes now its correct, the question did not specify anything other than what I wrote..its fully compete and nothing is missing thats why it took me a lot to know that the ball moves in opposite direction..
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Offline HUSH1994

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #506 on: April 13, 2011, 04:49:37 pm »
Quote from: Master_Key on April 13, 2011, 03:07:34 pm
Q15 - Should be clockwise.

as 160N acting anti-clockwise

to get in equilibruim

160 - 160 = 0
 

-160N is 160 N in clockwise direction.

And you correctly got 40 N so i am not doing it.


Q31

charge = 1.6 * 10-19

4.8A = 4.8C

Rate of Flow =       4.8
                   ------------
                    1.6 * 10-19

Rate of Flow = 3 * 1019 s-1

Thanks dude +rep ;)
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Offline Master_Key

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #507 on: April 14, 2011, 08:50:55 am »
Quote from: mdwael on April 13, 2011, 04:04:41 pm
Yes now its correct, the question did not specify anything other than what I wrote..its fully compete and nothing is missing thats why it took me a lot to know that the ball moves in opposite direction..

I think the same happened with me.
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Offline joel

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #508 on: April 19, 2011, 04:31:14 am »
Hey Guys, I need help!

- May/June 2002

 Q6 (b) (iii) - where should capacitor be placed? Mark Scheme states "capacitor connected across SQ". Should two capictors be conected one on each side of the load after the bridge rectifier.

- October/November 2002

 Q3 (b) (i) and (ii) - Can someone explain the concept and solve the questions.
 Q6 (c) (i) Why is E upwards?. Mark Scheme states "arrow pointing up page"

- May/June 2003
 Q1 (c) (i) and (ii)
 Q2 (b). Also please explain the concept.
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Offline HUSH1994

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Re: ALL CIE PHYSICS DOUBTS HERE !!!
« Reply #509 on: April 19, 2011, 04:52:29 pm »
Which paper?
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