Physics: Post your doubts here!

[Abhishek Neogi]

Help

A motorist travelling at 10 m/s
can bring his car to rest in a distance of 10 m.
If he had been travelling at 30 m/s
, in what distance could he bring the car to rest using the same
braking force?

I have no idea how to solve this... Plz help!

Tobi will help you solve it!
Alright
u = 10 m/s
v = 0 m/s
s = 10 m

That's the first case
So...
You gotta find the retardation of the car first [Basically negative acceleration]
So....
Yeah
* = Multiply

V*V = U*U - 2*a*s
0 = 100 - 20a
-100 = -20a
a = 100/20 =( 5m/s*s)

That's the acceleration of your car! CAR FTW ! XD

Now the question is IF he were to travel at 30m/s,what'd be the distance


Therefore use the same equation again nacho

But in this case:
u = 30 m/s
v = 0m/s
s = ??? m
a = 5 m/s*s


V*V = U*U - 2*a*s
0 = 30*30 - 2*5*s
0 = 900 - 10s
- 900 = -10s
s = -900/-10 = 90m

90m OMG. That's the answer lol xD


btw doesn't a motorist drive a motorcycle instead of a car?How can a biker drive a car at the same time when he's driving a bike?:/

 

[raamish]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
how do we solve 2d and 5c. I dont have any clue. Help me out please.

 

[Yousif Mukkhtar]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_13.pdf
Can someone explain Q2.

 

[maleeha123]

hey i want the link for phys june 2012 papers........guys guide me

 

[autumnsakura]

hey i want the link for phys june 2012 papers........guys guide me

http://olevel.sourceforge.net/papers/
Look for subject code 9702

 

[autumnsakura]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_13.pdf
Can someone explain Q2.

The easiest way to do this is to substitute V with a number such as 10. So X would be 10 cos angle and Y would be 10 sin angle. Substitute in two angles, for example 30 degrees and 60 degrees. Then compare the X and Y obtained from the two different angles.

 

[autumnsakura]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_2.pdf
how do we solve 2d and 5c. I dont have any clue. Help me out please.

2(D) Use s=ut + 1/2 at^2. Sub in time from c(ii) and acceleration from c(i). Your answer should be s=12.71cm. But the length of the metal plates is only 12cm. Since s calculated is longer than metal plates, electron will emerge from plates.

5(c) I'm not so sure. My answer for you would be from the marking scheme so if you're looking for explanation, it wouldn't be helpful..

 

[Mirna]

Hi everyone
Can anyone help me find a website where I can find classified pastpapers ??

 

[autumnsakura]

Can anyone help clarify? For Q4, why is the answer D not B? For Q10, Why is it not A but D? Thanks!!

 

[Ange12]

Does anyone has an idea about the physics practical where we don't have to draw any graph?

 

[elbeyon]

Can anyone help clarify? For Q4, why is the answer D not B? For Q10, Why is it not A but D? Thanks!!

For Q no. 4 the total range with which the values are fluctuating is +- 0.04 not +- 0.02 . So the range is +- 0.04. That's my reason . And for question no. 10 we know that in case of elastic collision u1 - u2 = v2 - v1 ( i.e Speed of approach = Speed of separation ) . Since the body are initially travelling in opposite direction so one velocity must be in negative direction. Since u1, v1 and v2 are in same direction so it is better to take u2 as negative ( But you can alter that and take u2 as positive and others as negative. You will end up with the same answer in either case. ) Now since u2 is negative so the above eqn. becomes (u1 + u2 = v2 - v1 ) So D not A. I guess you are clear now.

 

[elbeyon]

Does anyone has an idea about the physics practical where we don't have to draw any graph?

Yes for the question no. 2 of Paper 3 . You have to write the errors and ways to solve them and you usually don't have to draw graph in those questions.

 

[Ange12]

well, thank you

 

[autumnsakura]

For Q no. 4 the total range with which the values are fluctuating is +- 0.04 not +- 0.02 . So the range is +- 0.04. That's my reason . And for question no. 10 we know that in case of elastic collision u1 - u2 = v2 - v1 ( i.e Speed of approach = Speed of separation ) . Since the body are initially travelling in opposite direction so one velocity must be in negative direction. Since u1, v1 and v2 are in same direction so it is better to take u2 as negative ( But you can alter that and take u2 as positive and others as negative. You will end up with the same answer in either case. ) Now since u2 is negative so the above eqn. becomes (u1 + u2 = v2 - v1 ) So D not A. I guess you are clear now.

Thank you soo much

 

[autumnsakura]

For Q no. 4 the total range with which the values are fluctuating is +- 0.04 not +- 0.02 . So the range is +- 0.04. That's my reason . And for question no. 10 we know that in case of elastic collision u1 - u2 = v2 - v1 ( i.e Speed of approach = Speed of separation ) . Since the body are initially travelling in opposite direction so one velocity must be in negative direction. Since u1, v1 and v2 are in same direction so it is better to take u2 as negative ( But you can alter that and take u2 as positive and others as negative. You will end up with the same answer in either case. ) Now since u2 is negative so the above eqn. becomes (u1 + u2 = v2 - v1 ) So D not A. I guess you are clear now.

Oh, I forgot to ask. For speed of approach=speed of separation, the formula is u1-u2 = v2-v1 right? As you showed previously.. So I can't use m1u1 + m2u2 = m1v1 + m2v2 ?

 

[yumichikabyakuya renji]

Tobi will help you solve it!
Alright
u = 10 m/s
v = 0 m/s
s = 10 m

That's the first case
So...
You gotta find the retardation of the car first [Basically negative acceleration]
So....
Yeah
* = Multiply

V*V = U*U - 2*a*s
0 = 100 - 20a
-100 = -20a
a = 100/20 =( 5m/s*s)

That's the acceleration of your car! CAR FTW ! XD

Now the question is IF he were to travel at 30m/s,what'd be the distance


Therefore use the same equation again nacho

But in this case:
u = 30 m/s
v = 0m/s
s = ??? m
a = 5 m/s*s


V*V = U*U - 2*a*s
0 = 30*30 - 2*5*s
0 = 900 - 10s
- 900 = -10s
s = -900/-10 = 90m

90m OMG. That's the answer lol xD


btw doesn't a motorist drive a motorcycle instead of a car?How can a biker drive a car at the same time when he's driving a bike?:/

haha, no the motorist drives a vehicle powered by a motor,here, a car

 

[yumichikabyakuya renji]

hey guys plz help me out with ques 9
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf

 

[elbeyon]

Oh, I forgot to ask. For speed of approach=speed of separation, the formula is u1-u2 = v2-v1 right? As you showed previously.. So I can't use m1u1 + m2u2 = m1v1 + m2v2 ?

Its not for elastic collision. That's for inelastic collision. In case of elastic collision the kinetic energy of the body is conserved so we have to use the another equation that I mentioned earlier. I guess you are clear now.

 

[InnocentAngel]

For Q1, cube root 2.7 x 10 ^ 25 m ^-3 and you get 6.46 x 10^-9 m. Divide the answer by two again and you get the final answer. I don't know why divide by two though...
For Q2, use P1/T1 =P2/T2 to get the pressure when box is 200 degree celsius. Remember that temp should be in Kelvin. You should get 1.65 x 10^ 5 Pa. Since it says net force, find its net pressure. Outside the box there's atmospheric pressure of 1.01 x 10^5 Pa already. Inside the box, there's 1.65 x 10^5 Pa. So minus that and you get 6.5 x 10^4 Pa. That times the area which is 0.09m squared, you'll get the final answer

Heyyyyy!! Thanks loads !

 

[InnocentAngel]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_13.pdf
Can someone explain Q2.

Resolve v :
1. You get v cos theta along X . So magnitude of X = v cos theta
But using mathematical knowledge :
Increasing theta form 0 to 90 degrees causes cos theta to decrease . Remember the graph for cos theta ? Do refer to it.
Hence along X , v cos theta decreases, so X decreases.

2. You get v sin theta along Y .So magnitude of Y = v sin theta .
But again applying mathematics:
Increasing theta from 0 to 90 degrees causes sin theta to increase . Refer graph of sin theta .
Hence Y increases.

Hope it helped!