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Physics: Post your doubts here!

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Can someone please explain how to do q30
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_13.pdf

because that is not on the AS syllabus

I have found a site to help me but I don't get anything from it
http://www.physicsclassroom.com/Class/sound/u11l5d.cfm equation for closed end tube: f=v/4l
http://www.physicsclassroom.com/class/sound/u11l5c.cfm equation for open end tubes: f=v/2l

does the frequency have to be like greater than 0.5 for stationary waves??

They wouldn't set a question outside of the syllabus. You just need to apply your knowledge to the situation.
What you need to do is find out if you can draw a stationary wave in both the tubes P and Q. If you are able to draw a stationary wave in a tube, then a stationary wave can be formed in it.
Closed end of a pipe is always a node, and the open end is always an anti-node.
For P you have an anti-node at the open end, and a node at the closed end. You are able to draw a stationary wave 35cm long with 4 nodes, and 4 anti-nodes so P works.
For Q, both ends have an anti-node. You are able to draw a stationary wave 50cm long with 5 nodes, and 6 anti-nodes, so Q works too.
Final answer is A!
 
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Hey people i have an important question.. Does number of significant figures matter in phy paper 2? If answer scheme gives 2sf for final answer but i gave in 3sf, do i lose any mark?
 
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i just read something about simple harmonic motion and was wondering is that covered in our AS Physics syllabus?
 
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They wouldn't set a question outside of the syllabus. You just need to apply your knowledge to the situation.
What you need to do is find out if you can draw a stationary wave in both the tubes P and Q. If you are able to draw a stationary wave in a tube, then a stationary wave can be formed in it.
Closed end of a pipe is always a node, and the open end is always an anti-node.
For P you have an anti-node at the open end, and a node at the closed end. You are able to draw a stationary wave 35cm long with 4 nodes, and 4 anti-nodes so P works.
For Q, both ends have an anti-node. You are able to draw a stationary wave 50cm long with 5 nodes, and 6 anti-nodes, so Q works too.
Final answer is A!

Thanks : D

but they do sometimes bring in questions form A2 syllabus i remember a question asking us to find a phase difference between two points 0.17m apart on a wave and there is a formula for it in A2
 
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Hey people i have an important question.. Does number of significant figures matter in phy paper 2? If answer scheme gives 2sf for final answer but i gave in 3sf, do i lose any mark?

you should not because you give an answer to same no of sig figs as used in question or 1 better ie 2 or 3
 
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a
Hi everyone, assalamoalaikum!! :)

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your AS/A2 PHYSICS doubts in this thread. InshaAllah other people here will help me and you all. :D ;)

NOTE: If any doubts in the pastpapers, please post the link! You can find links here!

Any Physics related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites! :)
Thanks!
Jazak Allah Khair!

Physics Notes:

Some links & Notes - by 'destined007'

Physics Practical Tips - by arlery

Notes for A2 Direct Sensing (Applications) - shared by sweetiepie

Physics Summarised Notes (Click to download)

AS and A-Level Physics Definitions

A2: Physics Revision notes - by smzimran

Paper:5 Finding uncerainty in log - by XPFMember

Physics Paper 5 tips - by arlery

Physics Compiled Pastpapers:

any tips how to prepare for phy ppr2 .. should we do book or past papers ?
 
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A ball is kicked towards goal posts from a position 20 m from and directly in front of the posts.
The ball takes 0.60 s from the time it is kicked to pass over the cross-bar, 2.5 m above the ground.
The ball is at its maximum height as it passes over the crossbar. You may ignore air resistance.
a. Calculate the ball’s horizontal component of velocity.
b.Calculate the vertical component of the velocity of the ball immediately after it is kicked.
c.Determine the magnitude of the initial velocity of the ball immediately after it is kicked.
d.Determine the angle above the horizontal at which the ball is kicked.

Can anyone please solve it? Thanks.
 
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A ball is kicked towards goal posts from a position 20 m from and directly in front of the posts.
The ball takes 0.60 s from the time it is kicked to pass over the cross-bar, 2.5 m above the ground.
The ball is at its maximum height as it passes over the crossbar. You may ignore air resistance.
a. Calculate the ball’s horizontal component of velocity.
b.Calculate the vertical component of the velocity of the ball immediately after it is kicked.
c.Determine the magnitude of the initial velocity of the ball immediately after it is kicked.
d.Determine the angle above the horizontal at which the ball is kicked.

Can anyone please solve it? Thanks.
can you tell the year plz?
 
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A ball is kicked towards goal posts from a position 20 m from and directly in front of the posts.
The ball takes 0.60 s from the time it is kicked to pass over the cross-bar, 2.5 m above the ground.
The ball is at its maximum height as it passes over the crossbar. You may ignore air resistance.
a. Calculate the ball’s horizontal component of velocity.
b.Calculate the vertical component of the velocity of the ball immediately after it is kicked.
c.Determine the magnitude of the initial velocity of the ball immediately after it is kicked.
d.Determine the angle above the horizontal at which the ball is kicked.

Can anyone please solve it? Thanks.

a) The horizontal component of the balls velocity remains constant throughout it's motion as the air resistance is negligible.
so, speed = distance/ time
= 20/ .6 = 33.3 m/s

b) v^2 = u^2 +2as
the final velocity will be 0m/s when the ball is at it's max. height.
so, 0 = u^2 + (2x -9.8x 2.5)
u= 7m/s

c) initial velocity = root over (7^2 + 33.3^2)
= 34.0 m/s

d) for angle tan(theta)= 7/33.3
therefore, angle = 11.90
 
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