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Author Topic: Edexcel PHYSICS Doubts  (Read 132088 times)

Offline blackberry

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Re: Edexcel PHYSICS Doubts
« Reply #420 on: January 11, 2011, 03:15:34 pm »
evn i selected B, vampire are you sure the answer is A??
i tried finidng the answers to these questions online but i cudnt find them.
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Offline astarmathsandphysics

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Re: Edexcel PHYSICS Doubts
« Reply #421 on: January 11, 2011, 03:15:47 pm »
The projectile is fired from a height of 21m
Working vertically, v^2 =u^2 +2as
s=-21
u=0
v=?
a=-9.8
t=?
s=1/2at^2 so t=sqrt(2s/g) =sqrt(2*21/9.7)=2.07s
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Offline astarmathsandphysics

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Re: Edexcel PHYSICS Doubts
« Reply #422 on: January 11, 2011, 03:17:08 pm »
Stokes Law doesnt apply to turbulent flow. The flow is definitely turbulent cos the speed is about 200m/s.
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Offline blackberry

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Re: Edexcel PHYSICS Doubts
« Reply #423 on: January 11, 2011, 04:20:46 pm »
Quote from: Vampire-Love4ever on January 11, 2011, 02:57:32 pm
how to do part c???

consider the 2nd image here, so the vertical distance from which it is launched is 16+4+1=21m
s=ut + .5at^2
21=(0xt) + (.5x9.81xt^2)
t=2.07s
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Offline Vampire-Love4ever

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Re: Edexcel PHYSICS Doubts
« Reply #424 on: January 11, 2011, 04:42:50 pm »
Quote from: blackberry on January 11, 2011, 03:15:34 pm
evn i selected B, vampire are you sure the answer is A??
i tried finidng the answers to these questions online but i cudnt find them.

Yupe, it is A!
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Offline Vampire-Love4ever

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Re: Edexcel PHYSICS Doubts
« Reply #425 on: January 11, 2011, 04:44:29 pm »
PLEASE HELP WITH THE OTHER QUESTIONS I POSTED.


AND OTHERS, THANKS FOR UR HELP =]
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Offline The Golden Girl =D

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Re: Edexcel PHYSICS Doubts
« Reply #426 on: January 11, 2011, 06:16:07 pm »
Quote from: Vampire-Love4ever on January 11, 2011, 02:54:27 pm
Q-A hot air ballon is rising vertically at a speed of 10ms^-1. An object is released from the balloon. Graph shows how the velocity if the object varies with time from when it leaves the balloon to when it reaches the ground 4s later. It's assumed that the air resistance is neglected.

(i) Determine the object's final displacement from it's point of release from the balloon. [2]


Mark scheme answer:

rise 5m and fell 5m [1]

displacement 40m before point of release [1]

i] area/distance = 0.5*b*h
                      = 0.5 * 1*10
                      =  5m

ii] total distance = A1 + A2 = 5 + (0.5*3*30*)
                                      = 5 + 45
                                      = 50 m

b] -45 + 5 = -40 m

hence 40 m below the level of release.
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Offline Vampire-Love4ever

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Re: Edexcel PHYSICS Doubts
« Reply #427 on: January 11, 2011, 07:29:21 pm »
Quote from: Miss Invisible on January 11, 2011, 06:16:07 pm
i] area/distance = 0.5*b*h
                      = 0.5 * 1*10
                      =  5m

ii] total distance = A1 + A2 = 5 + (0.5*3*30*)
                                      = 5 + 45
                                      = 50 m

b] -45 + 5 = -40 m

hence 40 m below the level of release.

Thanks. I didnt get this point: rise 5m and fell 5m [1]

How can it be rising and falling 5m? :S

Plus, please help with this question:

Why do we add the drag force here? Shouldn't we minus it?
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Offline Vampire-Love4ever

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Re: Edexcel PHYSICS Doubts
« Reply #428 on: January 11, 2011, 11:43:32 pm »
Quote from: astarmathsandphysics on January 11, 2011, 03:15:47 pm
The projectile is fired from a height of 21m
Working vertically, v^2 =u^2 +2as
s=-21
u=0
v=?
a=-9.8
t=?
s=1/2at^2 so t=sqrt(2s/g) =sqrt(2*21/9.7)=2.07s

where is s= 21m? =S
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Offline The Golden Girl =D

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Re: Edexcel PHYSICS Doubts
« Reply #429 on: January 12, 2011, 04:40:41 am »
they said in the part before it that it's 20 m  ,but the WHOLE thing is 1 m from the ground hence 20 + 1 = 21 m ;)

drag Q : I think we should minus it *since drag ALWAYS opposes the motion ,that's weird. :-X *

I think.
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Offline blackberry

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Re: Edexcel PHYSICS Doubts
« Reply #430 on: January 12, 2011, 01:48:22 pm »
hey vampire i found out why drag is added instead of substracted. There are 2 ways to do this:

a) F= ma so find out F. Now F here is actually the resultant force (RF) NOT the driving force, since the general formula for finding RF=ma. So, RF= Driving Force - Drag. Rearrange the equation to find driving force: Driving Force= RF + Drag

b) Find F using F= ma. Here we consider F as the Forward Force but NOT the Resultant Driving Force. To find the Resultant Driving  Force use RF= Forward Force - Drag. Notice that Drag is acting in the opposite direction and so drag will be taken as a negative value. In other words RF= Forward Force - ( - Drag) which is equal to RF= Forward Force + Drag.

However, i am not too sure if the 2nd method is acceptable so just stick to the 1st method. But if you get stuck you could use the 2nd method to help you out!!! :) :)
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Re: Edexcel PHYSICS Doubts
« Reply #431 on: January 12, 2011, 01:58:39 pm »
Quote from: Vampire-Love4ever on January 11, 2011, 07:29:21 pm
Thanks. I didnt get this point: rise 5m and fell 5m [1]

How can it be rising and falling 5m? :S

Plus, please help with this question:

Why do we add the drag force here? Shouldn't we minus it?


Using F = ma

Driving force - Drag = mass * acceleration

X - 430 = 520*3.5

X = 520*35 + 430

X = 2250 N
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Offline Vampire-Love4ever

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Re: Edexcel PHYSICS Doubts
« Reply #432 on: January 12, 2011, 04:06:01 pm »
Quote from: blackberry on January 12, 2011, 01:48:22 pm
hey vampire i found out why drag is added instead of substracted. There are 2 ways to do this:

a) F= ma so find out F. Now F here is actually the resultant force (RF) NOT the driving force, since the general formula for finding RF=ma. So, RF= Driving Force - Drag. Rearrange the equation to find driving force: Driving Force= RF + Drag

b) Find F using F= ma. Here we consider F as the Forward Force but NOT the Resultant Driving Force. To find the Resultant Driving  Force use RF= Forward Force - Drag. Notice that Drag is acting in the opposite direction and so drag will be taken as a negative value. In other words RF= Forward Force - ( - Drag) which is equal to RF= Forward Force + Drag.

However, i am not too sure if the 2nd method is acceptable so just stick to the 1st method. But if you get stuck you could use the 2nd method to help you out!!! :) :)

Thanks. Though I exam got over today already. ;)
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Offline blackberry

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Re: Edexcel PHYSICS Doubts
« Reply #433 on: January 13, 2011, 05:23:16 pm »
yeaaa....pfft they always experiment on meh by posing all the tough questions. the paper was reallly bleh!!!
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Offline 3bood

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Re: Edexcel PHYSICS Doubts
« Reply #434 on: January 15, 2011, 06:25:43 pm »
a student connects two pieces of copper wire, A and B, in series with each other and a battery.
the diameter of wire A is twice that of B.  calculate the ratio of the drift velocity in wire A to the drift velocity in wire B and explain your answer.
urgent plz because phy unit 2 is on monday  :( :(
thx in advance   ;D ;D ;D ;D
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