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CIE AS Physics questions

Need help with these questions
Reply 1
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NotYourType
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Would it be possible for you to explain the 10th question.I know its done on the book,but I can't quite seem to figure it out.
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You work out the time taken to get back to the same level vertically and then use this to work out the horizontal distance moved. You use the fact that time is the same in each dimension.

Vertical:

u=20sin⁡30=10ms−1u = 20 \sin 30 = 10ms^{-1}
s=0ms = 0m
a=−9.8ms−2a = -9.8ms^{-2}


s=ut+12at2s = ut + \frac{1}{2}at^2
0=10t−4.9t2 0 = 10t - 4.9t^2
0=t(10−4.9t) 0 = t(10 - 4.9t)
∴t=0s,2.04s\therefore t = 0s, 2.04s (3sf)
∴t=2.04s\therefore t = 2.04s (3sf)


Horizontal:

v=u=20cos⁡30=17.3ms−1v = u = 20\cos 30 = 17.3ms^{-1} (3sf)
a=oms−2a = o ms^{-2}
t=2.04st = 2.04s


s=vts = vt
s=(17.3)(2.04)s = (17.3)(2.04)
s=35.3ms = 35.3m (3sf)

Edit: I'll do 9 too if you want. Tell me if there's anything you're unsure on or want me to do 9.
(edited 9 years ago)
Reply 3
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NotYourType
OP
2
Original post by morgan8002
You work out the time taken to get back to the same level vertically and then use this to work out the horizontal distance moved. You use the fact that time is the same in each dimension.

Vertical:

u=20sin⁡30=10ms−1u = 20 \sin 30 = 10ms^{-1}
s=0ms = 0m
a=−9.8ms−2a = -9.8ms^{-2}


s=ut+12at2s = ut + \frac{1}{2}at^2
0=10t−4.9t2 0 = 10t - 4.9t^2
0=t(10−4.9t) 0 = t(10 - 4.9t)
∴t=0s,2.04s\therefore t = 0s, 2.04s (3sf)
∴t=2.04s\therefore t = 2.04s (3sf)


Horizontal:

v=u=20cos⁡30=17.3ms−1v = u = 20\cos 30 = 17.3ms^{-1} (3sf)
a=oms−2a = o ms^{-2}
t=2.04st = 2.04s


s=vts = vt
s=(17.3)(2.04)s = (17.3)(2.04)
s=35.3ms = 35.3m (3sf)

Edit: I'll do 9 too if you want. Tell me if there's anything you're unsure on or want me to do 9.


I'm okay with the 9th one, had it explained already :tongue: Thank you for asking though.
Anyway , I was wondering why a(gravity) would be -9.81 and not positive? Since, the ball goes up , it defies gravity , so it a= - , but then when it comes back down ,a= + , so won't the + and - charge nullify each other , leaving it with a positive charge?
Original post by NotYourType
I'm okay with the 9th one, had it explained already :tongue: Thank you for asking though.
Anyway , I was wondering why a(gravity) would be -9.81 and not positive? Since, the ball goes up , it defies gravity , so it a= - , but then when it comes back down ,a= + , so won't the + and - charge nullify each other , leaving it with a positive charge?


Always keep positive vectors in the same direction. You can choose arbitrarily at the start of the question which direction is positive, but you have to keep this constant throughout the question.

I chose positive to be up.

The weight (and hence gravitational acceleration) is acting downwards (away from up), so the acceleration is negative. As the positive direction never changes, the acceleration is always negative.

Acceleration is the rate of change of velocity, not the rate of change of speed.
When the object goes up, its speed decreases and velocity decreases, hence its acceleration is negative.
When it goes down, its speed increases but its velocity still decreases, so its acceleration is still negative.
(edited 9 years ago)
Reply 5
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NotYourType
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Original post by morgan8002
Always keep positive vectors in the same direction. You can choose arbitrarily at the start of the question which direction is positive, but you have to keep this constant throughout the question.

I chose positive to be up.

The weight (and hence gravitational acceleration) is acting downwards (away from up), so the acceleration is negative. As the positive direction never changes, the acceleration is always negative.

Acceleration is the rate of change of velocity, not the rate of change of speed.
When the object goes up, its speed decreases and velocity decreases, hence its acceleration is negative.
When it goes down, its speed increases but its velocity still decreases, so its acceleration is still negative.


Sorry for being a pain , but how would we know which vector to keep as positive?and in the case of weight acting downwards , isn't that the case for all objects , then why does gravity remain positive for other objects , whilst in this case it is negative? I know you must really be getting frustrated with me ..Sorry
Original post by NotYourType
Sorry for being a pain , but how would we know which vector to keep as positive?and in the case of weight acting downwards , isn't that the case for all objects , then why does gravity remain positive for other objects , whilst in this case it is negative? I know you must really be getting frustrated with me ..Sorry



Just talking about vertical components here.
In my answer to question 10, I chose upwards to be positive.
Because initial velocity was upwards, it is positive.
Because acceleration due to gravity acts downwards, it is negative.
Because the velocity decreases, eventually (at the top of the parabola) it will be zero, then become negative. When it is negative, velocity is now downwards.

I could just have easily chosen downwards to be the positive direction, which would have made the initial velocity negative, the acceleration positive and the final velocity positive.
Question 9 takes down to be positive, so acceleration due to gravity is positive. In this case the displacement is also positive because the object falls down during the movement.

No problem.
(edited 9 years ago)
Reply 7
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NotYourType
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Omg , I totally understood that :smile:
Thank you so much !
Honestly wish , you were my physics teacher !
Reply 8
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NotYourType
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Uhmm..hey , sorry for annoying you again , but need help with these questions
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Original post by NotYourType
Posted from TSR Mobile
Uhmm..hey , sorry for annoying you again , but need help with these questions

For the first question:
The wire is 800x2 = 1600m. (goes there and back)
Let R be the resistance of the wire. R=1600(0.005)=8ΩR = 1600(0.005) = 8\Omega

Therefore the voltage lost to heating the wires is given by v = IR. The current is constant around the circuit (I = 0.60A).
v = (0.60)(8)
v = 4.8V


The emf has to supply the relay with 16.0V and also has to supply the voltage that heats up the wires.
∴v=4.8V+16.0V=20.8V\therefore v = 4.8V + 16.0V = 20.8V


For the second question:
v2=u2+2asv^2 = u^2 + 2as

v=0  ⟹  0=u2+2asv = 0 \implies 0 = u^2 +2as

u2=−2asu^2 = -2as

s=−u22as = -\frac{u^2}{2a}

Increase by 20% means that u is 1.2 times greater.
Because s∝u2s \propto u^2, s is 1.22=1.441.2^2 = 1.44 times greater.
Posted from TSR Mobile

Thanks again for your help , you must really find me annoying..But it's good to know that there are people out there who are ready to help you :smile:
Posted from TSR Mobile
Another help , if you don't mind.
For this question , I'd like to know , how we'd know which angle we have to calculate ..I mean why can't it be the angle I've marked with the pencil ?
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Original post by NotYourType
Posted from TSR Mobile
Another help , if you don't mind.
For this question , I'd like to know , how we'd know which angle we have to calculate ..I mean why can't it be the angle I've marked with the pencil ?




It has to be the angle between the base of the velocity and another vector. The other vector is usually either the horizontal or the vertical axis.

As shown in the diagram, in some questions you could answer with 90−θ90-\theta, taking the angle with the horizontal. It's always a good idea to say whether you are taking the angle from the horizontal or vertical (eg, say θ=14o\theta = 14^o right of the vertical.)

In the case of questions where you have compass directions (North, East, etc.), it's standard to always take the angle with the vertical rather than the horizontal, but in most questions you can decide which angle to take.

Can't change the size of the picture for some reason.
(edited 9 years ago)
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Posted from TSR Mobile
Oh ok. Thank you :smile:
Posted from TSR Mobile
Must be really annoying with me asking questions,but please help me out if you could...
The 5th one from the book
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Original post by NotYourType
Posted from TSR Mobile
Must be really annoying with me asking questions,but please help me out if you could...
The 5th one from the book

first:
R=ρlAR = \frac{\rho l}{A}
R=ρlπr2R = \frac{\rho l}{\pi r^2}
∴R∝l\therefore R \propto l and R∝1r2R \propto \frac{1}{r^2}
From this you can use the scale factors to work it out.


second:
From the triangle, θ=90−45=45o\theta = 90-45=45^o

Then you work out the plane components of the non-driving forces.

Then use E = Fx to calculate the work done.
(edited 9 years ago)

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