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Physics: Post your doubts here!

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Question Paper: https://papers.xtremepape.rs/CAIE/AS and A Level/Physics (9702)/9702_s11_qp_23.pdf

Answer 4c(i)
Since all the energy is converted to kinetic, we have
\(W=\frac{1}{2}mv^{2}\)
\(0.81=\frac{1}{2}mv^{2}\)
\(v=8.05ms^{-1}\)

Answer 4c(ii)
\(\text{speed }v_{2}=8.05\times2=16.1ms^{-1}\)
Therefore, \(W=3.24J\)
Now for compression, we can use
\(W=\frac{1}{2}kx^{2}\)
and with W=3.24J, we get \(x=72mm\)

Answer 4c(iii)
At maximum height. all KE would be converted to GPE, so
\(\frac{\frac{1}{2}k{x_{1}}^{2}}{\frac{1}{2}k{x_{2}}^{2}}=\frac{36^{2}}{72^{2}}=\frac{1}{4}\)

Hope this helps!
thank you so much :D
 

PlanetMaster

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From a velocity-time graph, the acceleration at time t is given by the gradient at time t. This gives the instantaneous acceleration at that time.
In this graph, the gradient and hence the acceleration is changing.

Note that the gradient at time t = 3s is obtained by drawing a tangent at that point and calculating its gradient. We cannot just take the coordinate at that point (3, 6) to obtain acceleration (as done for choice D).

A tangent at time t = 3s would pass almost at/close to points (0, 2) and (6, 10).
\(\text{Gradient}=\frac{(10–2)}{(6 – 0)}=\frac{8}{6}=1.33ms^{-1}\)

P.S. This is taken from an earlier post
 
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From a velocity-time graph, the acceleration at time t is given by the gradient at time t. This gives the instantaneous acceleration at that time.
In this graph, the gradient and hence the acceleration is changing.

Note that the gradient at time t = 3s is obtained by drawing a tangent at that point and calculating its gradient. We cannot just take the coordinate at that point (3, 6) to obtain acceleration (as done for choice D).

A tangent at time t = 3s would pass almost at/close to points (0, 2) and (6, 10).
\(\text{Gradient}=\frac{(10–2)}{(6 – 0)}=\frac{8}{6}=1.33ms^{-1}\)

P.S. This is taken from an earlier post
Thank you
 
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1572538446832.png
so i solved this and got B as the answer but ms says the right answer is 8cm/s. Can anyone please help me with this? :)
 

PlanetMaster

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View attachment 65045
so i solved this and got B as the answer but ms says the right answer is 8cm/s. Can anyone please help me with this? :)
Since the collision is perfectly elastic, we can use
\(v_{1}i+v_{1}f=v_{2}i+v_{2}f\) (You might have a different variant of this formula depending on your board)

Substituting in our values, we get
\(6+2=0+v_{2}f\)

And therefore,
\(v_{2}f=8\)

Hope this helps!
 
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Since the collision is perfectly elastic, we can use
\(v_{1}i+v_{1}f=v_{2}i+v_{2}f\) (You might have a different variant of this formula depending on your board)

Substituting in our values, we get
\(6+2=0+v_{2}f\)

And therefore,
\(v_{2}f=8\)

Hope this helps!
i thought the formua was mu1+mu2=mv1+mv2? and if the directions were opposite do we have to change the signs?
 
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1572583042619.png
Is the 6.3 s for the whole graph or just beween the pulses? i attempted this and got 2.1. The answer should actually be C. Im sorry for asking a lot xD My cie is within two weeks so help would be appreciated :D
 

PlanetMaster

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i thought the formua was mu1+mu2=mv1+mv2? and if the directions were opposite do we have to change the signs?
The formula I used earlier is derived from this under special conditions for elastic collisions.
Since velocity is a vector quantity, the signs need to be changed with directions.

View attachment 65046
Is the 6.3 s for the whole graph or just beween the pulses? i attempted this and got 2.1. The answer should actually be C.
The pulse sent by transmitter is reflected and returns to the transmitter so it takes it \(6.3\mu s\times2\) to travel from the transmitter to the reflector and back to transmitter. So we'll need a time-base setting of \(12.6\mu s/3cm=4.2\mu s\ cm^{-1}\).

Im sorry for asking a lot xD My cie is within two weeks so help would be appreciated :D
You can never ask too many questions. Keep em coming.. (terms and conditions apply! 🤪 )
 
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How to solve this?

A moving body undergoes uniform acceleration while travelling in a straight line between points X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and from Y to Z is 6.0 s.

What is the acceleration of the body?

A) \(0.37 m s^{–2}\)
B) \(0.49 m s^{–2}\)
C) \(0.56 m s^{–2}\)
D) \(1.1 m s^{–2}\)
 
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PlanetMaster

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How to solve this?

A moving body undergoes uniform acceleration while travelling in a straight line between points X, Y and Z. The distances XY and YZ are both 40 m. The time to travel from X to Y is 12 s and from Y to Z is 6.0 s.

What is the acceleration of the body?

A) \(0.37 m s^{–2}\)
B) \(0.49 m s^{–2}\)
C) \(0.56 m s^{–2}\)
D) \(1.1 m s^{–2}\)
There are two unknowns here, starting speed \(u\) and acceleration \(a\). Using \(s=ut+\frac{1}{2}at^2\), we’re given that in 12 seconds from X to Y:
\(\qquad 40=12u+\frac{1}{2}12^2a=12u+72a\)
and in 18 seconds from X to Z:
\(\qquad 80=18u+\frac{1}{2}18^2a=18u+162a\)
Multiplying the first equation by \(\frac{3}{2}\) we get:
\(\qquad 60=18u+108a\)
and subtracting this from the second equation gives:
\(\qquad 20=54a\)
Therefore the answer is that the acceleration \(a=\frac{10}{27}\) m/s\(^2\). We’re not asked about \(u\) so we needn’t calculate it.
 
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There are two unknowns here, starting speed \(u\) and acceleration \(a\). Using \(s=ut+\frac{1}{2}at^2\), we’re given that in 12 seconds from X to Y:
\(\qquad 40=12u+\frac{1}{2}12^2a=12u+72a\)
and in 18 seconds from X to Z:
\(\qquad 80=18u+\frac{1}{2}18^2a=18u+162a\)
Multiplying the first equation by \(\frac{3}{2}\) we get:
\(\qquad 60=18u+108a\)
and subtracting this from the second equation gives:
\(\qquad 20=54a\)
Therefore the answer is that the acceleration \(a=\frac{10}{27}\) m/s\(^2\). We’re not asked about \(u\) so we needn’t calculate it.
Thankyou :)
 
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Can anyone explain this?


I know it's quite late, but I hope the explanation helps nonetheless.
The spring is moving vertically up and down, with equilibrium point being in the centre. At "lowest point of its motion" the mass has MAXIMUM displacement downwards. And at maximum displacement, we know that potential energy is maximum and kinetic energy is minimum(=0) therefore velocity=0 at lowest point and answer is D :)
 
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Is there any whatsapp group for this group and if yes can someone kindly add me over there
 
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