Physics: Post your doubts here!

[PlanetMaster]

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!

 

[fdshfdahda]

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?

 

[fdshfdahda]

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

 

[Kar345]

https://papers.xtremepape.rs/CAIE/AS and A Level/Physics (9702)/9702_s19_qp_52.pdf

Can someone help me calculate the uncertainties in the table for number 2?? I can't get the correct answer given in the mark scheme. Please help

 

[PlanetMaster]

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

You can never ask too many questions. Keep em coming.. (terms and conditions apply! 🤪 )

 

[Rid@Ahmed]

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) \(1.1 m s^{–2}\)

 

[PlanetMaster]

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.

 

[Rid@Ahmed]

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

 

[Rid@Ahmed]

How to sketch this velocity-time graph?

 

[Rid@Ahmed]

Can anyone explain this?

 

[Mushraf altaf]

How to sketch this velocity-time graph?

some help with kinematics graphs
https://notesnsolutions.com/cie-a-level/cie-a-level-physics/cie-a-level-kinematics-as/

 

[KashishV]

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

 

[muhammad afzal1970]

Any one can help in physics practical paper 32 may/ June 19
i dont get the value of x in question 1

 

[A*(a*)]

Is there any whatsapp group for this group and if yes can someone kindly add me over there

 

[A*(a*)]

Should I tell my number ?

 

[Rid@Ahmed]

How to solve part b?

 

[KashishV]

How to solve part b?

Hi, firstly this is a physics forum and your question is chemistry-related. Please be mindful henceforth
As to your question, it's quite a memory based answer. You need to know that MgCl2 dissolves in water to give Mg(OH)2+HCl; Since one product is basic and the other is acidic, your best guess is pH 7.
in case of MgO, you need to know that it hardly dissolves/is slightly soluble. It is more of a surface reaction that occurs in which Mg(OH)2 is produced. As a result, you predict a high pH. Dont go till 14, Mg(OH)2 is not as strong of an alkali.

 

[asra_070]

pls help with this

 

[Rid@Ahmed]

Hi, firstly this is a physics forum and your question is chemistry-related. Please be mindful henceforth
As to your question, it's quite a memory based answer. You need to know that MgCl2 dissolves in water to give Mg(OH)2+HCl; Since one product is basic and the other is acidic, your best guess is pH 7.
in case of MgO, you need to know that it hardly dissolves/is slightly soluble. It is more of a surface reaction that occurs in which Mg(OH)2 is produced. As a result, you predict a high pH. Dont go till 14, Mg(OH)2 is not as strong of an alkali.

Sorry. I didn't notice at that time and thanks.

 

[dasbutt]

how will the diagram be drawn for this?