Suggested Solutions

[Apink]

I started with the entire system: mass=8kg, force causing acceleration=2*10N, Friction = 0N.
Next, calculate that the acceleration of the system a=F/m=20N/8kg=2.5ms^-1.
The tension is equal to the net force acting on mass 2, which is F=T=ma=6kg*2.5ms^-1, which is equal to 15N.

Hope that helps.

Thanks!

[Robert123]

Doesn't work function equal KE max - hf

 
I thought it is hf- KE.
This is because the energy of a photon is equal to the sun of kinetic energy and the work function so work function equal the energy of a photon take away KE

[sin0001]

Anyone else get around ~2eV for the work function? o.O

[randomposter]

Anyone else get around ~2eV for the work function? o.O

yeah 2.29 I think.

[lzxnl]

What did you guys put for 'describe the intensity at the centre of the interference pattern'

I said it would be twice as intense as the original laser beam due to constructive interference - it's an antinodal point.
Wasn't sure what they were looking for.

It's actually four times as intense there; the amplitude doubles, but the intensity is proportional to the square of the amplitude (that's not part of this course). But you have the right idea.

[~T]

20.a) An atom is at 3.19eV state? That’s interesting wording…
          f = 3.19/4.14e-15 = 7.71e+14 Hz.  Wavelength = 3e8 / 7.71e+14 = 3.89e-7 m
     b) 588.63nm, so f = 3e8/588.63e-9 = 5.10e+14 Hz. So E = 5.10e+14 x 4.14e-15 = 2.11 eV
         This exactly corresponds to electron falling (emission spectra) from 2.11eV level to ground state
         So, there is a spectral line at 588.63 nm

For 20b, do you think it would matter if I went the other way? I started with the 2.11eV to ground state transition and then showed the wavelength of light emitted by this was 588.63nm. I'm presuming not, but I'm being pedantic about my solutions just in case.

[Professor Polonsky]

     b) Work function  = KE max / hf = 2.96e-19/(6.63e-34 x 1e15) = 0.446

hmmm? Isn't it ?

[AsianNerd]

U scared the shit outa me for first question... So both ways are acceptable?

[silverpixeli]

U scared the shit outa me for first question... So both ways are acceptable?

both methods are technically correct so they should both be accepted

hmmm? Isn't it ?

*

[Professor Polonsky]

Yes, sorry, that. Shouldn't be typing physics at 2am

[~T]

22.d) Particle model only predicts one central light band (one mark)
         Wave model predicts bands formed by interference patterns (one mark)
         Young’s double slit experiment showed constructive and destructive interference so supports
         the wave model (one mark)
        Moto: never trust uncyclopedia (bonus ten marks)

Hang on, doesn't the particle model predict two bright bands, with the light propagating linearly through the two slits?

[Alwin]

Hang on, doesn't the particle model predict two bright bands, with the light propagating linearly through the two slits?

yup, you're 100% correct. now you know why I didn't get any 50s last year... and probs never will

dem silly mistakes *shakes fist at the sky*

[~T]

Thanks for the reference Alwin 
I'm feeling 10B for Synchrotron. I think. I'm fairly sure undulator light is spatially coherent.

[Phy124]

Worked solutions to Materials and their use in structures:

Worked Solutions:

1. The young's modulus is given by the gradient of a stress-strain graph in its linear region



D

2. The force which the cable can support at its breaking point is given by



Therefore the mass is given by



C

3. The strain energy of the sample can be estimated by the area under the graph from the origin to the stretched strain value.



B

4. It is important for a cable in a crane not to undergo plastic deformation, this sample R would be chosen as it deforms elastically for the greatest range of strain values.

D

5. We know that , and

From this we can derive the equation

We have;







and from the linear region of the graph for cable R;





A

6. The torque about point P is given by the force provided by the mass multipled by the perpendicular distance.



C

7. To find the tension in SR we cneed to sum the torques about point P and equate them to zero, as the system is in rotational equilibrium.





C

8. To find the force extered by KL downwards on MN we can sum the torques about point K and equate them to zero, as once again this system is in rotational equilibrium.

The additional forces acting are that of the weight of the beam and the mass at point L, these both cause torques.





D

9. Personally I don't think any of the options would be appropriate placement of the beams, however, option A is the most appropriate of those given as for the mostpart the top of the beam will be in tension.

A

10. It can be noted that the mass at point G will elongate the beam FG, in order to counteract the tension in this beam both EF and FH must be in compression.

B

11. The arch is in compression and the weight of the truck on the road will cause the road to deflect downwards and as such the cables will elongate and will be in tension.

B