OCR A Physics AS 15th May 2018 Unofficial Markscheme

alright, here’s the beginning of the unofficial markscheme for this morning's breadth paper! comment below all of your answers and i’ll add them to this first post.

MULTIPLE CHOICE

Impossible kinetic energy of electron - 3eV
Rate of change of momentum - $100kgms^{-3}$
Regular change of momentum - $30kgms^{-2}$
Drag forces diagram - B (line going to bottom left, mostly downwards)
Rod forces diagram - B (line going top right diagonal)
Correct definition without units - Electromotive force
Coherent waves constant - Phase difference
Greatest velocity from graph - B (steepest gradient)

SHORT ANSWER

Trolley slope angle - 12°
Trolley acceleration - $2.1ms^{-2}$

Spring energy - 0.24J

Waves in water vs air - frequency same, speed lower, wavelength lower
Upthrust acceleration - $42ms^{-2}$

Oscilloscope reading -
Drawing wave - in anti phase with example, amplitude is half of example, same frequency/wavelength
Intensity at interference point - there is destructive interference so the amplitude is lower, and as intensity is directly proportional to amplitude squared intensity is lower
Interference at Q - destructive as path difference is half wavelength

Component from V and I values - NTC thermistor

Voltage through component X (unknown current) - 4V

Current/resistivity question - 0.018A

Kinetic energy of photoelectron - $4.0\times10^{-19}J$

For calculating the final velocity, it was simply measuring the length of the ramp, start the timer when released, stop when it is at the end of the ramp, repeat multiple times, calculate v via (distance/time).

The acceleration one, was 2.1 ms^-2

I got something like 0.20J (0<answer<1) for the energy in the spring

For the wave one, part i was drawing the wave. It had to be in anti phase, with the lower amplitude.
Then in ii (I think), the waves are in anti phase so the amplitude is decreased, and therefore the intensity is reduced (as intensity is directly proportional to the square of the amplitude). You also had to work out the path difference was 1/2(wavelength), which meant destructive interference.

For determine the frequency from the oscilloscope, count the number of squares for the time period, multiply that by the time base, then do 1/T to get frequency. I drew a diagram also.

For the diffraction in water one, the frequency is the same in both media, however the wavelength decreases in water, hence the slower wave speed, and it changes direction as it changes speed, maybe some other stuff.

For the glass semicircle one, use a ray box and shine the light into the curved edge, measure the angle of incidence to the normal, and the angle of refraction. Repeat with different angles, until the angle of refraction, is 90 degrees. Calculate mean angle of incidence. Calculate the refractive index, via (1 / sin(C) ).

Then the diffraction one, was that the wavelength was larger than the gap size, so it could successfully diffract.

MCQ:

For the first one (greatest velocity), it was B
For the correct definition one, it was electromotive force equals energy per unit charge (D)
For the change in momentum one, I think it was 2.5 (so B), could be wrong though

For calculating the final velocity in your first answer, using v=d/t would give you the average velocity as the trolley travels down the ramp, not the velocity at the end of the ramp. I think using SUVAT would give you the final velocity. Something like s = (1/2)(u+v)t should work.

For the diffraction one, I always thought diffraction was most pronounced when the wavelength was similar to the gap size not larger. I'm not very confident on that topic though so I'll have to look that one up.

For calculating the final velocity in your first answer, using v=d/t would give you the average velocity as the trolley travels down the ramp since the trolley is accelerating at a uniform rate, not the velocity at the end of the ramp. I think using SUVAT would give you the final velocity. Something like s = (1/2)(u+v)t should work.

For the diffraction one, I always thought diffraction was most pronounced when the wavelength was similar to the gap size not larger. I'm not very confident on that topic though so I'll have to look that one up.

I'm not sure I remember the wording of the change in momentum question. I thought it asked for the final momentum so I added the change in momentum to the initial momentum and got 6.5.

What did you guys get for that MCQ on the wavelength the one with 60 cm

i got 80, you?

I have to disagree with the rod attached to wall one. I don't think the line goes diagonally I think it goes straight through the rod. This is because the forces must all cancel out to be in equilibrium which will be where the weight force, the tension in the rope and the force of the wall on the rod meet. Since the rope starts from the centre of the rod, the point of intersection is directly in the centre of the rod.

In other questions, the rope usually starts at the end and the forces cross diagonally like you said but I think this one was a trick question.

What was the depth in physics paper like?