AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

The Electric field strength should have been negative as the charge was negative. -1560 Vm^-1.

I got this wrong.

Also any clues on the damping question? I wrote there would be more damping without the ring as the total energy given to the system with the ring will be greater as GPE=mgh. The work done against air resistance is the same as time period is not affected - because the pendulum system has more energy with the ring it will take longer for the initial energy to be lost through doing work against air resistance. Thus the amplitude will decrease at a slower rate.

Did people get 1.4 for the height ?


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Surely the ring has no effect on damping... It was placed in such a way that no extra surface area was provided to the cone.. so no effect on air resistance..
time period is no different, so velocity is no difference and so again air resistance does not change..

Yep

Yes, and KE is proportional to v^2 since the factor of 0.5m is constant since both are electrons and therefore have the same mass m.

Kinetic energy of x is twice that of y, hence $v_x^2=0.5v_y^2$ by what I just said. Hence $\frac{v_x}{v_y}=\sqrt{0.5}$.

Now, the field exerts a force of F=BQv on a charge Q moving at right angles to the field with velocity v. BQ is the same for both electrons for obvious reasons, so F is proportional to v. Therefore, $\frac{F_x}{F_y}=\frac{v_x}{v_y}=\sqrt{0.5}$ which is the same as $\frac{1}{\sqrt{2}}$. (option B I think)

For Coulombs law, would I get any marks for quoting the formula and saying what each symbol meant?

Yes, and KE is proportional to v^2 since the factor of 0.5m is constant since both are electrons and therefore have the same mass m.

Kinetic energy of x is twice that of y, hence $v_x^2=0.5v_y^2$ by what I just said. Hence $\frac{v_x}{v_y}=\sqrt{0.5}$.

Now, the field exerts a force of F=BQv on a charge Q moving at right angles to the field with velocity v. BQ is the same for both electrons for obvious reasons, so F is proportional to v. Therefore, $\frac{F_x}{F_y}=\frac{v_x}{v_y}=\sqrt{0.5}$ which is the same as $\frac{1}{\sqrt{2}}$. (option B I think)

No, force = rate of change momentum look up newton's 2nd law

if you don't believe newton Force = Mass x Acceleration = Kg x Ms-2 = N!!!!!!

by saying unit of force = unit of acceleration you're saying that Mass has no unit, which is totally wrong

But the unit of rate of change of momentum is Newtons as it equals Force. So how can it be acceleration. I do not understand

yes



I said what you did but I realise its wrong

light damping has no effect on the time period because Amplitude is not related to time period, but with damping the amplitude generally increases

heavier damping should be with the ring on i think

Don't you get 4mJ from the battery and then minus 2mj to get the energy used up which is 2. So B couldn't havw been incorrect.

yes



I said what you did but I realise its wrong

light damping has no effect on the time period because Amplitude is not related to time period, but with damping the amplitude generally increases

heavier damping should be with the ring on i think

Same, and I know two others who got this

So the answer should have been weight?

First question 10000% was weight, by laws of dimensional analysis 😋

So, everyone agrees A was right (as in correct so this was not the answer) Between BCE. if C was wrong then so was D and vice versa (there were only two components not including the cell) So B was left

For the coil with a power supply and ammeter, which way should the needle have initially gone? And which way should it have gone in the next part?
I explained using lens's law and faradays law, did anyone else do this?