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Need help with physics exam questions

I'm on the AQA January 2013 Unit 4 paper, I've not had much trouble with most of the question but hoping someone can help with these three which I couldn't do...

9. Two pendulums, P and Q, are set up alongside each other. The period of P is 1.90 s andthe period of Q is 1.95 s. How many oscillations are made by pendulum Q between two consecutive instantswhen P and Q move in phase with each other?
A 19
B 38
C 39
D 78

12. The gravitational field strength at the surface of the Earth is 6 times its value at thesurface of the Moon. The mean density of the Moon is 0.6 times the mean density ofthe Earth. What is the value of the ratio (radius of the earth) / (radius of the moon)?
A 1.8
B 3.6
C 6.0
D 10

13. The diagram shows two points, P and Q, at distances r and 2r from the centre of aplanet. The gravitational potential at P is –16 kJ kg–1. What is the work done on a 10 kg masswhen it is taken from P to Q?
A 120 kJ
B 80 kJ
C + 80 kJ
D + 120 kJ

I know question 9 has something to do with phase difference but I don't know what the equation would be to find the answer...For question 12 would you equate GM/r^2 for the earth and 6GM/r^2 for the moon, and then find M for both using density = m/V? As for question 13 I have no idea how to work that out...

If anyone could point me in the right direction for each question I'd appreciate it
:biggrin:
(edited 9 years ago)
Original post by jf1994
I'm on the AQA January 2013 Unit 4 paper, I've not had much trouble with most of the question but hoping someone can help with these three which I couldn't do...

9. Two pendulums, P and Q, are set up alongside each other. The period of P is 1.90 s andthe period of Q is 1.95 s. How many oscillations are made by pendulum Q between two consecutive instantswhen P and Q move in phase with each other?
A 19
B 38
C 39
D 78

12. The gravitational field strength at the surface of the Earth is 6 times its value at thesurface of the Moon. The mean density of the Moon is 0.6 times the mean density ofthe Earth. What is the value of the ratio ?
A 1.8
B 3.6
C 6.0
D 10

13. The diagram shows two points, P and Q, at distances r and 2r from the centre of aplanet. The gravitational potential at P is –16 kJ kg–1. What is the work done on a 10 kg masswhen it is taken from P to Q?
A 120 kJ
B 80 kJ
C + 80 kJ
D + 120 kJ

I know question 9 has something to do with phase difference but I don't know what the equation would be to find the answer...For question 12 would you equate GM/r^2 for the earth and 6GM/r^2 for the moon, and then find M for both using density = m/V? As for question 13 I have no idea how to work that out...

If anyone could point me in the right direction for each question I'd appreciate it
:biggrin:


9) So, at the start both pendulums are in phase and the time for Q to perform one oscillation is 1.95s. After two oscillations, 3.9s will have elapsed. After three oscillations...

Apply this same principle for pendulum P. Find the time elapsed when both pendulums P and Q have made a complete number of oscillations.

12) Not sure what this is asking about, possibly the ratio of the gravitational field strength to density?

gEarth=6gMoong_{Earth} = 6g_{Moon}

0.6ρEarth=ρMoon0.6 \rho_{Earth} = \rho_{Moon}

13) What is the value of gravitational potential at a distance of \infty from the centre of mass of the planet? So, should the value of gravitational potential increase or decrease? How does gravitational potential vary with distance?
(edited 9 years ago)
Reply 2
Original post by pleasedtobeatyou
9) So, at the start both pendulums are in phase and the time for Q to perform one oscillation is 1.95s. After two oscillations, 3.9s will have elapsed. After three oscillations...

Apply this same principle for pendulum P. Find the time elapsed when both pendulums P and Q have made a complete number of oscillations.

12) Not sure what this is asking about, possibly the ratio of the gravitational field strength to density?

gEarth=6gMoong_{Earth} = 6g_{Moon}

0.6ρEarth=ρMoon0.6 \rho_{Earth} = \rho_{Moon}

13) What is the value of gravitational potential at a distance of \infty from the centre of mass of the planet? So, should the value of gravitational potential increase or decrease? How does gravitational potential vary with distance?


Thanks, I get the pendulum question now :smile:

For 12, it cut out the fraction. I've updated the OP to include it in.

For 13, isn't the gravitational potential at a distance of infinity from the centre of mass of a planet 0? So surely to go from P to Q, away from the planet, gravitational potential would increase towards 0?

e/ Just remembered it asks for work done to get it to Q, but since gravitational field strength decreases further away from the planet, how do you know how much work is done?
(edited 9 years ago)
Original post by jf1994
Thanks, I get the pendulum question now :smile:

For 12, it cut out the fraction. I've updated the OP to include it in.

For 13, isn't the gravitational potential at a distance of infinity from the centre of mass of a planet 0? So surely to go from P to Q, away from the planet, gravitational potential would increase towards 0?

e/ Just remembered it asks for work done to get it to Q, but since gravitational field strength decreases further away from the planet, how do you know how much work is done?



12) Use the equations that you know for gravitational field strength. M\bf{M} will be the density multiplied by the volume of a sphere. Do this for both the Earth and the Moon and then divide one by another to try and form a ratio.

The gravitational field strengths and densities are given as a scalar multiple of each other, so you can very easily cancel them out.

13) isn't the gravitational potential at a distance of infinity from the centre of mass of a planet 0? <-- Yes

Start from a simpler case. Say you have a horizontal plane and a mass resting on it. In this case, all the potential lines are parallel to the horizontal.

How much goes the gravitational potential change as you raise it a certain distance to another potential line? Thus, how much work is done?
Reply 4
Original post by pleasedtobeatyou
12) Use the equations that you know for gravitational field strength. M\bf{M} will be the density multiplied by the volume of a sphere. Do this for both the Earth and the Moon and then divide one by another to try and form a ratio.

The gravitational field strengths and densities are given as a scalar multiple of each other, so you can very easily cancel them out.

13) isn't the gravitational potential at a distance of infinity from the centre of mass of a planet 0? <-- Yes

Start from a simpler case. Say you have a horizontal plane and a mass resting on it. In this case, all the potential lines are parallel to the horizontal.

How much goes the gravitational potential change as you raise it a certain distance to another potential line? Thus, how much work is done?


But it doesn't say the lines are equipotentials and the field would not be uniform so the gravitational potential gradient would be different from the surface to r and r to 2r right? How do I relate the potential at r to the potential at 2r? And how do I know what the potential at the surface is?

Sorry if I'm being dim :frown:
Original post by jf1994
But it doesn't say the lines are equipotentials and the field would not be uniform so the gravitational potential gradient would be different from the surface to r and r to 2r right? How do I relate the potential at r to the potential at 2r? And how do I know what the potential at the surface is?

Sorry if I'm being dim :frown:


It should be fine to assume that the locus of all the points r away from the centre of mass form an equipotential line.

You don't need to need the potential at the surface. All you need to know if the potential at r and 2r away from the center of mass.

You should know that equation for gravitational potential. Use this to work out the value of the gravitational potential at the r and 2r away.

The gravitational potential is defined per unit mass. Hence, use this to determine to work done when moving the 10kg mass from P to Q.
Reply 6
Original post by pleasedtobeatyou
It should be fine to assume that the locus of all the points r away from the centre of mass form an equipotential line.

You don't need to need the potential at the surface. All you need to know if the potential at r and 2r away from the center of mass.

You should know that equation for gravitational potential. Use this to work out the value of the gravitational potential at the r and 2r away.

The gravitational potential is defined per unit mass. Hence, use this to determine to work done when moving the 10kg mass from P to Q.


But if the potential at the surface is not given, how do you work out the potential gradient between the surface and P?

I know that the gravitational potential energy at R would be -160kJ, and I know the answer is W = 80kJ so the potential energy at Q must be -80kJ...does this mean you have to assume that if distance is doubled, V is halved? Which makes perfect sense now I think about it, since V = -GM/r, but I thought that the potential gradient declined the further away you traveled from the center of mass...
Original post by jf1994
But if the potential at the surface is not given, how do you work out the potential gradient between the surface and P?

I know that the gravitational potential energy at R would be -160kJ, and I know the answer is W = 80kJ so the potential energy at Q must be -80kJ...does this mean you have to assume that if distance is doubled, V is halved? Which makes perfect sense now I think about it, since V = -GM/r, but I thought that the potential gradient declined the further away you traveled from the center of mass...


Yes and what should and does happen is consistent with this. You don't need to know what the potential is at the surface because the potential is defined from the centre of mass - which is not the surface,

You know that: VP=GMr=16[kJkg] V_{P} = - \frac{GM}{r} = -16 \left [ \frac{kJ}{kg} \right ] and VQ=GM2r= ???[kJkg]V_{Q} = - \frac{GM}{2r} = \ ??? \left [\frac{kJ}{kg} \right ].

So, you can work out Δ(VPQ)\Delta(V_{PQ}). Therefore, you know the change in potential from point P to Q per unit mass.

It should be easy to get work done from here.
(edited 9 years ago)
Reply 8
Original post by pleasedtobeatyou
Yes and what should and does happen is consistent with this. You don't need to know what the potential is at the surface because the potential is defined from the centre of mass - which is not the surface,

You know that: VP=GMr=16[kJkg] V_{P} = - \frac{GM}{r} = -16 \left [ \frac{kJ}{kg} \right ] and VQ=GM2r= ???[kJkg]V_{Q} = - \frac{GM}{2r} = \ ??? \left [\frac{kJ}{kg} \right ].

So, you can work out Δ(VPQ)\Delta(V_{PQ}). Therefore, you know the change in potential from point P to Q per unit mass.

It should be easy to get work done from here.


Ah I think I understand, so is it basically just the fact that V will be halved if r doubles? So the potential at Q would be half of at P, and then multiplied by 10kg for the work done gives 80J. I can't believe I missed that, I was approaching the question conpletely wrong!

Thanks for your help!

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