On part a3, why do you treat the acceleration like simple harmonic motion instead of using F = ma and m = 1 to say F = a, which would make the acceleration graph the same as the force graph?
(I suppose the graph for velocity would then be the same as kinetic energy. But, a = v(dv/dx) if I understand M3 kinematics, so integral of a dx would be 0.5v^2 — which is the same as the energy graph, since KE = 0.5mv^2 and m = 1, so KE =0.5v^2 = a … I don’t know if that makes sense 😦 )
Thank you!

Your reasoning in the first paragraph is correct, but the graph that the question shows is force v distance, and the graph they are asking you to sketch is force v time. So you need to go from the differential equation d²x/dt² = -x to be able to find dx/dt and d²x/dt² in terms of t. That’s when it helps to spot that this is just SHM.

How rigorous do they expect answers to be in general? I didn’t notice the SHM property of this question and consequently got the shock of my life when I saw this..

I wouldn’t worry too much about this question — it is by far the hardest question that they’ve ever asked. It took me a couple of hours to get the answer out properly. If you look at the examiners’ report for that year you’ll notice that nobody got more than 83% on the paper as a whole — my guess is that nobody got more than about 10/20 on this question.

Having said that, though, the more rigorous your answers the better. The one thing the examiners are looking for is that you can turn vague words into accurate mathematical content (e.g. equations, diagrams), and then do something useful with it. Every line of maths that you can get down (as long as it’s right) is worth something.

Hi please could you explain your first line of working for a1. The paticular bit where you say Ek = fFdx = 100 I dont understand that line but the other lines i do

The graph is really two graphs, one for x > 0 and one for x < 0, with a discontinuity at x = 0. Ignore the x > 0 part and you can see that it does look like SHM when x < 0. Specifically, when x < 0, the equation of the graph is F = -(x + 10), which looks like a mass/spring equation.

Try working out the graph of force against height for a 1kg mass on a spring that has spring constant 1N/m, given these conditions: the acceleration due to gravity g is 10m/s/s; when the height is positive the spring is slack; when the height is zero the spring is just about to start extending; and the more negative the height the more extended the spring gets. You should find that you get the same graph that you see in the question.

I’m really sorry to be a pain, but I cannot see how you generated the equation relating xi+1 and xi, and also where the 9/11 has come from. I presume it relates to the force acting?

When the mass is stationary at xi, the PE is 10xi, and the KE is zero.

The mass then moves a distance xi (with friction acting) and enters the x<0 zone. Some time later is comes bouncing back from the x<0 zone, travels a further distance xi+1 (with friction acting), and ends up stationary again at xi+1.

When the mass is stationary again at xi+1, the PE is 10xi+1 and the KE is zero.

In moving from the stationary point xi to the stationary point xi+1, the mass did work in overcoming friction by travelling xi (from the first stationary point to the x=0 point), and then travelling xi+1 (from the x=0 point to the second stationary point). So, by conservation of energy:

10 xi – xi – x(i+1) = 10 x(i+1), which means that 9xi = 11xi+1, so xi+1 = (9/11) xi

OK, I’ll describe each of the bullet points in turn:

When x<0 there is no friction, so the mass doesn’t lose any energy. So if it crosses the x=0 boundary on the way to the x<0 zone with velocity -v, it will eventually come bouncing back across the x=0 boundary with velocity +v. And also it is worth noting that this means the section of the graph for x<0 is symmetric about the x axis.

In the x>0 zone, the mass is stationary at the point of maximum x, it then accelerates back to the x0 zone, where is decelerates to a new point of maximum x. But because the mass loses energy to friction each maximum x point has a smaller value of x than before. We write the initial maximum as x0, and sebsequent ones as x1, x2, … xi

When the mass is at +x it has PE of 10x, and when it moves against friction a distance x it loses energy x. So we can generate an equation relating xi+1 to xi, and when we plug in the numbers above we get xi+1 = (9/11) xi

After a long time the maximum x value tends to zero, so ultimately we just end up with the mass bouncing up and down on the string endlessly.

Having considered the KE it should be easy to work out what this implies for v, which is proportional to the square root of KE. Just be careful to the the sign of v right — it starts negative.

Finally in part b2, you should be able to see that the mass only disspates energy when x>=0 and it does this by moving against a frictional force of 1N. It starts with energy of 100J, which all has to be dissipated, so it must move 100m in the x>=0 zone.

Just for completeness, I have also summed up the infinite series to demonstrate that you get the same result. But this isn’t necessary in order to answer the question.

On part a3, why do you treat the acceleration like simple harmonic motion instead of using F = ma and m = 1 to say F = a, which would make the acceleration graph the same as the force graph?

(I suppose the graph for velocity would then be the same as kinetic energy. But, a = v(dv/dx) if I understand M3 kinematics, so integral of a dx would be 0.5v^2 — which is the same as the energy graph, since KE = 0.5mv^2 and m = 1, so KE =0.5v^2 = a … I don’t know if that makes sense 😦 )

Thank you!

Your reasoning in the first paragraph is correct, but the graph that the question shows is force v distance, and the graph they are asking you to sketch is force v time. So you need to go from the differential equation

d²x/dt² = -xto be able to finddx/dtandd²x/dt²in terms oft. That’s when it helps to spot that this is just SHM.Ohh, thank you! 🙂

How rigorous do they expect answers to be in general? I didn’t notice the SHM property of this question and consequently got the shock of my life when I saw this..

I wouldn’t worry too much about this question — it is by far the hardest question that they’ve ever asked. It took me a couple of hours to get the answer out properly. If you look at the examiners’ report for that year you’ll notice that nobody got more than 83% on the paper as a whole — my guess is that nobody got more than about 10/20 on this question.

Having said that, though, the more rigorous your answers the better. The one thing the examiners are looking for is that you can turn vague words into accurate mathematical content (e.g. equations, diagrams), and then do something useful with it. Every line of maths that you can get down (as long as it’s right) is worth something.

Hi please could you explain your first line of working for a1. The paticular bit where you say Ek = fFdx = 100 I dont understand that line but the other lines i do

The line says that the energy is the integral of force with respect to distance, i.e. the area between the force-distance curve and the x-axis.

Could you explain how you get SHM for part a3? The force vs. position graph doesn’t look anything like SHM, right? Thank you!!

Hi Francis, thanks for your question.

The graph is really two graphs, one for

x > 0and one forx < 0, with a discontinuity atx = 0. Ignore thex > 0part and you can see that it does look like SHM whenx < 0. Specifically, whenx < 0, the equation of the graph isF = -(x + 10), which looks like a mass/spring equation.Try working out the graph of force against height for a 1kg mass on a spring that has spring constant 1N/m, given these conditions: the acceleration due to gravity

gis 10m/s/s; when the height is positive the spring is slack; when the height is zero the spring is just about to start extending; and the more negative the height the more extended the spring gets. You should find that you get the same graph that you see in the question.I’m really sorry to be a pain, but I cannot see how you generated the equation relating xi+1 and xi, and also where the 9/11 has come from. I presume it relates to the force acting?

When the mass is stationary at xi, the PE is 10xi, and the KE is zero.

The mass then moves a distance xi (with friction acting) and enters the x<0 zone. Some time later is comes bouncing back from the x<0 zone, travels a further distance xi+1 (with friction acting), and ends up stationary again at xi+1.

When the mass is stationary again at xi+1, the PE is 10xi+1 and the KE is zero.

In moving from the stationary point xi to the stationary point xi+1, the mass did work in overcoming friction by travelling xi (from the first stationary point to the x=0 point), and then travelling xi+1 (from the x=0 point to the second stationary point). So, by conservation of energy:

10 xi – xi – x(i+1) = 10 x(i+1), which means that 9xi = 11xi+1, so xi+1 = (9/11) xi

Please could you explain the (Xi)notation for the CoE part onwards? Thanks for all your help!

Or if you could, just to clarify I understand, a step-by-step explanation of part b would be great. Thank you!

OK, I’ll describe each of the bullet points in turn:

When the mass is at +x it has PE of 10x, and when it moves against friction a distance x it loses energy x. So we can generate an equation relating xi+1 to xi, and when we plug in the numbers above we get xi+1 = (9/11) xi

Finally in part b2, you should be able to see that the mass only disspates energy when x>=0 and it does this by moving against a frictional force of 1N. It starts with energy of 100J, which all has to be dissipated, so it must move 100m in the x>=0 zone.

Just for completeness, I have also summed up the infinite series to demonstrate that you get the same result. But this isn’t necessary in order to answer the question.