Since the paper states that answers must be given to 2 s.f. in part B physics, I don’t think they would be expecting answers in which you need to divide by root 2. Doesn’t this imply that they were obviously looking for the more simple method, for the first few parts, that others people have mentioned?
Well dividing by root 2 is just multiplying by root 2 and dividing by 2 so I don’t really see a problem, and you’re allowed a calculator now regardless. A general rule of thumb is to use the method you think is simplest though!
Hello, sorry but I’m a bit confused about the first part of the question. Here you have used the equation : work done = force x distance, but why aren’t you using (1/2)Fx? On another website (PMT) this equation was used and gave an answer of 50 ms^-1 and so the rest of the question was differently answered? Please could you clear up some of this confusion? Thanks in advance.
Have a look further down the comments where we discuss the force-draw-curve for bows. The work done is ∫Fdx. If we are pulling back a spring where F = kx then this equals kx²/2, which equals Fx/2. But here we are talking about a bow, not a spring. Bows are designed to have a force draw curve which is as close to flat as possible (there is a link to one in the comments below). In my answer I am assuming that we have an ideal bow where the force draw curve is flat (and hence work done = Fx), and not something with the behaviour of a spring, which would be the worst bow ever made.
Wish E=Fx, F is the average force. The average force would be 120+0/2 = 60. As the string’s extending, the force applies slowly increases as x is increasing. You should be using (1/2)Fx getting 60 ms-1 as the answer.
Hi, for part 4 “to account for the effects of gravity”: Why can we simply use the value for the time calculated in part 3? This time was calculated with the assumption that the arrow travels purely in a horizontal direction. But as the archer aims higher, the velocity gets a vertical component and the horizontal component of the velocity decreases as the total velocity has to stay the same which would then lead to a change in time? Thanks
HI Uri. That’s a good question and what you way is correct. But we are asked for an estimate, and if you do a quick calculation you can see that the effect you describe is very small. If you suppose the value of 2.5m is roughly right then let’s see what effect this has on the speed of the arrow over its trajectory. At the midway point it needs have risen by roughly 1.25m, so as you say it has slowed down a bit (before speeding up again as it falls downwards while heading towards the target). If it started at speed v and has speed u at the mid point and is distance h higher at the midpoint then by conservation of energy we can see that mgh = 0.5m(v² – u²), so 2gh = v² – u², so u/v = √(1 – 2gh/v²). Plugging in numbers h = 1.25 and v = 50√2 we get u/v = √(1 – 25/5000) = √(1 – 5/1000) = √0.995 = roughly 0.9975. So it’s not a big effect and because we are estimating it’s OK to ignore it.
Have a look at the discussion below. Summary — I bet you wouldn’t lose marks for putting 0.5Fx, but actually the question is a silly one because bows to not work like springs.
(I’m afraid the Message-Reply structure is a bit messed up below for some reason, but its possible to follow the exchange I think).
You see for part 5, can we use a different approache: by using F=change of momentum divide by time. Time =s/v where s =5×10^-3 and v=[v1+v2]/2 where v1=50 and v2=0 then we can just plot the value of average speed into the formula F=change of momentum divide by time.
Do you think it is a reasonable approach or we can’t find the average speed by this way?
Thanks
If you do this you will get the right answer, but I don’t really like this solution because you are assuming that the arrow has a constant (negative) acceleration in order to calculate the answer. In general this assumption won’t be true. If you use the energy method that I’ve used, you don’t have to make this assumption.
To see why this matters, and why your assumption might not be correct, imagine a case where the arrow very quickly penetrates the first 4mm of target and then comes up against a very hard bit of target for the final 1mm. In this case the deceleration is small for the first 4mm and then much bigger for the final 1mm, so the time taken to travel 5mm is much less than in your calculation, but the average force is still the same.
I have a similar method to George above, in mine I used the kinematic formula v^2 = u^2 + 2as, I understand that it will run into the same problem of the assumption of constant acceleration not being valid. However, can’t we just say that this is the average acceleration and then calculate average force from there? Since the question is asking for an average force anyways, where the force also is not constant throughout.
This should probably work, and is again probably ok in this scenario. In general it’s more a matter of taste; working in terms of energies or forces. If you imagine a world in which you had total information of the materials, velocities etc, you could formulate the equation of motion in terms of forces or in terms of energies and if you did everything right the motion should be exactly the same.
(If you find yourself with nothing to do over summer, look up Lagrangian and Hamiltonian mechanics for a reformulation of Newtonian mechanics in terms of energies. It gets pretty advanced though)
Thanks for your reply! Yep I’ve delved a bit into the Lagrangian and the action before, will look into it more if i have the time! Pretty cool how action can be seen as quite a fundamental quantity
Good question. See the exchange below, where James asked essentially the same question two years ago.
The simple answer is that the PE does equal Fx/2 if the bow works like a perfect spring, in which the force is proportional to the distance that the arrow is pulled back. But that is not how bows work. In fact the perfect bow would have a near constant draw force as it is extended (see this high quality bow for an example http://archeryreport.com/wp-content/uploads/2011/02/envy_draw_force_curve.jpg). So I have assumed that the bow in the question is some kind of ‘ideal bow’, in which case the PE equals Fx.
There’s no right answer here, I’m afraid, because it is really a silly question. My guess is that the person who set the question was probably a physicist rather than an engineer, naively assumed that a bow works like a perfect spring, and was expecting the answer that you have gone for. But I’ve gone for a different answer to bring out the fact that the draw force curve of the bow hasn’t been described.
Probably the perfect answer would be to state explicitly that the PE is the integral of force with respect to distance but we can’t work out the exact PE here because the force/distance curve hasn’t been given, so we are going to assume … (then put in whatever assumption you want to make, either that the bow obeys Hooke’s law or that its force is constant, or whatever …)
Hi James — thanks a lot for your comment — there are two very interesting points in it.
You are absolutely right that the force clearly isn’t going to be constant for the whole 0.6m. But, as I assume you know very well given your question, the ‘elastic potential energy method’ that you have used is not going to give the correct answer either. In reality, bows obey a complex force/distance curve which archers call the ‘draw force curve’, which lies somewhere between the ‘Hooke’s Law’ curve you have used and the ‘constant force’ curve that I have used; the energy is the integral of the draw force curve. But in this case the examiner hasn’t told us the shape of the curve! So the best one can do is to state an assumption and go with it. However I think it might well be true that your answer was the expected/desired one, and t he very least my answer above should have included some statement of what I have just said in this comment, to make it clear that there was some reasoning behind it.
Is there error carry forward for PAT? I’m afraid nobody knows the answer to that question. I kind of assume not, because it would be daft, but I guess it is a bit daft to set a question which requires students to integrate a function that you don’t give them, so who knows? I suppose the motto is to be rigorous: show your working clearly, and clearly state any assumptions you have made and why you have made them.
Hope this helps, and the very best of luck tomorrow!
For the first part i don’t think it is wise to assume that the force is constant throughout 0.60m. I did the elastic potential energy method and get an alternative solution. Anyway, the answer for the first part will affect the rest of the answers in the last question….. is there error carry forward for PAT? If not does that mean that the whole thing will be wrong if we chose the wrong technique to calculate?
Oxford PAT 
Since the paper states that answers must be given to 2 s.f. in part B physics, I don’t think they would be expecting answers in which you need to divide by root 2. Doesn’t this imply that they were obviously looking for the more simple method, for the first few parts, that others people have mentioned?
Well dividing by root 2 is just multiplying by root 2 and dividing by 2 so I don’t really see a problem, and you’re allowed a calculator now regardless. A general rule of thumb is to use the method you think is simplest though!
For the fifth part, shouldn’t you consider the vertical velocity of gt to find the speed?
When calculating the stopping distance? If so, then no, all the energy is being dissipated on impact, so we should consider that F = dE/dx
Hello, sorry but I’m a bit confused about the first part of the question. Here you have used the equation : work done = force x distance, but why aren’t you using (1/2)Fx? On another website (PMT) this equation was used and gave an answer of 50 ms^-1 and so the rest of the question was differently answered? Please could you clear up some of this confusion? Thanks in advance.
Have a look further down the comments where we discuss the force-draw-curve for bows. The work done is ∫Fdx. If we are pulling back a spring where F = kx then this equals kx²/2, which equals Fx/2. But here we are talking about a bow, not a spring. Bows are designed to have a force draw curve which is as close to flat as possible (there is a link to one in the comments below). In my answer I am assuming that we have an ideal bow where the force draw curve is flat (and hence work done = Fx), and not something with the behaviour of a spring, which would be the worst bow ever made.
Wish E=Fx, F is the average force. The average force would be 120+0/2 = 60. As the string’s extending, the force applies slowly increases as x is increasing. You should be using (1/2)Fx getting 60 ms-1 as the answer.
Very nice, explanation! Keep it up!
Hi there, for part six, how come we do not have to take into account the vertical velocity of the arrow? Is it negligible?
Yes I think so. Have a look at the discussion with Uri from yesterday.
Hi, for part 4 “to account for the effects of gravity”: Why can we simply use the value for the time calculated in part 3? This time was calculated with the assumption that the arrow travels purely in a horizontal direction. But as the archer aims higher, the velocity gets a vertical component and the horizontal component of the velocity decreases as the total velocity has to stay the same which would then lead to a change in time? Thanks
HI Uri. That’s a good question and what you way is correct. But we are asked for an estimate, and if you do a quick calculation you can see that the effect you describe is very small. If you suppose the value of 2.5m is roughly right then let’s see what effect this has on the speed of the arrow over its trajectory. At the midway point it needs have risen by roughly 1.25m, so as you say it has slowed down a bit (before speeding up again as it falls downwards while heading towards the target). If it started at speed v and has speed u at the mid point and is distance h higher at the midpoint then by conservation of energy we can see that mgh = 0.5m(v² – u²), so 2gh = v² – u², so u/v = √(1 – 2gh/v²). Plugging in numbers h = 1.25 and v = 50√2 we get u/v = √(1 – 25/5000) = √(1 – 5/1000) = √0.995 = roughly 0.9975. So it’s not a big effect and because we are estimating it’s OK to ignore it.
Alright, Thanks!
For part5, shouldn’t the work done be 0.5Fx since it is the average force that we are calculating.
Have a look at the discussion below. Summary — I bet you wouldn’t lose marks for putting 0.5Fx, but actually the question is a silly one because bows to not work like springs.
(I’m afraid the Message-Reply structure is a bit messed up below for some reason, but its possible to follow the exchange I think).
You see for part 5, can we use a different approache: by using F=change of momentum divide by time. Time =s/v where s =5×10^-3 and v=[v1+v2]/2 where v1=50 and v2=0 then we can just plot the value of average speed into the formula F=change of momentum divide by time.
Do you think it is a reasonable approach or we can’t find the average speed by this way?
Thanks
If you do this you will get the right answer, but I don’t really like this solution because you are assuming that the arrow has a constant (negative) acceleration in order to calculate the answer. In general this assumption won’t be true. If you use the energy method that I’ve used, you don’t have to make this assumption.
To see why this matters, and why your assumption might not be correct, imagine a case where the arrow very quickly penetrates the first 4mm of target and then comes up against a very hard bit of target for the final 1mm. In this case the deceleration is small for the first 4mm and then much bigger for the final 1mm, so the time taken to travel 5mm is much less than in your calculation, but the average force is still the same.
That makes sense. Thank you!
I have a similar method to George above, in mine I used the kinematic formula v^2 = u^2 + 2as, I understand that it will run into the same problem of the assumption of constant acceleration not being valid. However, can’t we just say that this is the average acceleration and then calculate average force from there? Since the question is asking for an average force anyways, where the force also is not constant throughout.
This should probably work, and is again probably ok in this scenario. In general it’s more a matter of taste; working in terms of energies or forces. If you imagine a world in which you had total information of the materials, velocities etc, you could formulate the equation of motion in terms of forces or in terms of energies and if you did everything right the motion should be exactly the same.
(If you find yourself with nothing to do over summer, look up Lagrangian and Hamiltonian mechanics for a reformulation of Newtonian mechanics in terms of energies. It gets pretty advanced though)
Thanks for your reply! Yep I’ve delved a bit into the Lagrangian and the action before, will look into it more if i have the time! Pretty cool how action can be seen as quite a fundamental quantity
Why is the formula for elastic potential energy in the 1st part not 0.5Fx ?
Good question. See the exchange below, where James asked essentially the same question two years ago.
The simple answer is that the PE does equal Fx/2 if the bow works like a perfect spring, in which the force is proportional to the distance that the arrow is pulled back. But that is not how bows work. In fact the perfect bow would have a near constant draw force as it is extended (see this high quality bow for an example http://archeryreport.com/wp-content/uploads/2011/02/envy_draw_force_curve.jpg). So I have assumed that the bow in the question is some kind of ‘ideal bow’, in which case the PE equals Fx.
There’s no right answer here, I’m afraid, because it is really a silly question. My guess is that the person who set the question was probably a physicist rather than an engineer, naively assumed that a bow works like a perfect spring, and was expecting the answer that you have gone for. But I’ve gone for a different answer to bring out the fact that the draw force curve of the bow hasn’t been described.
Probably the perfect answer would be to state explicitly that the PE is the integral of force with respect to distance but we can’t work out the exact PE here because the force/distance curve hasn’t been given, so we are going to assume … (then put in whatever assumption you want to make, either that the bow obeys Hooke’s law or that its force is constant, or whatever …)
Hi James — thanks a lot for your comment — there are two very interesting points in it.
You are absolutely right that the force clearly isn’t going to be constant for the whole 0.6m. But, as I assume you know very well given your question, the ‘elastic potential energy method’ that you have used is not going to give the correct answer either. In reality, bows obey a complex force/distance curve which archers call the ‘draw force curve’, which lies somewhere between the ‘Hooke’s Law’ curve you have used and the ‘constant force’ curve that I have used; the energy is the integral of the draw force curve. But in this case the examiner hasn’t told us the shape of the curve! So the best one can do is to state an assumption and go with it. However I think it might well be true that your answer was the expected/desired one, and t he very least my answer above should have included some statement of what I have just said in this comment, to make it clear that there was some reasoning behind it.
Is there error carry forward for PAT? I’m afraid nobody knows the answer to that question. I kind of assume not, because it would be daft, but I guess it is a bit daft to set a question which requires students to integrate a function that you don’t give them, so who knows? I suppose the motto is to be rigorous: show your working clearly, and clearly state any assumptions you have made and why you have made them.
Hope this helps, and the very best of luck tomorrow!
Alright, thank you so much!
For the first part i don’t think it is wise to assume that the force is constant throughout 0.60m. I did the elastic potential energy method and get an alternative solution. Anyway, the answer for the first part will affect the rest of the answers in the last question….. is there error carry forward for PAT? If not does that mean that the whole thing will be wrong if we chose the wrong technique to calculate?