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!! LC Applied Maths 2016 Before and after
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It's probably too late for most people at this stage to be even bothered, but I have attempted the HL paper now and got the answers below.
Q1 (a)
(i) v-t graph
(ii) 5 m/s
(iii) 107.5 m
Q1 (b) proof of relationship between h and u,g,t hinges on realising that when the collision takes place between the particles the displacement above ground for 1st particle = h = ut-0.5gt^2 while displacement for 2nd particle is also h = u(t-2T)-0.5g(t-2T)^2, as it has been traveling for 2T seconds less at the time of collision. Once these are equated find that time of collision is u/g +T and subbing this into original displacement equation for particle 1 proves h = (u^2-g^2T^2)/2g
Q2 (a)
(i) least possible value of v is 11.49km/h
(ii) 2 possible directions to ensure interception are West 72.89 North and West 17.11 North
(b) (i) 150 seconds (ii) 187.5 seconds
Q3 (a)
(i) 1 second
(ii) 0.65m
(iii) 30.87 m/s
(b) k = 0.1435 = 0.1 to 1 decimal place
Q4 (a)
(i) acceleration of p = 2.98 m/s/s, acceleration of q is 5.96 m/s/s
(ii) 1.337 m/s
(b) (i) Tension = 70g/11 N (ii) time to reverse direction = 0.3605 seconds
Q5 (a)
(i) Law of Conservation of Energy used here. PE + KE when A is pulled back 60 degrees = PE + KE when it collides with B. Take collision point as zero height reference here => KE when pulled back = 0 and PE at collision point is zero => PE when pulled back = KE on collision => 2g(2-2Cos60) = 0.5(2)v^2 => v = root(2g) as expected.
(ii) Speed of A after collision is 0.2213 m/s and speed of sphere B is 3.099 m/s
(iii) theta = 40.974 degrees
(b)
(i) velocity of P before collision = 3i + 2j
velocity of Q before collision = 2i - 2j
(ii) 128 degrees, to the nearest degree
will post on my solutions to Q6-Q10 tomorrow (later today!!)
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As promised, my solutions to the remaining 5 questions!
Q6 (a)
(i) minimum speed of projection = 8.8543 m/s
(ii) angle to vertical when it goes slack is 75.52 degrees
(b)
(i) To = kx = mg
T1 = mg -k(x+dx) = mg -kx - kdx (where dx is extension from equilibrium position)
=> T1 = -kdx = ma => a = -(k/m)dx
as k and m are constants, acceleration is directly proportional to the extension in the string (displacement from equilibrium position) and is always pointing towards that position (hence the - sign above). Therefore P is moving with SHM.
(ii) string first goes slack when displacement from equilibrium position is -0.2m as natural length of string is 1m. This happens when t = 0.2992 seconds so speed at this time is 2.425 m/s
(iii) back to Q1-type equations here as string is slack so P is just moving under the influence of gravity directly. Speed at highest point is zero. Initial speed is 2.425, so time taken is 0.2474 seconds. Add this to time taken for string to become slack to get 0.546 seconds in total.
Q7 (a)
(i) T = 1185.25N
(ii) R = 980N
(b)
(i) take moments about z for rod xz, then note forces up=forces down => R=W in this case, subbing for R in moments equation yields T=W/tan(theta)
(ii) take moments about z for both rod XZ and rod YZ separately. Forces up = forces down => R1+R2 = 4W here. Turns out T = 1.25W/tan(theta) so tension has increased on earlier case by 25%
Q8 (a) Usual proof here using integration
(b)
(i) 1.488 kg m2
(ii) total KE here is rotational + translational = 0.5Iw*w + 0.5mv*v
= 117.92 J
(iii) Use Law of Conservation of Energy here to find distance traveled up incline is 11.35m
Q9 (a) h = 4M/(Pi*Ro*d*d)
(b) Buoyancy = (x/2l)(W/s) where (x/2l) is fraction of rod immersed in water, W is weight of rod, 2l is length of rod
also taking moments about B => Buoyancy = [l/(2l-x/2)]W
equating the 2 gives s = [2lx-(x*x/2)]/(2l*l) = x/l - (x*x/4l*l) = x/l - (x/2l)*(x/2l) => s = 2f - f*f where f is fraction submerged i.e. x/2l
solving quadratic for f gives f = 1 +/- root(1-s) but fraction submerged cannot be > 1 => f = 1-root(1-s)
Q10 (a)
(i) integrate to find v = 4t*t + 4t -24. Analysing this you find that v is always increasing as t increases and that at time t=2, v = 0 so at this point the direction of the velocity changes so the direction of motion also changes at this point. From that point onwards v is only getting more and more positive and will increase for all time so we conclude that P only changes its direction of motion once in subsequent motion.
(ii) integrate to find displacement s = (4/3)t*t*t + 2t*t -24t
in order to find distance traveled between t=0 and t=3 need to find the displacement from t=0 to t=2 and then the displacement from t=2 to t=3 and add these together. This is vital to ensure we get the correct answer because the direction of motion changes at time t=2 seconds.
displacement at t=2 is -29.333m i.e. in the -ve direction .... this is the most negative the displacement value gets. From this point on the direction of motion changes and so the displacement gets less and less negative. displacement at t=3 is -18m so this means that between t=2 and t=2 the particle traveled a distance of 11.333m in the poitive direction. Adding this to the 29.333m traveled in the -ve direction between t=0 and t=2 gives a total distance traveled of 40.666m
(b)
(i) v = w* root(A*A-x*x)
(ii) x = ASin(wt+pi/2)
people may recognise these as formulas used in SHM questions
That's me done for another 12 months 0 -
It's probably too late for most people at this stage to be even bothered, but I have attempted the HL paper now and got the answers below.
Q1 (a)
(i) v-t graph
(ii) 5 m/s
(iii) 107.5 m
Q1 (b) proof of relationship between h and u,g,t hinges on realising that when the collision takes place between the particles the displacement above ground for 1st particle = h = ut-0.5gt^2 while displacement for 2nd particle is also h = u(t-2T)-0.5g(t-2T)^2, as it has been traveling for 2T seconds less at the time of collision. Once these are equated find that time of collision is u/g +T and subbing this into original displacement equation for particle 1 proves h = (u^2-g^2T^2)/2g
Q2 (a)
(i) least possible value of v is 11.49km/h
(ii) 2 possible directions to ensure interception are West 72.89 North and West 17.11 North
(b) (i) 150 seconds (ii) 187.5 seconds
Q3 (a)
(i) 1 second
(ii) 0.65m
(iii) 30.87 m/s
(b) k = 0.1435 = 0.1 to 1 decimal place
Q4 (a)
(i) acceleration of p = 2.98 m/s/s, acceleration of q is 5.96 m/s/s
(ii) 1.337 m/s
(b) (i) Tension = 70g/11 N (ii) time to reverse direction = 0.3605 seconds
Q5 (a)
(i) Law of Conservation of Energy used here. PE + KE when A is pulled back 60 degrees = PE + KE when it collides with B. Take collision point as zero height reference here => KE when pulled back = 0 and PE at collision point is zero => PE when pulled back = KE on collision => 2g(2-2Cos60) = 0.5(2)v^2 => v = root(2g) as expected.
(ii) Speed of A after collision is 0.2213 m/s and speed of sphere B is 3.099 m/s
(iii) theta = 40.974 degrees
(b)
(i) velocity of P before collision = 3i + 2j
velocity of Q before collision = 2i - 2j
(ii) 128 degrees, to the nearest degree
will post on my solutions to Q6-Q10 tomorrow (later today!!)
Thanks Japester for all your hard work. Could you checK the answers on the following. I got:
3 (b) k = 0.5
4 (a) (i) 1.92 m/s2, 3.84 m/s2, (ii) 1.07 m/s; (b) (ii) 44/49 s
5 (b) (i) 3i + 0.5j, 2i - 0.5j, (ii) 28 degrees
Thanks0 -
Thanks for the feedback lostatsea! Just checked over my work on Q3 (b) there and you are dead right, the answer I'm now getting for k is 0.53589 which rounds to 0.5 to 1 d.p. Numerical slip on my part
We're not a million miles apart on Q4 (a) (i) and at least one acceleration is twice the other so our methods must be similar. Also the value for v will differ as a result of our different acceleration values but at least we have both used the same formula to calculate it. There could easily be something wrong with my equations of motion here, but I didn't spot anything wrong with my number crunching. For your information the equations I had were:
Block P: 40gSin30 - 2T - 0.25(40gCos30) = 40a
Block Q: T-30g = 30b
I determined (like yourself) that a = b/2 in this case looking at the lengths of the strings.
Eliminating T from the equations left me with a = 2.98 m/s/s and b = 5.96 m/s/s
Do let me know what equations you came up with here.
On Q4 (b) I spotted another numerical slip I had made
- once I corrected this I ended up with t = 0.8979 seconds, same as yourself.
I looked through my solution to Q5 (b) again and I couldn't spot any numerical errors. I know you can't judge directly on the diagram given but (assuming it's semi-accurate?) if you have a look at it you'll see that the velocity vectors before collision for both particles appear to make an angle aroound the 45 degrees with the horizontal/vertical which would tend to agree with the 3i+2j and 2i-2j which I am coming up with. Also the angle through which the direction of motion of P is deflected I am taking as the sum of the angle of incidence and the angle of reflection. Again, if you just look at the diagram for this, you'll see it looks to be over 90 degrees. I am getting 128 degrees which is reasonable but 28 degrees looks too small? For me the angle of incidence is 56.3 degrees while the angle of reflection is 71.6 degrees = 128 degrees to nearest degree.
You might let me know again what equations you are using here for PCM and NEL, I basically had:
PCM along line of centres at impact (j-axis): mu1SinA + mu2SinB = m(-1) + m(1) = 0 => mu1SinA = mu2SinB
NEL: 1-(-1)/(u2SinB-u1SinA) = -e = -0.5
Basically it turns out for me that u1SinA = 2 and u2SinB = -2 and knowing the i-components are unchanged I end up with my "before" velocities as stated earlier. I could be doing something very basic wrong though, it definitely has been known to happen
hope this helps you out and thanks for the feedback, knowing the slips and errors I've made is good for me too!0 -
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Thanks for the feedback lostatsea! Just checked over my work on Q3 (b) there and you are dead right, the answer I'm now getting for k is 0.53589 which rounds to 0.5 to 1 d.p. Numerical slip on my part
We're not a million miles apart on Q4 (a) (i) and at least one acceleration is twice the other so our methods must be similar. Also the value for v will differ as a result of our different acceleration values but at least we have both used the same formula to calculate it. There could easily be something wrong with my equations of motion here, but I didn't spot anything wrong with my number crunching. For your information the equations I had were:
Block P: 40gSin30 - 2T - 0.25(40gCos30) = 40a
Block Q: T-30g = 30b
I determined (like yourself) that a = b/2 in this case looking at the lengths of the strings.
Eliminating T from the equations left me with a = 2.98 m/s/s and b = 5.96 m/s/s
Do let me know what equations you came up with here.
On Q4 (b) I spotted another numerical slip I had made - once I corrected this I ended up with t = 0.8979 seconds, same as yourself.
I looked through my solution to Q5 (b) again and I couldn't spot any numerical errors. I know you can't judge directly on the diagram given but (assuming it's semi-accurate?) if you have a look at it you'll see that the velocity vectors before collision for both particles appear to make an angle aroound the 45 degrees with the horizontal/vertical which would tend to agree with the 3i+2j and 2i-2j which I am coming up with. Also the angle through which the direction of motion of P is deflected I am taking as the sum of the angle of incidence and the angle of reflection. Again, if you just look at the diagram for this, you'll see it looks to be over 90 degrees. I am getting 128 degrees which is reasonable but 28 degrees looks too small? For me the angle of incidence is 56.3 degrees while the angle of reflection is 71.6 degrees = 128 degrees to nearest degree.
You might let me know again what equations you are using here for PCM and NEL, I basically had:
PCM along line of centres at impact (j-axis): mu1SinA + mu2SinB = m(-1) + m(1) = 0 => mu1SinA = mu2SinB
NEL: 1-(-1)/(u2SinB-u1SinA) = -e = -0.5
Basically it turns out for me that u1SinA = 2 and u2SinB = -2 and knowing the i-components are unchanged I end up with my "before" velocities as stated earlier. I could be doing something very basic wrong though, it definitely has been known to happen
hope this helps you out and thanks for the feedback, knowing the slips and errors I've made is good for me too!
Just looked at 4 (a): Particle P is moving up the hill as its horizontal component of weight (20g) is less than the 30g weight of particle Q. This means the frictional force on P will be in the opposite direction to the tensions in the string acting on P. I'd imagine your calculation would have given you a negative value for the acceleration indicating a problem with the direction of movement you chose.
Normally you are told its direction of movement in the problem. If not you need to determine this movement from the weights as all other forces are determined by the weights. Friction changes direction depending on its movement up or down the hill.0 -
Thanks for that lostatsea, I did indeed end up with negative accelerations for P and Q!! Never spotted it when I was rechecking, I must have just used "poetic licence" in the first instance and dropped the minus sign, hoping for the best I didn't realise that if the hanging weight exceeds the weight component of P along the slope then P would be moving upwards, that's very good to know, I just looked at the vertical weights for P and Q, said 40g>30g so it must be moving down the slope!
Any update on Q5(b)? Can you see if I've done something wrong? All feedback gratefully accepted.0 -
Thanks for that lostatsea, I did indeed end up with negative accelerations for P and Q!! Never spotted it when I was rechecking, I must have just used "poetic licence" in the first instance and dropped the minus sign, hoping for the best I didn't realise that if the hanging weight exceeds the weight component of P along the slope then P would be moving upwards, that's very good to know, I just looked at the vertical weights for P and Q, said 40g>30g so it must be moving down the slope!
Any update on Q5(b)? Can you see if I've done something wrong? All feedback gratefully accepted.
I have attached a pdf of my solution to 5 (b)0 -
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In relation to the angle of deflection, I can see where you are getting your slopes alright, and for P with the velocity vector = 3i+2j before collision I would say that if you were to get the slope of this vector with respect to the x-axis then yes its slope is certainly 2/3. However I am looking at the angle it is making with the vertical here (the y-axis), to me this is the angle of incidence as such here. So for me the slope is 3/2 which corresponds to an angle of incidence = 56.3 degrees roughly. Similarly, when I look at the velocity vector of P after collision I am looking at the angle of reflection with respect to the y-axis again, so my slope is -3 in this case corresponding to an angle of reflection = 71.6 degrees roughly. I am (rightly or wrongly) treating the angle of deflection as the sum of these angles, basically the angle between the incoming and outgoing trajectories. This gives me 128 degrees to nearest degree. This explains why your angle is 180 - my angle here. I'd love to know which of us is right on this, I guess we'll have to wait until the marking scheme is released but I might be able to find something online that has a similar set up with given solutions - if I do I'll get back to you.
Thanks for taking the time to get back to me, your solutions look very professional, are you using adobe writer for them? I'm doing mine the old fashioned way with pen and paper 0 -
Hi again lostatsea, I've been checking online and I've come across a number of problems that deal with angles of deflection (LC 2012 involves this) and from what I have seen your interpretation of events is the correct one, in this case you basically "project" the vector before collision onwards (using a dotted-line say) so that you end up with 2 vectors whose ends (without arrows) are meeting - the angle between these then becomes the angle of deflection. So the angle of deflection here would be your 52 degrees rather than my 128 degrees. I am learning a lot from this stuff! Its all good 0
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Hi again lostatsea, I've been checking online and I've come across a number of problems that deal with angles of deflection (LC 2012 involves this) and from what I have seen your interpretation of events is the correct one, in this case you basically "project" the vector before collision onwards (using a dotted-line say) so that you end up with 2 vectors whose ends (without arrows) are meeting - the angle between these then becomes the angle of deflection. So the angle of deflection here would be your 52 degrees rather than my 128 degrees. I am learning a lot from this stuff! Its all good
Well Japester I was looking to see if there is a difference between an angle of deflection and the angle through which a body turns which there obviously is - though I think the angle through which P turned in our example is 360 - 52 degrees. As you say, it is good to be forced to think beyond what we take for granted.0 -
I'm pretty sure I got it right in the exam but cant remember the paper very much any more, be grand sure haha.Thanks for that lostatsea, I did indeed end up with negative accelerations for P and Q!! Never spotted it when I was rechecking, I must have just used "poetic licence" in the first instance and dropped the minus sign, hoping for the best I didn't realise that if the hanging weight exceeds the weight component of P along the slope then P would be moving upwards, that's very good to know, I just looked at the vertical weights for P and Q, said 40g>30g so it must be moving down the slope!
Any update on Q5(b)? Can you see if I've done something wrong? All feedback gratefully accepted.0
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