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Applied Maths Exam Paper Hints?
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02-04-2015 5:31pm#1Does anyone have the Applied Maths exam paper hints?
It looks like this:
Does anyone have it, or know where I can buy it?0
Comments
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This saves you from downloading the file:) :
Answering higher level exam questions
There are usually many ways to answer these questions; I went with the methods outlined in the marking schemes.
2009 (a)
Use v = u + at for both particles. Then use v2 = u2 + 2as for both particles to get the required expression.
2009 (b)
(i) Velocity-time graph
(ii) You need to play around with lots of algebra. Get an expression for total time in terms of v and f, then use the fact that average speed = total distance/total time to get an answer of f = 1 m s-2.
2008 (a)
(i) Straightforward. Ans: t = 4 s.
(ii) Need to distinguish between the concepts of distance and displacement.
On the way up, distance and displacement will be the same, but in this question after 5 seconds the ball will have been on its way down for 1 sec (how do we know this?), so we need to establish how far it will have travelled in the fifth second and add this on to the maximum height.
Ans: total distance = 83.3 m
2008 (b)
(i) You know v, u and t for Q (t is when Q passes P), so use this to work out a.
Now use this to work out s (distance from the beginning to where Q passes P).
Now for P you have this same s, plus u and t, so use this to work out a for P.
Ans: aQ = 0.5 m s-2, aP = 5/24 m s-2.
(ii) Use v = u + at to find v for P. Ans: vP = 48 m s-2.
(iii) “When” they are moving at equal speeds Þ vP = vQ so get an expression for both and equate.
Use this to find t. Sub into expressions for SP and SQ and subtract one from the other to find the distance between them.
Ans: distance = 525 m
2007 (a)
(i) It’s not obvious, but this is a ‘Train-track’ type question.
During the third second of its motion the particle travels 29.9 metres.
We can get an expression for h - the distance travelled in the first 2 seconds: we know u, a and t.
We can then do the same for the distance travelled in the first 3 seconds. S in this case is (h + 29.9) m.
Ans: u = 5.4 m s-1
(ii) Straightforward. Ans: s = 100 m.
2007 (b)
(i) Straightforward.
(ii) Tricky: One car – two accelerations
You need to remember that , so we need to find an expression for this in relation to the information supplied, and then compare that to the expression they give us.
For each section write down the relationship between velocity, distance and time
To find the total time you need to find an expression for each of the 3 individual times (in terms of velocity and distance).
Then it’s just messy algebra to finish it out.
Ans: b = 1
2006 (a)
Acceleration / deceleration type question.
(i) Straightforward
(ii) Messy algebra. Ans: d = 3/8 ft2
2006 (b)
(i) Straightforward.
(ii) Straightforward once you draw a diagram to help you verify that SP + SQ = 159. Then solve. Ans: t = 6 s.
(iii) Straightforward. Ans: t = 3.1 s.
2005 (a)
(i) Straightforward (once you note that the acceleration is minus). Ans: s = 33.3 m
(ii) Note that there are two distances to take into account here. The first is to do with reaction-time distance, and the second is the normal stopping distance. Add these together but remember that you have to subtract the first distance of 33.3 m because the first car will have moved on by this distance. Ans: s = 43.3 m.
2005 (b)
(i) Straightforward once you are familiar with the concept of Fnet = ma.
To find out the forces acting on the mass in the sand you will need to work out its acceleration. Before you can do this you will first need to note that its initial velocity for the second stage (when it’s in the sand) will be the same as its final velocity in the air. So that gives you u; you know v = 0 and t = 0.01. From that you can work out acceleration a.
Then it’s just Forcedown – Forceup = ma, where Forcedown = mg, and Forceup is due to resistance of the sand, which is what you are looking for.
Ans: R = 39278.4 N.
(ii) Straightforward. Ans: s = 0.245 m.
2004 (a)
(i) Straightforward. s1 = s2. Note that first ball is in the air for 3 seconds and second ball is in the air for 2 seconds.
Ans: u = 17.75 ms-1.
(ii) You first have to establish for each ball whether it will be on the way up or the way down. Hint: look at their initial velocities and the time they taken to reach max. height. Draw a diagram for each ball to help you.
Ans: Ball A = 25 m, ball B = 16 m.
2004 (b)
(i) Straightforward once you are familiar with Fnet = ma.
Ans: f = 1.1 m s-2
(ii) Similar to (i), except in this case the car and caravan are going uphill so you will have to and also resolve the weight into components along the plane and perpendicular to the plane, and proceed accordingly.
Ans: a = 60.
2003 (a)
(i) Straightforward train-track question.
(ii) Straightforward once you remember that all numbers must be in relation to point p, so total distance travelled before coming to rest is (250 + s). Remember also that a = -3.
Ans: s = 51 m.
2003 (b)
(i) Man just catches bus, so at this time vMan = vbus. Also when the man catches up with the bus he will hae travelled 40 m more than the bus, so sMan = (40 + sBus).You will need to play around with the various equations of motion and use lots of algebra to solve.
Ans: u = 4 m s-1.
(ii) Find the distance between them means find get an expression for the distance travelled by both and subtract one from the other. In this case you are asked to find the minimum distance, so anytime maximum or minimum is asked for it usually means you have to differentiate and let your answer equal 0 to find t (remember from maths how to find the maximum or minimum point on a curve? – this is one of the most important applications of differentiation).
Ans: s = 17.5 m.
2002 (a)
(i) Straightforward. Note s = -30. Ans: u = 18.5 ms-1.
(ii) Straightforward. Ans: speed = 30.5 m s-1 (remember strictly speaking ‘speed’ implies magnitude only, not direction, so we should ignore the minus sign).
2002 (b)
(i) Train-track type question. Straightforward.
(ii) Straightforward.
2001 (a)
(i) Velocity-time graph. Lots of algebra. See notes for answering this type of question above.
Ans: t = 75 seconds.
(ii) Acceleration-deceleration. Algebra. See notes above.
Ans: t = 60 seconds.
2001 (b)
(i) Two balls are thrown up and collide in the air so remember that the key is s1 = s2. Remember also that if the second ball is in the air for t1 seconds, then the first ball (which is obviously in the air for longer) is in the air for (T + t1) seconds. Note also that you were asked for the time taken in relation to when the first particle was projected, so you may have to adjust your answer accordingly.
(ii) Sub value for time into expression for s.
2000 (a)
This involves a stone projected upwards, but is actually a type of train-track type question because all information must be with reference to the initial point of projection.
Ans: u = 56 m s-1.
2000 (b)
(i) Velocity-time diagram
The cars are moving in opposite directions so when they meet the total distance travelled will be 10,000 m, i.e. sp + sq = 10,000 m.
Ans: t = 215 s, sp = 4875 m, sq = 5125 m.
(ii) Cars meet halfway Þ sp = sq.
Ans: t = 10 s.
1999 (a)
(i) Straightforward Fnet = ma question using the line of slope as the x-axis.
Ans: F = 744 N
(ii) Power = Force × velocity
Ans: P = 14880 W
1999 (b)
(i) Velocity-time graph
(ii) You need to play around with lots of algebra.
Ans: u = 4/t
(iii) Again, play around with the equations
Ans: Total distance = 75.76 m.
1998 (a)
(i) Velocity-time graph. Very difficult algebra. Remember
1998 (b)
(i) Straightforward in principle, although the algebra gets a little messy.
Ans: u = 0.1 m s-1, b = 1
(ii) Straightforward.
Ans: s = 94.5 m
1997 (a)
Straightforward if you begin by throwing down all the relationships between acceleration, velocity, time and distance as per revision notes. Then just play around with the equations.
(i) Answer: v = 7.5 m/s
(ii) Answer: distance = 21 m
1997 (b)
(i) Nice question, but could cause difficulty in that the questions wouldn’t be familiar. When particles collide then S1 + S2 = d.
(ii) Tricky to decipher the significance of the information. You’ve just worked out an expression for the time for collision – call this t1. Now work out an expression for the time for the second particle to come to instantaneous rest – call this t2.
Now let t1 < t2 to obtain the required expression.
(iii) If the second particle has returned to q then S2 = 0, so get the time for this and let it equal to t1 (time at which collision would have taken place) and obtain the required expression.
1996 (a)
(i) Straightforward train-track type problem
Answer: a = 3.5 m s-2
(ii) Straightforward
Answer: s = 7 m
1996 (b)
(i) Straightforward
(ii) Straightforward if you begin by throwing down all the relationships between acceleration, velocity, time and distance as per revision notes. Then just play around with the equations.
1995 (a)
A variation on the train-track problem. Use v2 = u2 + 2as for stage pq and stage qr.
Answer: v = 5u.
Straightforward. t1 = 4u/f and t2 = 2u/f.
1995 (b)
(i) Straightforward.
Answer: u = Ö6 g
(ii) Straightforward.
Answer: t = 0.26 seconds
(iii) Strightforward.
Answer: S6 = S2 = 5/3, S5 = S3 = 8/3, S4 = 30
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