Mechanics M2 statics and work-energy questions

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1 Mechanics M statics and work-energy questions Tuesday 9 June A ladder AB, of weight W and length l, has one end A resting on rough horizontal ground. The other end B rests against a rough vertical wall. The coefficient of friction between the 5. ladder and the wall is 1. The coefficient of friction between the ladder and the ground is. Friction is limiting at both A and B. The ladder is at an angle to the ground, where tan θ = 5. The ladder is modelled as a uniform rod which lies in a vertical plane perpendicular to the wall. Find the value of. (9) Figure A particle P of mass 10 kg is projected from a point A up a line of greatest slope AB of a fixed rough plane. The plane is inclined at angle to the horizontal, where tan = 1 5 and AB = 6.5 m, as shown in Figure. The coefficient of friction between P and the plane is. The work done against friction as P moves from A to B is 45 J. (a) Find the value of. The particle is projected from A with speed 11.5 m s 1. By using the work-energy principle, (b) find the speed of the particle as it passes through B. P4487A 1

2 Monday June A truck of mass 1800 kg is towing a trailer of mass 800 kg up a straight road which is 1 inclined to the horizontal at an angle α, where sin α =. The truck is connected to the 0 trailer by a light inextensible rope which is parallel to the direction of motion of the truck. The resistances to motion of the truck and the trailer from non-gravitational forces are modelled as constant forces of magnitudes 00 N and 00 N respectively. The truck is moving at constant speed v m s 1 and the engine of the truck is working at a rate of 40 kw. (a) Find the value of v. 7. As the truck is moving up the road the rope breaks. (b) Find the acceleration of the truck immediately after the rope breaks. Figure A uniform rod AB of weight W has its end A freely hinged to a point on a fixed vertical wall. The rod is held in equilibrium, at angle θ to the horizontal, by a force of magnitude P. The force acts perpendicular to the rod at B and in the same vertical plane as the rod, as shown in Figure. The rod is in a vertical plane perpendicular to the wall. The magnitude of the vertical component of the force exerted on the rod by the wall at A is Y. W (a) Show that Y cos. Given that θ = 45 : (b) find the magnitude of the force exerted on the rod by the wall at A, giving your answer in terms of W. (6)

3 Thursday 6 June 01. A particle P of mass kg moves from point A to point B up a line of greatest slope of a fixed rough plane. The plane is inclined at 0 to the horizontal. The coefficient of friction between P and the plane is 0.4. Given that AB = 15 m and that the speed of P at A is 0 m s 1, find (a) the work done against friction as P moves from A to B, (b) the speed of P at B. () 5. Figure A uniform rod AB, of mass m and length a, is freely hinged to a fixed point A. A particle of mass m is attached to the rod at B. The rod is held in equilibrium at an angle θ to the horizontal by a force of magnitude F acting at the point C on the rod, where AC = b, as shown in Figure. The force at C acts at right angles to AB and in the vertical plane containing AB. (a) Show that F = amg cos. b (b) Find, in terms of a, b, g, m and θ, (i) the horizontal component of the force acting on the rod at A, (ii) the vertical component of the force acting on the rod at A. Given that the force acting on the rod at A acts along the rod, (c) find the value of a b.

4 Thursday 1 May 01. Figure 1 A uniform rod AB, of mass 5 kg and length 4 m, has its end A smoothly hinged at a fixed point. The rod is held in equilibrium at an angle of 5 above the horizontal by a force of magnitude F newtons applied to its end B. The force acts in the vertical plane containing the rod and in a direction which makes an angle of 40 with the rod, as shown in Figure 1. (a) Find the value of F. (b) Find the magnitude and direction of the vertical component of the force acting on the rod at A. 6. A car of mass 100 kg pulls a trailer of mass 400 kg up a straight road which is inclined to the horizontal at an angle, where sin = 141. The trailer is attached to the car by a light inextensible towbar which is parallel to the road. The car s engine works at a constant rate of 60 kw. The non-gravitational resistances to motion are constant and of magnitude 1000 N on the car and 00 N on the trailer. At a given instant, the car is moving at 10 m s 1. Find (a) the acceleration of the car at this instant, (b) the tension in the towbar at this instant. The towbar breaks when the car is moving at 1 m s 1. (c) Find, using the work-energy principle, the further distance that the trailer travels before coming instantaneously to rest.

5 Question Number 4. June 015 Marks Resolve horizontally or vertically: 1 R N or W R N Take moments about A or B. N M(A): ln sin l cos Wl cos A M(B): l cos R Wl cos Rl sin 10 N N W or 5 R W R 1 4N W 4N R N d 11 R R d ( ) (9 marks)

6 Question Number Scheme 5. (a) Max friction 10g cos B1 Work done against friction gcos 45 Marks Equation in : g 45, d or 0.4 (b) g6.5sin 10 v or equivalent v 5.85 (5.9) (m s -1 ) A (9 marks) Question Number June 014 Marks 4 F 00 N Tk 00 N Tr 1800g α (a) Constant speed F g sin g sin g Fv 1774v v.5 "1774" gsin a A F F ma : a (m s - ) 1800 [9] (b) A

7 7(a) Resolving vertically: Y P cos W Moments about A: Wl cos lp W cos W cos W cos ** P Y W D (6) (b) NB W Y Pcos with correct conclusion is possible W 45 Y B1 4 X Psin 45 Wcos45 W D =.sin 45 4 Resultant at A = W W 10 1 (0.79W) 4 4 D (6) June 01 a Work done = 15R15 0.4gcos0 b = 18g cos (J) Energy: WD against F + GPE + final KE = initial KE 1 1 their WD gsin 015 v 0 Aft v 1.7 (m s -1 ) () [7]

8 5a F B V C mg θ A H Moments about A: mg bf acos mg acos mg( a cos mg) A 5b : amg cos F *Answer given* b amg cossin H Fsin b : mg V F cos amg cos amg cos V mg cos mg b b 5c amg cos mg b tan amg cossin b b acos sin a cos sin cos b a cos a sin b a, a b D [1]

9 Question Number 6. (a) (b) May 01 F 6000 B1 F 100g sin 400g sin a - a. (m s ) T 400gsin x. ft ft T 1400 Marks (c) 1 d gd sin d 60 (m) D 14 marks (a) Taking moments about A F. 4sin 40 = 5g. cos 5 F=5 (b) F cos75 + Y = 5g Y = 40 newtons, up

10 June 015 Question 4 This was a straightforward question testing a basic understanding of ladders. Many candidates produced a correct moments 5 equation, usually about A but occasionally about B. The question gives tan, which gives awkward values for cos and sin. Several candidates divided by cos so that they could substitute directly for tan - this makes the resulting equations easier to work with. Seeing and 5 led some candidates to the incorrect assumption that they were working with a, 4, 5 triangle. Other common errors included having the friction force on the wall acting downwards rather than upwards, making a sign or trig error in the moments equation, multiplying W by g, and processing the equations inaccurately Although there were many clear and concise solutions to this question, candidates could help themselves (and the examiners!) by adopting logical labelling of forces such as R A, F A, R B, F B or by limiting themselves to two forces and using R, μr, N, N. Question 5 (a) There were many correct answers to this part of the question. The method for finding the work done against the friction was well understood, but some candidates also included the gain in gravitational potential energy in their answer. In the course of the working an approximate value for g is used, so an exact answer of 5 1 is inadmissible. (b) The majority of candidates followed the request to answer this using the work-energy principle. There were many correct answers. Common errors involved sign errors in the work-energy equation, or omitting a term (often the work done against friction) from the equation. As the question specifies the method to be used, attempts to answer the question using suvat equations scored no marks. June 014 Question 4 Many students gave fully correct answers to this question. In part (a) the majority of students treated the truck and trailer as a single system from the start but some considered the equations of motion for the truck and trailer separately and solved their two equations simultaneously to derive the equation of motion for the complete system. They went on to use correctly to find the speed. A common error was to give the final answer as.55 m either forgetting or not realising that having used the approximation a maximum of significant figures would be accepted. In part (b) most students realised that a new equation of motion for the truck was now required. Common errors were to subtract the wrong resistance from the driving force or to omit the weight component. Some students did not realise that the driving force on the truck would be unchanged at this instant and recalculated it, often incorrectly. Question 7 In part (a) the stronger students identified the two equations independent of the horizontal force at A very quickly, and they had little difficulty in obtaining the required result. A few students chose alternative points to take moments about, although generally these attempts were not fruitful because they had introduced another variable and could not see how to eliminate it. Another cause of difficulties was the poor use of variables, for example using X both as a force and a length, and then confusing the two. In part (b) the first mark was achieved by the majority of students. A handful of students ignored the 45 and assumed that the resultant force at A should act along the rod. In general, students who answered part (a) were able to correctly resolve to find X in terms of P. Some students who did not score many marks in part (a) nevertheless managed to answer part (b) successfully. Most students who had found the horizontal and vertical components found the resultant correctly. June 01 Question This proved to be a straightforward question for the majority of candidates. In part (a) the majority of candidates were able to resolve correctly, and almost all understood that finding the work done involved multiplying force by distance. The question was very specific in asking for the work done against the friction, and

11 too many candidates thought that this needed to include the work done against the weight. It was common to see the final answer given to an inappropriate level of accuracy. The majority of candidates attempted part (b) by forming a work/energy equation. Most attempts included all of the required terms, but there were frequently sign errors, either in placing the work done against friction or in the change in kinetic energy. There are still many candidates including both the work done against the weight and the change in gravitational potential energy, not recognising that this is the same thing. Some energy equations did not include the work done against friction at all. Candidates using the alternative approach via the suvat formulae often muddled the signs in their equations. Several did not realise that the acceleration up the plane was actually a deceleration. Question 5 In part (a) the majority of candidates took moments and combined their results correctly. Candidates who overlooked the weight of the rod frequently fiddled the given result rather than look for their error. This commonly led to further difficulties in part (b). In part (b) most candidates made sensible attempts to resolve horizontally and vertically following the request for horizontal and vertical components of the force at A. Candidates who resolved parallel and perpendicular to the rod at this stage usually found the subsequent algebra beyond them. Some candidates offered no attempt to part (c), but following part (b), the most common approach was to use tan. Some candidates were not able to work through the resulting algebra and trigonometry to reach a conclusion. A number of very elegant solutions, involving moments about B or C or resolving perpendicular to the rod, did not depend on part (b) at all. May 01 Question In part (a) many students found the value of F correctly, usually by following the most direct method of taking moments about A. There were a few errors with sin/cos confusion and some candidates omitted one or more lengths from their moment s equation. For part (b), having used moments in part (a), many candidates tried the same approach here. The majority tried to take moments about B, but they usually omitted the horizontal component of the force acting on the rod at A. The other common approach was to resolve vertically, but this also proved difficult because many candidates considered the component of F perpendicular to the rod rather than the vertical component of F. Some candidates did give completely correct answers to this part, but many were confused by the simplicity of what they were being asked and preferred to give the magnitude and direction of the force acting at A, rather than work on just the vertical component of that force. Question 6 In part (a), students who considered the system as a whole were often more successful than those who considered the car and trailer separately. The most common error was to omit one or more of the forces acting, often the weight of one of the objects, or one or both of the resistances. Although many correct answers were seen in part (b), some candidates struggled to write down a correct equation for the motion of the car or the trailer alone. It was common to see the acceleration ignored, or the weight or resistance missing from the equation. Part (c) was answered well. The majority of candidates now accept that if the question specifies that they must use the workenergy principle then they will earn no marks for an alternative approach. The most common error is still to see candidates double counting the change in potential energy by considering both the gain in GPE and the work done against the weight. Some candidates forgot to include the work done against the resistance. There did seem to be some confusion about what happens when the towbar breaks. Although the question clearly asks for the additional distance travelled by the trailer, some candidates considered both vehicles as if they were still attached, and some considered only the car.

Time: 1 hour 30 minutes

Paper Reference(s) 6678/01 Edexcel GCE Mechanics M2 Gold Level G3 Time: 1 hour 30 minutes Materials required for examination Mathematical Formulae (Pink) Items included with question papers Nil Candidates