the Further Mhemics network www.fmnetwork.org.uk V 07 1 REVISION SHEET MECHANICS 1 MOTION GRAPHS The main ideas are AQA Edx MEI OCR Displacement-time graphs M1 M1 M1 M1 Distance-time graphs M1 M1 M1 M1 Velocit-time graphs M1 M1 M1 M1 Interpreting the graphs M1 M1 M1 M1 Using differentiion and integrion M M M1 M1 Before the exam ou should know The difference between a distance-time graph and a displacement-time graph, and the difference between a speed-time graph and a velocit-time graph. The gradient of a displacement time graph gives the velocit and the gradient of a distance time graph gives the speed. The displacement from the starting point is given b the area under the velocit-time graph (where area below the x-axis is given a negive sign) In the case of a particle moving with a variable accelerion, know how to use differentiion and integrion to calcule accelerion, velocit and displacement via the gradients and areas described in and 3 above. Displacement-time graphs and distance-time graphs A displacement time graph plots the displacement of the object from some fixed origin against time. For example, the journe described in the graph to the right can be described as follows, Phase 1: The object moves awa from the origin ( a constant speed) for seconds Phase : It remains motionless for 3 seconds. Phase 3: It starts moving back towards the origin a constant speed returning to the origin after about 4 seconds and then continues in the same direction another second. displacement (metres) 5 time (secs) Here is a distance-time graph of the same journe. Notice th the gradient of the distance-time graph is never negive (as time increases it is impossible for the total distance ou have travelled to decrease). The gradient is positive whenever the object is moving and it is zero whenever the object is stionar. In fact the gradient of the distance-time graph is the modulus of the gradient of the displacement-time graph all times. Think about this! distance (metres) 5 time (secs)
the Further Mhemics network www.fmnetwork.org.uk V 07 1 Velocit-time graphs and speed-time graphs A velocit time graph plots the velocit of the object against time, whereas a speed-time graph plots speed against time. Remember an objects speed is just the modulus of its velocit. For example for the journe we looked in the displacementtime graph over the page the velocit-time graphs is drawn to the right as a dotted line and the speed time graph is drawn in as a solid line. Phase 1: Object moving in a direction (which has been designed the positive direction a constant speed of ms. Phase : Stionar, zero velocit. Phase 3: Moving in the opposite direction to phase 1(the negive direction) a constant speed of 1 ms, and therefore with a velocit of 1 ms. ms 5 time (secs) speed Interpreting the feures of a graph. velocit Ke Points The gradient of a velocit time graph gives the accelerion. The area underneh (taking area below the x-axis to have a negive sign) a velocit time graph gives the displacement. The gradient of a displacement-time graph gives the velocit The gradient of a distance-time graph gives the speed. In the case of non-constant accelerion the gradients and areas can be calculed using differentiion and integrion. Here is an example involving a particle moving with a non-constant accelerion. (In fact the accelerion time t is 6t, can ou see wh? Hint differentie the expression for the displacement twice.) Example. A particle moves so th its displacement, x, from a fixed origin time t where 0 t 3is given b x = t( t 1)( t 3). i) Sketch a displacement-time graph of this journe. ii) Wh is the velocit of the particle when t = 1 and when t =?. i) displacement ii) To calcule the velocit the times given firstl differentie the displacement 4 with respect to time. x = t( t 1)( t 3) so the easiest wa to do this is probabl to multipl out the brackets so th there is a polnomial in t to differentie. 3 Hence, x = tt ( 1)( t 3) = t 4t + 3t. 0.0 3.0 time dx And so = 3t 8t+ 3. dt dx When t = 1: 3 8 3 ms dt = + = and when t = : dx -4 ( 3 4) ( 8 ) 3 1ms dt = + =.
the Further Mhemics network www.fmnetwork.org.uk V 07 1 1 REVISION SHEET MECHANICS 1 NEWTON S LAWS APPLIED ALONG A LINE The main ideas are Before the exam ou should know: Tpe of Problem AQA Edx MEI The tpe of problem th will be looked on this sheet is largel concerned with objects which are moving in a horizontal or vertical straight line. Note th in man of these problems the resistive force could be taken to represent friction. However, friction is discussed in the revision sheet vectors and Newton s laws in D (and is not required for the MEI M1 specificion) Moving in the horizontal plane (e.g a car, a train, a ship) OCR Motion in a horizontal plane M1 M1 M1 M1 Motion in a vertical plane M1 M1 M1 M1 Pulles M1 M1 M1 M1 Connected bodies M1 M1 M1 M1 1. You should know Newton s second law, th the change in motion is proportional to the force, or as an equion F = ma where F is the resultant force, m is the mass of the object and a is its accelerion.. Newton s first law th ever particle continues in a ste of rest or uniform motion in a straight line unless acted on b a resultant external force. 3. Newton s third law, th when one object exerts a force on another there is alwas a reaction, which is equal and opposite in direction, to the acting force. 4. The five SUVAT equions, wh each letter stands for and when to use them. The are: u+ v v = u+, s = t, 1 1 s = ut+, v = u + as, s = vt When the vehicle is moving along a horizontal road, the reaction force, R reaction force and the weight force (labeled mg in the breaking force, B driving diagram) are equal and opposite i.e. R = mg and as the force, D are perpendicular to the horizontal motion the do not contribute (unless Friction is involved, which is reled to resistance, F R). So the resultant force acts in the horizontal direction mg and equals (taking right as positive): D B F Moving in the vertical plane (e.g balls thrown verticall in the air, cres hanging from ropes, lifts) tension, T The onl forces acting on the cre are in the vertical direction. Hence the resultant force acts in the vertical direction. If the tension is greer than mg the cre will accelere upwards (or decelere if the cre is alread moving down). If the tension equals mg the cre will maintain its current velocit (or remain stionar). If the tension is less than mg the cre will accelere downwards (or decelere if the cre is alread moving upwards). mg We also need to know how to deal with situions of this tpe in which objects are connected. It is possible to have an object moving in a straight line in the horizontal plane connected to an object moving in a straight line in the vertical plane (via a pulle). We need to know how to deal with this too. The common theme is th Newton s second law will alwas be used, which when written as an equion, reads: resultant force = mass accelerion.
the Further Mhemics network www.fmnetwork.org.uk V 07 1 1 Pulles The tension in a rope passing over a pulle is constant throughout the length of the rope (assuming no friction in the pulle, which is alwas the case in Mechanics 1). Consider a tractor tached to a cre b rope which passes over a pulle. Forces acting on tractor (ignoring vertical forces which are perpendicular to the motion and hence do not contribute to the horizontal motion) driving force, D tension, T Forces acting on cre tension, T mg Forces acting on pulle (of negligible mass) (assuming the pulle is fixed, these will cancel each other out) reaction, R tension, T tension, T Ke Point: The value of T is the same in all three of the above force diagrams. Tpical Standard Problem: A child of mass 50kg is in a lift of mass 00kg. Calcule the tension in the lift cable and the reaction force (of the lift floor on the child) when a) the lift is stionar, b) the lift - - - going up and accelering 5 ms, c) the lift is going down and decelering 5 ms. (let g = 10 ms ). : (a) When the lift is stionar the accelerion is zero, therefore we have tension, T reaction, R (taking up to be positive), considering the lift: T mg = T 500 = resultant force = ma= m 0 = 0 T = 500N And b considering the child: R mg = R 500 = resultant force = ma= m 0 = 0 R= 500N Forces acting on lift mg = (00 + 50) 10 = 500 Forces acting on child mg = 50 10 = 500 (b) When the lift is accelering upwards 5 ms, we have (taking up to be positive), considering the lift: T mg = T 500 = resultant force = ma= 500 5 = 1500 T = 1500 + 500 = 15000N And b considering the child: R mg = R 500 = resultant force = ma= 500 5 = 500 T = 500 + 50 = 750N - (c) When the lift is going down and decelering 5 ms, as the accelerion is in the opposite direction to the - motion, the accelerion is upwards. Thus, we have an upwards accelerion of 5 ms and so our answers are exactl as in (b). Connected Bodies Example: A car, mass 800kg is pulling a caravan, mass 1000kg along a straight, horizontal road. The caravan is connected to the car with a light, rigid tow bar. The car is exerting a driving force of 170N. The resistances to forward motion are 400N on the car and 600N on the caravan. These resistances remain constant. Calcule the accelerion of the car and caravan. : There are two possible approaches: Approach 1: Consider forces on the car and the caravan separel. tension + resistance driving force 170N = (T + 400)N CAR resistance tension, T = 600N CARAVAN Considering the car 170 T 400 = 800a, and b considering the caravan T 600 = 1000a. B adding these equions together we get 70 70 = 1800a a = = 0.15ms 1800 - - Approach : Tre the sstem as a whole. The forces are then as follows: resistance driving force 170N = (600+400)N This gives 170 1000 = 1800a 70 a = = 0.15ms 1800 Notice the equion we arrive immediel here is exactl the same as the one obtained b adding the two equions in approach 1, where the internal tension forces cancel. To calcule the tension in the tow bar, we could use this value of a in either of the equions in approach 1. -
the Further Mhemics network www.fmnetwork.org.uk V 07 1 REVISION SHEET MECHANICS 1 PROJECTILES The main ideas are Finding the maximum height of a projectile Example A ball is kicked with a speed of 15 ms over level ground an angle of 40 to the horizontal. Wh is the maximum height reached? We are concerned with the vertical component of the ball s motion. u a v = 15sin 40 = g = 9.8 = 0 ( maximum height) AQA Edx MEI OCR The maximum height of a projectile M1 M M1 M The range of a projectile M1 M M1 M The ph of a projectile M1 M M1 M =? (this is the maximum height we wish to find) Choose the approprie equion of motion, based on the informion ou have and wh ou need to calcule: v = u + as v u = a ( ) 0 15sin40 = 9.8 = 4.74m (3 s.f.) so the maximum height of the ball is 4.74m (3 s.f.) Before the exam ou should know: You must be completel familiar and fluent with all of the constant accelerion equions, especiall: v= u+ 1 r = ut+ You must be fluent with the use of vectors and resolving into horizontal and vertical components. The onl force which acts on a projectile is gravit and we assume: - a projectile is a particle - it is not powered - the air has no effect on its motion A projectile experiences a constant accelerion of - g = 9.8ms verticall downwards The horizontal component of accelerion is 0 for a projectile, so its horizontal component of velocit is constant. If a projectile has an initial speed u, an angle of θ to the horizontal, its initial velocit is u cosθ u = u sinθ and its accelerion is 0 g At maximum height, the vertical component of a projectile s velocit,, is 0. v Know how to derive the equion of the ph of a gx projectile: = xtanθ ( 1+ tan θ ) u
the Further Mhemics network www.fmnetwork.org.uk V 07 1 Finding the range of a projectile Continuing the example used for maximum height, if the ball is kicked over level ground, the point where it lands will have vertical displacement,, of 0. Its range is its horizontal displacement, x, from its starting point the point where it lands. To calcule the ball s range, ou can calcule the time it takes to return to the ground and then use this time to calcule the horizontal displacement th time. Question: Wh is the range of the ball? Considering the vertical motion: u = 15sin 40 a = 9.8 = 0 (when the ball returns to the ground) t =? (when the ball returns to the ground) 1 = t u + at 0 = t( 15sin 40 4.9t) 15sin 40 t = 0 or t = = 1.968 seconds (4 s.f.) 4.9 Now considering the horizontal motion: u = 15cos40 a x x = 0 x =? (the range) t = 1.968 (when the ball returns to the ground, calculed above) 1 x= t ux + axt x= 1.968( 15cos 40 + 0) =.6m (3s.f.) Finding the ph of a projectile Sticking with the same example: Choose the approprie equion of motion, based on the informion ou have and wh ou need to calcule: 1 s= ut+ t = 0 is when the ball left the ground, so it lands when t = 1.968 seconds (4s.f.) Choose the approprie equion of motion, based on the informion ou have and wh ou need to calcule: 1 s = ut+ Question: Derive the equion of the ph of the ball, assuming it starts the origin. Using s ut 1 = + with ux = 15cos 40 and u = 15sin 40 gives: ( ) ( ) x= t = t t 15cos 40 [1] and 15sin 40 4.9 [] x [1] t =. Substituting for t in [] gives 15cos 40 x = xtan 40 4.9 ( 15cos 40 ) The range of the ball is.6m (3 s.f.) The ph of the ball is the parabola = 0.839x 0.0371 x (3 s.f.)
the Further Mhemics network www.fmnetwork.org.uk V 07 1 1 REVISION SHEET MECHANICS 1 CONSTANT ACCELERATION & SUVAT EQUATIONS The main ideas are AQA Edx MEI OCR Introduction to the variables M1 M1 M1 M1 Using the variables M1 M1 M1 M1 Problems involving bodies or sstems acted upon b constant forces often begin b calculing the accelerion using Newton s second law; once the accelerion has been found, the become suv problems. The variables which appear in the SUVAT equions are: u = initial velocit v = velocit after t seconds a = accelerion t = time s = displacement (from the initial displacement) time t. Before the exam ou should know: 1. The five SUVAT equions, wh each letter stands for and when to use them. The are: v= u+, u+ v s = t, v = u + as 1 s = ut+, 1 s = vt. You should know the units of all the quantities in the SUVAT equions. Important things to remember: The SUVAT equions can onl be used for objects moving under a constant accelerion. This occurs whenever a constant force is applied to a bod for a period of time. If, over the course of a journe, the accelerion changes from one constant re to another constant - - re (e.g 5ms for seconds followed b -3ms for 5 seconds) the SUVAT equions must be applied to each leg of the journe separel. The final velocit for the first leg will be the initial velocit for the second leg. Be careful with units. Make sure th our units are consistent with one another. e.g. if accelerion is - given in ms it might be wise to give all displacements in m, all times in seconds and all velocities in ms. Get our signs right. In these problems the particle is alwas moving in a straight line. From the starting position ou should decide which wa along the line ou are going to specif as the positive direction and which wa the negive. From then on be consistent with our choice. Remember it is possible for a particle to have a positive velocit and a negive accelerion. Select the equion ou use appropriel. Work out which variables ou know and which variable ou need to do this. Tr to have a clear picture in our mind of wh is going on. Here are some examples of this. In them we ve specified right as positive and left as negive.
the Further Mhemics network www.fmnetwork.org.uk V 07 1 1 Example 1. If u = 3ms and v = 6ms when t = 3 seconds we expect the particle to have a positive accelerion negive direction positive direction - Example 3. If u = 4ms with a = -6ms then plugging s = 3m into the equion s = ut+ gives 3= 4t 3t or 3t 4t 3 = (3t 1)( t 3) =0. Which means th 1 t = or t = 3. This is because there are two occasions when the 3 particle is 3m to the right of the starting point. The first time when the particle has positive velocit, the second time after the particle has slowed to a stop and began moving in the opposite direction due to the leftwards accelerion. Standard Questions 1. A car deceleres from 4ms to rest in 5s. Assuming the decelerion is constant, calcule how for the car travels in this time. We know u = 4ms, v = 0ms, t = 5s. We want s. u+ v The equion to use is therefore s = t. 4 This gives s = 5 = 60m.. A ball is thrown verticall in the air with a speed of 0m s. It acceleres (downwards) - a re of 9.8m s. How long does it take to hit the ground? In this we have u = 0ms. (I ve designed up as being positive here.) We have a = 9.8m/s. We would like to know wh time t, s = 0. It looks as though s = ut+ 0.5 is the right choice of equion. We have 0 = 0t 4.9t = t( 0 4.9t). The values of t for which this is true are t = 0 and 0 t = = 4.08 to two decimal places. Clearl t = 0 4.9 is when the ball is thrown so the time we require is 4.08 seconds. Example. If u = 6ms and a = 3ms the particle begins b moving to the right. Its accelerion is in the opposite direction to this movement and so it s actuall a decelerion. After two seconds its velocit is zero (wh?). From then on it moves to the left accelering in th direction. Harder Questions In earl examples using the SUVAT equions ou are given three of the quantities u, v, a, t and s and are asked to calcule one of the other two. It s eas to do this b selecting the approprie equion. A more difficult question is one where given enough informion simultaneous equions involving two of the variables can be set-up. The following is an example of this: A ball is dropped from a building and falls with - accelerion10m s. The distance between floors is constant. The ball takes 0.5 secs to fall from floor 8 to floor 7 and onl 0.3 secs to fall from floor 7 to floor 6. Wh is the distance between the floors? The trick is to see th we have informion about two journes both starting floor 8. One is the journe from floor 8 to floor 7 which takes 0.5s the other is twice as long, the journe from floor 8 to floor 6 which takes 0.5 + 0.3 = 0.8s If we let u be the velocit floor 8. Then we have using s = u t+ 0.5, Journe 1: s = u 0.5 + 0.5 10 0.5 = 0.5u+ 1.5 Journe : s = u 0.8 + 0.5 10 0.8 = 0.5u+ 3. This gives simultaneous equions s 0.5u = 1.5 and s 0.5u = 3. which can be solved to discover s. (i.e. s = 1.95 m)
the Further Mhemics network www.fmnetwork.org.uk V 07 1 1 REVISION SHEET MECHANICS 1 VARIABLE ACCELERATION USING DIFFERENTIATION AND INTEGRATION The main ideas are AQA Edx MEI OCR Differentiion M M M1 M1 Integrion M M M1 M1 Differentiion in dimensions M M M1 M1 Integrion in dimensions M M M1 M1 There are two main ideas in this topic. It is: Using differentiion and integrion to obtain expressions for the displacement, velocit and accelerion from one another. You should be able to do this in one, two or three dimensions. Obtaining values of associed quantities such as speed and distance travelled. Using differentiion - a particle travelling in a straight line. Example: An object moves in a straight line, so th its displacement relive to some fixed origin time t is given b s = t 3 5t + 4. 1. Find expressions for its velocit and accelerion time t.. Calcule the velocit and accelerion of the object when t = 0 and when t = 1. 3. Wh is the displacement of the object when its velocit is zero?. ds 1. 3 s = t 5t +4so th v = = 3t 10t and dt dv a = = 6t 10. dt -. When t = 0, v = 3 0 10 0 = 0ms and a = 6 0 10 = 10ms and when - t = 1, v = 3 1 10 1 = 7ms and a = 6 1 10 = 4ms. Before the exam ou should know: Velocit is the re of change of displacement. Therefore to obtain an expression for a particle s velocit time t ou should differentie the expression for its displacement. Accelerion is the re of change of velocit. Therefore to obtain an expression for a particle s accelerion time t ou should differentie the expression for its velocit. Reversing the two ideas above, a particle s velocit can be obtained b integring the expression for its accelerion and a particle s displacement can be obtained b integring the expression for its velocit. In both cases this will introduce a constant of integrion whose value can be found if the particle s displacement or velocit is known some particular time. The above facts appl to both: 1. particles travelling in one dimension. In this case each of the displacement, velocit and accelerion is a (scalar valued) function of time, all of which can be differentied and integred in the usual wa.. particles travelling in two and three dimensions, when the displacement, velocit and accelerion are all vectors with components dependent on t (time). We differentie and integre such expressions in the usual wa, dealing with each component separel. There are several examples of this on this sheet. You should be comfortable with both column vector and i, j, k notion for vectors. 10 3. The velocit of the object time t is 3t 10 t = t(3t 10). This is zero when t = 0 or when t =. The 3 10 displacement of the particle when t = 0 is s = 4m and then displacement of the particle when t = is 3
the Further Mhemics network www.fmnetwork.org.uk V 07 1 1 Using integrion a particle travelling in a straight line. Example: An object is moving in a straight line with accelerion time t given b a = 10 6t. Given th when t = 1, s = 0 and v = 5, where s is the object s displacement and v is the object s velocit, find an expression for v and s in terms of t. Hence find out the displacement of the particle when it first comes to rest. v = a dt = ( 10 6t) dt = 10t 3t + c. But when t = 1, v = 5= 10 3+ c c=. So v = 10t 3t. c 3 s = v d t = (10t 3t ) dt = 5t t t+. But when t = 1, s= 0= 5 1 + c c=. So 3 s 5t t t = Using differentiion an example in two dimensions using column vector notion. Example: A girl throws a ball and, t seconds after she releases it, its position in metres relive to the point where she is standing is modelled b x 15t = + 16t 5t where the directions are horizontal and vertical. 1. Find expressions for the velocit and accelerion of the ball time t.. The vertical component of the velocit is zero when the ball is its highest point. Find the time taken for the ball to reach this point. 3. Wh is the speed of the ball when it hits the ground. 1. The velocit is obtained b differentiing (with respect to t) the components in the vector giving the 15 ball s position. This gives v =. The accelerion is obtained b differentiing (with respect to 16 10t 0 t) the components in the vector giving the ball s velocit. This gives a =. 10 16 8. The vertical component of the velocit is 16 10t. This is zero when t = = = 1.6 seconds. 10 5 3. The ball hits the ground when the vertical component of the balls position is zero. In other words when + 16t 5t = 0. Rearranging this as 5t 16t = 0and then using the formula for the solutions of a quadric we see th the solutions of this are t = 0.1 and t = 3.3(to s.f). Clearl the value we require 15 is t = 3.3. The velocit of the ball when t = 3.3 is and so the speed is 17 15 + ( 17) =.67 ms. Using Integrion and Newtons nd Law an example in dimensions with i, j notion. Example: A particle of mass 0.5 kg is acted on b a force, in Newtons, of F= t i+ tj. The particle is initiall rest and t is measured in seconds. 1. Find the accelerion of the particle time t.. Find the velocit of the particle time t. Newton s second law, F = ma gives th F= t i+ tj = 0.5a so th a= t i+ 4tj. 3 t We have th v = a d t = + c i + ( t + d) j where c and d are the so-called constants of integrion. 3 3 t We are told th the particle is rest when t = 0 and so c = d = 0. This gives v = i+ t j. 3
the Further Mhemics network www.fmnetwork.org.uk V 07 1 1 REVISION SHEET MECHANICS 1 VECTORS AND NEWTON S LAWS IN DIMENSIONS The main ideas are Example AQA Edx MEI OCR Resolving forces into components M1 M1 M1 M1 Motion on a slope M1 M1 M1 M1 Motion on a slope (including friction) M1 M1 M M1 Two people pull on separe light inextensible ropes, which are tached to the front of a stionar car of mass 900kg. The first pulls the rope an angle of 30 to the right of the direction of motion and the tension in the rope is 700N. The second pulls an angle of 0 to the left of the line of motion. Before the exam ou should know: 1. You must be confident with the use of vectors.. You must be able to draw clear diagrams showing forces. 3. You must be confident resolving forces into components, both horizontall and verticall, or parallel and perpendicular to a slope. a j i j R F i W = mg Wsinθ F 0 Wcosθ + 0 + R a = m 0 (a) With wh force is the second person 4. You must be confident using Newton s second law, F = ma pushing? (b) Wh is the accelerion of the car? (c) Assuming the forces remain constant, how fast will the car be moving after 3 seconds? θ Begin with a clear diagram: a F Diagram is a view from above 900kg 0 30 Direction of motion (a) There can be no force perpendicular to the direction of motion so, resolving perpendicular to the direction of motion: 700sin 30 Fsin 0 = 0 F = 100N (3s.f.) (b) Resolving in the direction of motion: From Newton s second law, 700cos30 + 103cos 0 = 900a a = 1.74ms (c) u = 0, a = 1.74, t = 3, v =? v = u+ v = 5.3ms 700N -
the Further Mhemics network www.fmnetwork.org.uk V 07 1 1 The previous example was solved b resolving parallel and perpendicular to the direction of motion separel. In the next example vector equions are used. Both methods are equall good. You should be happ with both. Example A block of mass 10kg is rest on a plane which is inclined 0 to the horizontal. A light, inelastic string is tached to the block, passes over a smooth pulle and supports a mass m which is hanging freel. The part of the string between the block and the pulle is parallel to the line of greest slope of the plane. A friction force of 0N opposes the motion of the block. (a) (b) (c) (a) Draw a diagram and mark on all the forces on the block and the hanging mass, including the tension in the string. Calcule the value of m when the block slides up the plane a constant speed and find the tension in the string and the normal reaction between the block and the plane. Calcule the accelerion of the sstem when m = 6 kg and find the tension in the string in this case. R T T T T j i 0N 0 W = 98N mg N (b) When the block is moving with constant velocit, the forces on the block must be in equilibrium. Using Newton s second law: 0 0 98sin 0 T 0 T = 53.5 N (3 s.f.) (from the i component) + + + = 10 0 R 98cos 0 0 0 R = 9.1 N (3 s.f.) (from the j component) Considering the forces on the hanging mass, 53.5 T = mg m= = 5.46 N 9.8 (c) As the block and the mass are connected bodies, the both experience the same accelerion. Let this accelerion be a. Using Newton s second law: Considering the 10 kg block, resolving parallel to the slope: T 0 98sin 0 = 10 a [1] Considering the 6 kg hanging mass: 6g T = 6 a [] Solving [1] and [] simultaneousl gives: - a = 0.330 ms (3 s.f.), T = 56.8 N (3 s.f.)