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1 . A particle of mass m is projected vertically upwards, at time t =, with speed. The particle is mv subject to air resistance of manitude, where v is the speed of the particle at time t and is a positive constant. (a) Show that the particle reaches its reatest heiht above the point of projection at time tan. (6) (b) Find the reatest heiht above the point of projection attained by the particle. (6) (Total mars). At time t =, a particle P of mass m is projected vertically upwards with speed, where is a constant. At time t the speed of P is v. The particle P moves aainst air resistsance whose manitude is modelled as bein mv when the speed of P is v. Find, in terms of, the distance travelled by P until its speed first becomes half of its initial speed. (Total 9 mars) 3. At time t =, a particle of mass m is projected vertically downwards with speed from a point above the round. At time t the speed of the particle is v and the manitude of the air resistance is modelled as bein mv, where is a constant. Given that speed. <, find, in terms of, and, the time taen for the particle to double its (Total 8 mars) Edexcel Internal eview

2 4. A lorry of mass M moves alon a straiht horizontal road aainst a constant resistance of manitude. The enine of the lorry wors at a constant rate, where is a constant. At time t, the lorry is movin with speed v. (a) Show that Mv = ( v). dt (3) At time t =, the lorry has speed 4 and the time taen by the lorry to attain a speed of 3 M is, where is a constant. (b) Find the exact value of. (7) (Total mars) Edexcel Internal eview

3 . (a) v m( + ) = m dt M A T dt = ( + v ) DM T = = tan v tan ) A DM A 6 (b) v m( ) mv dx M A H vv d dx= ( + v ) DM H ln( v + A ( ) H = ln + DM A 6 []. m mv = ma M A ( + v ) = v = dx M ± X v d DM A x = ( + v ) [ ln( + v )] (both previous) X = M A = 5 ln ln 4 = 8 ln 5 M A [9] Edexcel Internal eview 3

4 3. m mv = m d t M*AA dt = v DM* t = ln( v) + c Acao t =,v = u c = ln( u) M T = ln( u) ln( u) u = ln u DM A 8 [8] 4. (a) v M F u F = v B u ( ), = M v dt M (u v) = Mv * dt A 3 B Correct expression involvin the drivin force. M se of F = ma to form a differential equation. Condone sin errors. a must be expressed as a derivative, but could be any valid form. A earrane to iven form. Edexcel Internal eview 4

5 (b) T M dt = 3 T = M 4 3 v u v MA u + DM u v 4 M [ v u ln( u v)] 3 4 = A M u u u 3u Mu 3u = u ln + + u ln C = ln M Mu 9 = + ln 8 M 9 Hence = ln 8 A 7 M Separate the variables A Separation correct (limits not necessarily seen at this stae) DM Attempt a complete interation process A Interation correct M Correct use of both limits substitute and subtract. Condone wron order. M Simplify to find from an expression involvin a loarithm A Answer as iven, or exact equivalent. Need to see = lna + B [] Edexcel Internal eview 5

6 . This question followed a standard format and was answered very well by a lare proportion of candidates. A majority too the constant of interation approach rather than the use of definite interation, but enerally did so successfully in both parts (a) and (b). A small number of candidates were confused about the directions involved and often tried to justify a chane of sin part way throuh their worin in order to arrive at the iven answer in part (a). A small number of candidates attempted to use their answer to part (a) to enerate a differential dx equation in for part (b) rather than the simpler method of restartin from the beinnin and dt d v usin v. These candidates were rarely able to obtain an expression for v correctly, or to d x interate their expression successfully, and tended to rind to a halt with little proress made.. This proved to be a particularly accessible question, with many candidates offerin completely correct solutions. The most common errors involved incorrect sins in the initial differential equation, or problems in substitutin the limits and simplifyin the answer. Some candidates v interated to obtain ln( v ) and went on to wor with the loarithms of v neative quantities with no indication at all of any reconition that this should have concerned them. A few candidates started with a differential equation in v and t. Althouh it is possible to complete the solution by this route, very few of these attempts ot as far as a correct first interal, and almost all stopped wor before achievin a differential equation involvin x. 3. This was a standard question and candidates demonstrated a sound understandin of how to solve this sort of problem. Most were able to obtain a differential equation in v and t, reconised the need to separate the variables and interated to find an expression for t in terms of v. A few attempts considered upwards as positive but then failed to use and as the values of v, but most were competent and successful. Most attempts to solve the problem by writin the equation of motion as a second order differential equation in x and t and then oin on to differentiate x failed because they inored the need for a particular interal as part of the eneral solution. 4. The vast majority of candidates found the drivin force, correctly applied Newton s second law and were able to rearrane to obtain the iven result in part (a). In part (b) candidates were enerally able to separate the variables correctly. The most popular method for interatin the expression in v was to use partial fractions, althouh this caused a number of candidates to mae sin errors. The minority who used substitution (e.. x = u v) were enerally successful. There was a sinificant number of candidates who interated incorrectly, or who attempted to use interation by parts only to rind to a halt because they did not now how to complete the second stae of the interation, or worse still did not show any indication of the need for a second step in the interation. The vast majority of candidates preferred findin a constant of interation to usin definite interation, but those who interated correctly tended to obtain the final simplified result successfully. Edexcel Internal eview 6

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