PhysicsAndMathsTutor.com

Transcription

1 . A raindrop falls vertically under gravity through a cloud. In a odel of the otion the raindrop is assued to be spherical at all ties and the cloud is assued to consist of stationary water particles. At tie t = 0, the raindrop is at rest and has radius a. As the raindrop falls, water particles fro the cloud condense onto it and the radius of the raindrop is assued to increase at a constant rate. A tie t the speed of the raindrop is v. (a) Show that v + ( t + a) = g. (8) Find the speed of the raindrop when its radius is a. (7) (Total 5 arks). A spaceship is oving in a straight line in deep space and needs to increase its speed. This is done by ejecting fuel backwards fro the spaceship at a constant speed c relative to the spaceship. When the speed of the spaceship is v, its ass is. (a) Show that, while the spaceship is ejecting fuel, = c. (5) The initial ass of the spaceship is 0 and at tie t the ass of the spaceship is given by = 0 ( kt), where k is a positive constant. Find the acceleration of the spaceship at tie t. (4) (Total 9 arks) Edexcel Internal Review

2 . At tie t = 0 a rocket is launched fro rest vertically upwards. The rocket propels itself upwards by expelling burnt fuel vertically downwards with constant speed U s relative to the rocket. The initial ass of the rocket is M 0 kg. At tie t seconds, where t <, its ass is M 0( t) kg, and it is oving upwards with speed v s. (a) Show that U = 9.8 ( t) (7) Hence show that U > 9.6 () (c) Find, in ters of U, the speed of the rocket one second after its launch. (5) (Total 4 arks) 4. A otor boat of ass M is oving in a straight line, with its engine switched off, across a stretch of still water. The boat is oving with speed U when, at tie t = 0, it develops a leak. The water coes in at a constant rate so that at tie t, the ass of water in the boat is t. At tie t the speed of the boat is v and it experiences a total resistance to otion of agnitude v. (a) Show that. ( M + t) + v = 0. (6) Show that the tie taken for the speed of the boat to reduce to M U is. (6) The boat sinks when the ass of water inside the boat is M. Edexcel Internal Review

3 (c) Show that the boat does not sink before the speed of the boat is U. () (Total 4 arks) 5. A space-ship is oving in a straight line in deep space and needs to reduce its speed fro U to V. This is done by ejecting fuel fro the front of the space-ship at a constant speed k relative to the space-ship. When the speed of the space-ship is v its ass is. (a) Show that, while the space-ship is ejecting fuel, =. k (6) The initial ass of the space-ship is M. Find, in ters of U, V, k and M, the aount of fuel which needs to be used to reduce the speed of the space-ship fro U to V. (6) (Total arks) 6. At tie t = 0, a sall body is projected vertically upwards. While ascending it picks up sall drops of oisture fro the atosphere. The drops of oisture are at rest before they are picked up. At tie t, the cobined body P has ass and speed v. (a) Show that, while P is oving upwards, + v = g. d t (5) The initial ass of P is M, and = Me kt, where k is a positive constant. kt kt Show that, while P is oving upwards, ( ve ) ge d =. () Given that the initial projection speed of P is g k, (c) find, in ters of M, the ass of P when it reaches its highest point. (7) (Total 5 arks) Edexcel Internal Review

4 7. A rocket-driven car oves along a straight horizontal road. The car has total initial ass M. It propels itself forwards by ejecting ass backwards at a constant rate per unit tie at a constant speed U relative to the car. The car starts fro rest at tie t = 0. At tie t the speed of the car is v. The total resistance to otion is odelled as having agnitude kv, where k is a constant. Given that t < M, show that (a) = U kv, M t (7) v = U k k t. M (6) (Total arks) 8. A rocket is launched vertically upwards under gravity fro rest at tie t = 0. The rocket propels itself upward by ejecting burnt fuel vertically downwards at a constant speed u relative to the rocket. The initial ass of the rocket, including fuel, is M. At tie t, before all the fuel has been used up, the ass of the rocket, including fuel, is M( kt) and the speed of the rocket is v. (a) Show that ku = g. kt (7) Hence find the speed of the rocket when t =. k () (Total 0 arks) Edexcel Internal Review 4

5 9. A rocket-driven car propels itself forwards in a straight line on a horizontal track by ejecting burnt fuel backwards at a constant rate kg s and at a constant speed U s relative to the car. At tie t seconds, the speed of the car is v s and the total resistance to the otion of the car has agnitude kv N, where k is a positive constant. When t = 0 the total ass of the car, including fuel, is M kg. Assuing that at tie t seconds soe fuel reains in the car, (a) show that = U kv, M t (7) find the speed of the car at tie t seconds, given that it starts fro rest when t = 0 and that = k = 0. (6) (Total arks) 0. A spaceship is oving in deep space with no external forces acting on it. Initially it has total ass M and is oving with speed V. The spaceship reduces its speed to V by ejecting fuel fro its front end with a speed of c relative to itself and in the sae direction as its own otion. Find the ass of fuel ejected. (Total arks) Edexcel Internal Review 5

6 . (a) dr = r = t + a B ( + δ)( v + δv) v = gδt A v + = g D A = 4πr (ρ) A B d d 4 r g v v g π ρ 4πr ρ r D v + d t t+ a = g * A 8 t+a ln( t+a) ln( t+a) R= e = e = e = ( t+ a) A v( t+ a) = g ( t+ a) D v(t + a) = g(t + a)4 A 4 4 = 0, = 0 = D 4 t v C ga t+ a= a D 4 ga 0ag v= 4 g( a) 4 = A 7 7a 7 [5]. (a) v = (+δ)(v +δv) ( δ)(c v) A v = v + δv + vδ+ cδ vδ δv = cδ d v c D A 5 = 0k B = c = 0k c0k = kt D 0 ( kt) ( ) ck = A 4 [9] Edexcel Internal Review 6

7 . (a) g δt = ( + δ)(v + δv) + δ(u v) v g δt = v + δv + vδ + U δ vδ v A d v g = + U A = M 0 ( t) = M 0 B M 0 g( t) = M 0 ( t) M 0U U g( t ) = ( t) U 9.8 = * ( t) A 7 U > 0 when t = > 0 U > 9.6* A (c) U v = 9.8 ( t) = U ln( t) 9.8t + C A t = 0, v = 0: 0 = U ln + C C = U ln so, v = U ln 9. 8t ( t) t = : v = U ln 9.8 A 5 [4] 4. (a) ( + δ)(v + δv) v = vδt A + v = v = ; = M + t B; B ( M + t) + v = 0* DA 6 Edexcel Internal Review 7

8 = t) u [ ln v] [ ] T u = ln( M + t) 0 DA (M + T ) ln = ln M D M T = ( ) * DA 6 v ( M + (c) Sinks at T S = M M Reaches speed U at T = ( ) Since ) <, T < T S I.e. Reaches speed U before it sinks Acso ( [4] 5. (a) v ( + )( v+ v) + ( )(K + v+ v) A v = ~ v + δv + vδ Kδ vδ Kδ δv In the liit, as t 0, = K A 6 Edexcel Internal Review 8

9 V M = U K ln ln M = (V U) A K = ln (V U) M K = V U K M e A Aount of fuel = M = M e V U K A 6 [] 6. O δ v + δv v + δ (a) ( + δ)(v + δv) v = gδt A,, 0 / / v + δv + vδ + / δ / / δv/ / / v = gδt + v = g A 5 = Me kt = kme kt B Hence Me kt + kv. kme kt = Me kt g d ( ve kt kt ) = ge A Edexcel Internal Review 9

10 (c) ve kt = g e kt g kt = e (+c) k g g t = o, v = c = k k e kt = A = 0 A Hence = Me kt = M A 7 [5] 7. (a) v v + v v U kv + t t + t + ve Ipulse oentu: kvδt = ( + δ)(v + δv) + ( δ)(v µ) v A kvδt = v + δv + δv δµ + δµ v δ v δ kv = + µ δt δt liit as δt 0: kv = µ d t d t d = = M t B kv = (M ) µ µ kv = (*) A 7 M t Edexcel Internal Review 0

11 t v µ kv A t µ [ln( M t)] 0 = [ln( µ kv)] 0 k A k (ln(m t) lnm) = ln(µ kv) lnµ k M t µ kv = M µ µ t k v = ( ) k (*) M A 6 = 0 M t 0 [] 8. (a) v t + v + v v u ( + δ) (v + δ) + ( δ) (v u) v = gδt A ( ee) v + δv + vδ vδ + uδ v = gδt + u d t d t = g A = M( kt) = km B M( kt) + u( km) = M( kt)g d v ku = kt g (*) A 7 v = k ku kt g 0 u ln ( kt) gt 0 A = [ ] k = u ln ( ) g k A [0] Edexcel Internal Review

12 9. (a) ( + δ)(v + δv) + ( δ )(v U) v = kvδ A A + U = kv A = M t B d v (M t) = U kv U kv = M t (*) (no incorrect working seen) A cso 7 0 Separating variables: = U v M 0t or equivalent Integrating: ln (U v) = ln (M 0 t ) (+ c) A 0 0 Using liits correctly: [ ] v = [ ] t applied or t = 0, v = 0 to find c U [c = ln ] M U M Coplete ethod to find v [ ln = ln ] U v M 0t 0Ut v = M A 6 [] 0. v (Tie δt) v+ δv v + c + δ δ ( + δ)(v + δv) + ( δ)(v + c) v = 0 A,, 0 c = 0 reducing equation V V d v = c M separating variables Edexcel Internal Review

13 v V V = v = c ln c ln M M integration applying liits A = V c Me eliinating ln and = Fuel used = M = M( V - c e ) A [] Edexcel Internal Review

14 . Many candidates knew exactly what to do here and there were a good nuber of fully correct solutions. Part (a) required a standard technique which ost candidates followed successfully to the given answer. It was pleasing to see the steps required were clearly set out. A few were unable to ake a correct start and then attepted to fudge the given answer. Many copletely correct solutions to part were seen. Most were able to find the correct Integrating Factor, solve the differential equation correctly, reeber the constant and use initial conditions to find v correctly. Soe candidates attepted to put in the initial conditions at the start before attepting to solve their differential equation, a costly error, as this lost the all of the arks in this part of the question.. The ajority of candidates recognise that they ust start variable ass questions by considering the change in oentu over a sall tie interval (in this question the change in oentu was zero!). They ust put the total ass at the beginning of the interval as and that of the rocket at the end of the interval as + δ (in the case of rocket questions, δ is negative). Candidates who siply try to eorise the solution to one version of this question and then try to adapt it are alost invariably unsuccessful. Although a significant nuber struggled with the first part, there was ore success with part. The easiest ethod was to use the chain rule to split up /, but soe candidates successfully integrated the result fro part (a) by separating the variables and then differentiating the result with respect to t.. In order to gain all of the arks in part (a), it was necessary to consider the change in oentu in a tie δt. Soe candidates obviously consider this to be a standard piece of bookwork, whereas others are clearly trying to work it out in the exa, in soe cases considering M 0 to be the ass at a general tie t. Candidates trying to use the given answer to correct their signs should beware of altering the wrong ones! Edexcel Internal Review 4

15 In part, any candidates gave the very siple explanation involving the acceleration at tie t = 0 being positive. Others considered the general expression for acceleration and struggled to produce the given inequality. Most candidates were successful with part (c). 4. Candidates who were successful in part (a) could either consider the change of oentu in a tie δt or else use Newton s Second Law and consider the rate of change of oentu i.e. d( v). Soe candidates tried to do this part fro eory, often apparently considering the alternative proble of a rocket ejecting fuel. Other difficulties included taking as t or else putting U in the general equation for v and t. Part was usually successfully done, ostly by separating the variables. Many candidates argued part (c) correctly, but others copared appropriate values of t or or v but did not then state any conclusion. 5. Candidates who attepted part (a) of this question fro first principles were often successful, with the possible exception of their signs. Those who tried to fit it to a solution that they had et before had probles, often caused by the inclusion of an ipulse ter which should not have been there. Many candidates gained four arks for the second part of the question, starting fro the printed answer but issing out the subtraction of the final ass fro M. Soe however worked with the answer that they had obtained for part (a). Others got confused over which speed went with which tie and often introduced a speed of zero. 6. This was found to be reasonable straightforward as a question on variable ass. The derivation of the given differential equation was generally good with nearly all correctly considering a sall tie interval appropriately. Parts and (c) were also generally well done. In ost cases, the question proved to be a good source of arks. 7. In part (a), any candidates proved the result correctly, deriving the result fro first principles as required. Those who did not either oitted ters or confused signs, writing δ instead of (-δ) and / = (but then writing = M t since that was in the given answer) In part, since the differential equation was given, candidates were able to progress and the standard of integration was pleasing, with any correct solutions seen. 8. This was for the ost part done extreely well. It was very pleasing to see the derivation of the equation of otion done fro first principles accurately (with the δ ter having the correct sign). Also the solution of the differential equation in part was also generally done fully accurately. Edexcel Internal Review 5

16 9. The derivation of the differential equation was often very good, with the ajority of candidates working, often successfully, fro first principles. In part which was well answered by any of the candidates who had a correct strategy, a significant nuber of candidates ade it uch harder for theselves by not substituting for and k iediately. Soe candidates, too, used an integrating factor approach to solve the differential equation; in this case leaving and k ade it quite a challenge but it was good to see soe succeed. 0. No Report available for this question. Edexcel Internal Review 6