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1 . Two smooth niform spheres S and T have eqal radii. The mass of S is 0. kg and the mass of T is 0.6 kg. The spheres are moving on a smooth horizontal plane and collide obliqely. Immediately before the collision the velocity of S is m s and the velocity of T is m s. The coefficient of restittion between the spheres is 0.5. Immediately after the collision the velocity of S is ( i + j) m s and the velocity of T is (i + j) m s. Given that when the spheres collide the line joining their centres is parallel to i, (a) find (i), (ii). (6) After the collision, T goes on to collide with a smooth vertical wall which is parallel to j. Given that the coefficient of restittion between T and the wall is also 0.5, find the angle throgh which the direction of motion of T is deflected as a reslt of the collision with the wall, (5) (c) the loss in kinetic energy of T cased by the collision with the wall. () (Total 4 marks). A fixed smooth plane is inclined to the horizontal at an angle of 45. A particle P is moving horizontally and strikes the plane. Immediately before the impact, P is moving in a vertical plane containing a line of greatest slope of the inclined plane. Immediately after the impact, P is moving in a direction which makes an angle of 0 with the inclined plane, as shown in the diagram above. Find the fraction of the kinetic energy of P which is lost in the impact. (Total 6 marks) Edexcel Internal Review

2 . Two small smooth spheres A and B, of mass kg and kg respectively, are moving on a smooth horizontal plane when they collide. Immediately before the collision the velocity of A is (i + j) m s and the velocity of B is i m s. Immediately after the collision the velocity of A is j m s. (a) Show that the velocity of B immediately after the collision is j m s. () Find the implse of B on A in the collision, giving yor answer as a vector, and hence show that the line of centres is parallel to i + j. (4) (c) Find the coefficient of restittion between A and B. (6) (Total marks) 4. A small ball is moving on a horizontal plane when it strikes a smooth vertical wall. The coefficient of restittion between the ball and the wall is e. Immediately before the impact the direction of motion of the ball makes an angle of 60 with the wall. Immediately after the impact the direction of motion of the ball makes an angle of 0 with the wall. (a) Find the fraction of the kinetic energy of the ball which is lost in the impact. (6) Find the vale of e. (4) (Total 0 marks) 5. A smooth niform sphere S of mass m is moving on a smooth horizontal plane when it collides with a fixed smooth vertical wall. Immediately before the collision, the speed of S is U and its direction of motion makes an angle α with the wall. The coefficient of restittion between S and the wall is e. Find the kinetic energy of S immediately after the collision. (Total 6 marks) Edexcel Internal Review

3 6. U A m B m U L Figre Two small smooth spheres A and B, of eqal size and of mass m and m respectively, are moving initially with the same speed U on a smooth horizontal floor. The spheres collide when their centres are on a line L. Before the collision the spheres are moving towards each other, with their directions of motion perpendiclar to each other and each inclined at an angle of 45 to the line L, as shown in Figre. The coefficient of restittion between the spheres is. (a) Find the magnitde of the implse which acts on A in the collision. (9) d WALL L Figre The line L is parallel to and a distance d from a smooth vertical wall, as shown in Figre. Find, in terms of d, the distance between the points at which the spheres first strike the wall. (5) (Total 4 marks) Edexcel Internal Review

4 7. A small smooth sphere S of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point A which is at a distance a from a smooth vertical wall. The sphere S hangs at rest in eqilibrim. It is then projected horizontally 7ga towards the wall with a speed. 5 7ga (a) Show that it strikes the wall with speed 5 (4) mga Given that the loss in kinetic energy de to the impact with the wall is, 5 find the coefficient of restittion between S and the wall. (7) (Total marks) 8. Two smooth niform spheres A and B have eqal radii. Sphere A has mass m and sphere B has mass km. The spheres are at rest on a smooth horizontal table. Sphere A is then projected along the table with speed and collides with B. Immediately before the collision, the direction of motion of A makes an angle of 60 with the line joining the centres of the two spheres. The coefficient of restittion between the spheres is. (a) Show that the speed of B immediately after the collision is. 4( k + ) (6) Immediately after the collision the direction of motion of A makes an angle arctan ( ) with the direction of motion of B. Show that k =. (6) (c) Find the loss of kinetic energy de to the collision. (4) (Total 6 marks) Edexcel Internal Review 4

5 9. A small smooth ball of mass kg is falling vertically. The ball strikes a smooth plane which is inclined at an angle α to the horizontal, where tan α = 4. Immediately before striking the plane the ball has speed 0 m s. The coefficient of restittion between ball and plane is. Find (a) the speed, to significant figres, of the ball immediately after the impact, (5) the magnitde of the implse received by the ball as it strikes the plane. () (Total 7 marks) 0. Q P A smooth niform sphere P is at rest on a smooth horizontal plane, when it is strck by an identical sphere Q moving on the plane. Immediately before the impact, the line of motion of the centre of Q is tangential to the sphere P, as shown in the diagram above. The direction of motion of Q is trned throgh 0 by the impact. Find the coefficient of restittion between the spheres. (Total marks). Edexcel Internal Review 5

6 . m s A B.5 m s kg kg Two smooth niform spheres A and B of eqal radis have masses kg and kg respectively. They are moving on a smooth horizontal plane when they collide. Immediately before the collision the speed of A is.5 m s and the speed of B is. m s. When they collide the line joining their centres makes an angle α with the direction of motion of A and an angle β with the 4 direction of motion of B, where tan α = and tan β = 5 as shown in the diagram above. (a) Find the components of the velocities of A and B perpendiclar and parallel to the line of centres immediately before the collision. (4) The coefficient of restittion between A and B is. Find, to one decimal place, the speed of each sphere after the collision. (9) (Total marks). X 45 m A m B Edexcel Internal Review 6

7 The diagram above represents the scene of a road accident. A car of mass 600 kg collided at the point X with a stationary van of mass 800 kg. After the collision the van came to rest at the point A having travelled a horizontal distance of 45 m, and the car came to rest at the point B having travelled a horizontal distance of m. The angle AXB is 90. The accident investigators are trying to establish the speed of the car before the collision and they model both vehicles as small spheres. (a) Find the coefficient of restittion between the car and the van. (5) The investigators assme that after the collision, and ntil the vehicles came to rest, the van was sbject to a constant horizontal force of 500 N acting along AX and the car to a constant horizontal force of 00 N along BX. Find the speed of the car immediately before the collision. (9) (Total 4 marks). U T S d A small smooth niform sphere S is at rest on a smooth horizontal floor at a distance d from a straight vertical wall. An identical sphere T is projected along the floor with speed U towards S and in a direction which is perpendiclar to the wall. At the instant when T strikes S the line joining their centres makes an angle α with the wall, as shown in the diagram above. Each sphere is modelled as having negligible diameter in comparison with d. The coefficient of restittion between the spheres is e. (a) Show that the components of the velocity of T after the impact, parallel and perpendiclar to the line of centres, are U(l e) sin α and U cos α respectively. (7) Edexcel Internal Review 7

8 Show that the components of the velocity of T after the impact, parallel and perpendiclar to the wall, are U (l + e) cos α sin α and U[ ( + e) sin α] respectively. (6) The spheres S and T strike the wall at the points A and B respectively. Given that e = and tan α = 4, (c) find, in terms of d, the distance AB. (5) (Total 8 marks) 4. d W α B d C A small ball Q of mass m is at rest at the point B on a smooth horizontal plane. A second small ball P of mass m is moving on the plane with speed and collides with Q. Both the balls are smooth, niform and of the same radis. The point C is on a smooth vertical wall W which is at a distance d from B, and BC is perpendiclar to W. A second smooth vertical wall is perpendiclar to W and at a distance d from B. Immediately before the collision occrs, the direction of motion of P makes an angle α with BC, as shown in the diagram above, where tan α = 5. The line of centres of P and Q is parallel to BC. After the collision Q moves towards C with speed 5. (a) Show that, after the collision, the velocity components of P parallel and perpendiclar to CB are 5 and 5 respectively. (4) Find the coefficient of restittion between P and Q. (c) Show that when Q reaches C, P is at a distance 4 d from W. () () Edexcel Internal Review 8

9 For each collision between a ball and a wall the coefficient of restittion is. Given that the balls collide with each other again, (d) show that the time between the two collisions of the balls is 5d, (4) (e) find the ratio d :d. (5) (Total 8 marks) 5. A smooth sphere S is moving on a smooth horizontal plane with speed when it collides with a smooth fixed vertical wall. At the instant of collision the direction of motion of S makes an angle of 0 with the wall. The coefficient of restittion between S and the wall is. Find the speed of S immediately after the collision. (Total 6 marks) 6. A B 0 A smooth niform sphere A, moving on a smooth horizontal table, collides with a second identical sphere B which is at rest on the table. When the spheres collide the line joining their centres makes an angle of 0 with the direction of motion of A, as shown in the diagram above. The coefficient of restittion between the spheres is e. The direction of motion of A is deflected throgh an angle θ by the collision. Show that ( + e) tanθ =. 5 e (Total 0 marks) Edexcel Internal Review 9

10 . (a) S 0. kg T 0.6 kg v w 0.v 0.6w= 0. M A v w= + = M A v+ w= 4 w=, v= (i) = i+ j (ii) = + i j A A 6 v v = 0.5 B tan θ = 0.5 tan θ = their v M θ = 6.6 A their θ + 45 M o Defln angle = = 7.6 A 5 (c) KE Loss = x 0.6 x {( ) ( v )} + + M A = 0.5 J A [4]. CLM along plane: v cos 0 = cos 45 M A v = A Fraction of KE Lost = m m mv = m m m = M M A [6]. (a) CLM: (i + j) + i = j + v M A Edexcel Internal Review 0

11 v = j m s A I = (j (i + j)) M A = ( i j) Ns A Since I acts along l.o.c.c., l.o.c.c is parallel to i + j B 4 (c) Before After A: B : A: B : ( i + j). ( i + j) j. j. j. ( i + j) ( i + j) = + = ( i + j) + M A NIL: e = = DM A 6 5 [] 4. (a) 60 U 0 V cos 60 = v cos 0 MA = v A KE lost = m( v ) M v Fraction of KE lost = DM = = or at least sf ending in 7 A 6 M A or ( e ) 4 Resolve parallel to the wall Alt: reasonable attempt at eqation connecting two variables Correct as above or eqivalent eqation correct Edexcel Internal Review

12 A M in terms of v or v.v. not necessarily simplified. or ration of the two variables correct expression for KE lost DM expression in one variable for fraction of KE lost cold be /v as above A cao The first three marks can be awarded in if not seen in (a) v sin 0 e = sin 60 MA v =. DM = A 4 M Use NIL perpendiclar to the wall and form eqation in e A Correct nsimplified expression as above or e sin 60 = v sin 0 or eqivalent DM Sbstitte vales for trig fnctions or se relationship from (a) and rearrange to e =... A cao accept decimals to at least sf The first two marks can be awarded in (a) [0] 5. ( ) cosα = vcosθ M A ( ) esinα vsinθ M A ( cos α sin α) v = + e M ( α α ) KE = m cos sin + e A [6] Edexcel Internal Review

13 6. (a) Form: I = m(v + ) m m CLM mv mv ( ) : = + M A = v + v NIL : e = = v + v M A M A = v M A I = m( + ) = m A 9 v v = (separation speed) M Time to wall = d d M A d Separation = = d M A 5 [4] Edexcel Internal Review

14 7. (a) a v a Energy: 7ga m v = mg.a( cosθ) M A 5 Using θ = π 7ga & solve: v = M A 4 5 v 0 Impact: = evsin0 M A KE loss = m (v sin 0 e v sin 0) cos 0 mga + mv m = M A 5 [Using = v cos 0 if necessary & redcing to eqation in (m, g, a) e alone] mga 7ga = m.. ( e ) A Solve for e: e = M A 7 [] Edexcel Internal Review 4

15 8. (a) cos 60 0 sin 60 0 m km v w sin 60 CLM( ) : m cos 60 = mv + kmw NLI: cos 60 = w v Solve for w: ( + k)w = ( + ) M A M A M w = 4( k + ) A 6 v sin 60 Solve for v v = ( k) 4( k + ) M A tan θ = = sin 60 M A v 4( k + ) =. ( k) Solve k: k = M A 6 (c) k = v = 4, w = B KE loss = m m. + m. +. m. M A = m = m A 4 [6] Edexcel Internal Review 5

16 9. (a) sin v a cos sin Cpt along plane = 0sinα after impact = 0 5 I B = 6 V = e 0 cosα M A 4 = 0 = 4 5 Speed = = 7. ms ( sf) M A 5 I = (4 8) = 6Ns M A [7] 0. a sin θ = a θ = 0 M A sinθ O V V sinθ O cosθ O Edexcel Internal Review 6

17 v + v = cos θ M A v + v = e cos θ M A sinθ = tan (θ + 0 ) (or eqivalent) v M A Prodcing an eqation in e only M (dep. On all previos M s) e = A c.s.o []. (a) Before y v y x A B v x A: y =.5 sin α = = (ms ) either M x =.5 cos α =.5 5 =.5 (ms ) both A B: v y =. sin β =. =. (ms ) either M v x =. cos β =. 5 = 0.5 (ms ) both A 4 After A x B w LM x + w = 0.5 (=.5) M A ft NEL w x = (=) M A ft Solving x = 0.5, y =.5 M solving for either M A Speed of A is ( ) = 4.5. (ms ) M either M A Speed of B is (. +.5 ) =.69.9 (ms ) A 9 Note: Not d.p. loses maximm of one mark []. (a) Before After C V C V x v v LM 600 = 800x M A NEL x = e M A e = 0.75 A 5 Van NL 500 = 800a M 0 = x , x = 56.5 (x = 7.5) M, A Car NL 00 = 600a M Edexcel Internal Review 7

18 0 = v 0.5, v = M, A From (a) NEL = = 0 M V = 0 +, V = (ms ) cao M, A 9 [4]. (a) U sin α U cos α o o w v U cos α No implse perpendiclar to line of centres velocity perpendiclar to line of centres nchanged = U cos α (*) B : CLM U sin α = v + w M A NLI eu sin α = w v M A v = U sin α ( e) M A 7 ( ) Component perpendiclar to wall = v sin α + U cos α cos α M = U sin α ( e) + U cos α = U(sin α e sin α + sin α) M = U [ sin α ( + e)] (*) A Component parallel to wall = U cos α sin α v cos α M = U cos α sin α U sin α cos α ( e) A = U cos α sin α ( + e) (*) A 6 Edexcel Internal Review 8

19 (c) T S θ 5 α α T 0 A X B e =, tan α = 4 Component perpendiclar to wall = U ( ) = 5 0 Component parallel to wall = 4 5 U = 5 5 Distance of A from X = d cot θ = BX = d cot θ cot θ = = 7 7 Total distance AB = = 40d 4d 4d + 7 4d B B M A A 5 [8] 4. (a) 5 5 P before: cos α =, sin α = B, B 0 v 5 m m m m PCLM ( ) m = mv + m, v =, i.e. // CB M A NLI e = v v e =, i.e. e = M, A Edexcel Internal Review 9

20 d 5d (c) Q C t = = B / 5 5d P travels d = in direction CB M 5 d 4d P is d + = from w ( ) A c.s.o (d) A fter hitting w, Q has speed in direction CB 0 Velocity of Q relative to P in direction CB is 0 B M 4 4d Time for Q to travel d is: 40d 0 = A 40d Total time between collisions is: 5d + 5d = ( ) A c.s.o 4 (e) 5d For collision to occr P mst travel d and d in time d d d t = = 5 / 5 B 5 d d velocity is, 4d t = = 4 5 / 4 5 B, B 6d Total time is 5 =, 5 d M d = 5d, i.e. d :d = :5 A 5 [8] Edexcel Internal Review 0

21 5. 0 v v cos α = cos 0 v sin α = sin 0 sqaring and adding, v = ( + ) v = M A M A M A [6] 6. cos 0 0 v + v = cos 0 M A A m B m v + v = e cos 0 M A sin 0 0 v v sin 0 0 sbtracting, v = ( e) A 4 Edexcel Internal Review

22 tan θ = tan (α 0 ) = tanα tan0 + tanα tan0 M tan α = sin 0 = v ( e) M A tan θ = ( e) + ( e) M = ( + e) 5 e (*) A 0 [0] Edexcel Internal Review

23 . In part (a) many candidates were able to obtain correct eqations by applying the conservation of momentm and Newton s experimental law parallel to the line of centres. A common mistake, however, was to solve the eqations and give the speed of the spheres rather than the velocities as reqired in the qestion. The need to move from a qestion posed in vectors to scalar eqations cased difficlties for some candidates. Many prodced a momentm eqation in i and j, rather than confining themselves to consideration of the components parallel to the line of centres. The j component did not always cancel ot. Those candidates who introdced vectors to Newton s Experimental law were penalised for this significant error. In part the majority of candidates were able to se the coefficient of restittion to find the velocity of the sphere after the collision with the wall and to find the angle between the wall and the path after impact. Far fewer identified and fond the correct angle of deflection. In part (c) the majority of candidates fond the loss in kinetic energy correctly. Errors were sally de to failre to find v correctly from their vectors, or the se of v rather than v in attempting to find the kinetic energy.. Most candidates started by applying CLM parallel to the plane to obtain an eqation connecting the speed of approach and the speed of rebond, which enabled them to form an expression for the fraction of kinetic energy lost. Many candidates, who also considered the components of the speeds perpendiclar to the srface and sed the impact law to find the vale of e, did eventally reach a correct conclsion, bt in most instances this method took longer becase they were doing more work than necessary. Some candidates fond the amont of kinetic energy lost and did not go on to complete the qestion. There was also evidence of some confsion amongst candidates who worked only with the components of velocity perpendiclar to the plane and did not express their change in kinetic energy as a fraction of the total initial kinetic energy candidates shold be discoraged from thinking of energy as resolvable, as this has clearly led to some misconceptions.. In part (a) the majority of candidates demonstrated their dislike of vectors by calclating the i and j components of the velocity of B in two separate eqations. Many candidates lost marks becase they only considered the j component. In their attempts to derive the given reslt withot the se of vectors some candidates created some nconvincing notation and argments. Most candidates in part nderstood that implse cases a change in momentm, bt there were many sign errors in finding the implse, with large nmbers of candidates arriving at the negative of the reqired reslt when considering the implse on A. Eqally, some candidates started by finding the implse on B and did not take the final step to answer the qestion. When candidates had fond an implse acting in the right direction they were sally able to draw the correct conclsion abot the line of centres of the two spheres, althogh here too some of the presentation was rather mddled and the examiner was often left to interpret the candidate s se of the word it to apply to whatever the candidate had in mind. Part (c) challenged many candidates, and was a good indicator of the level of nderstanding of what was happening in this collision. Only the stronger candidates appreciated the need to resolve their velocities parallel to the line of centres, and for these candidates the initial velocity of A proved the most difficlt to deal with. This was largely becase the sal approach was to draw a diagram and se a trigonometry, rather than sing the scalar prodct of the vectors. Edexcel Internal Review

24 4. Candidates were generally able to obtain a correct relationship between the speed of the ball before and after the collision by resolving parallel to the wall. The vast majority then went on to find the loss in KE in one variable, bt a significant proportion of these candidates did not then go on to find the fraction of KE lost. Some candidates fond the loss in KE sing only the components perpendiclar to the wall. In most cases, however, these candidates then incorrectly went on to find the fraction of KE lost by sing the perpendiclar component of the initial velocity rather than the initial speed in their expression for the initial KE. Most candidates were able to obtain a correct eqation in e and sbstitte to find a correct vale for e. 5. Completely correct soltions to this qestion were common. Even those who initially decided that momentm might be conserved perpendiclar to the wall sally realised the error of their ways and corrected their soltions. A few candidates correctly sed conservation of momentm and Newton s experimental law bt then were nable to proceed to an expression for the kinetic energy. 6. (a) This was a standard qestion that many candidates completed sccessflly. Many however forced the problem to conform to their preconceived idea of this type of qestion. A diagram, preferably in the orientation given in the qestion, with velocity components clearly shown, wold have helped candidates to get the signs correct in their eqations. Some candidates sed the velocities in their calclation of implse, rather than the components along the line of centres. Despite problems with part (a), most candidates sed a correct method with often only a single mark being lost de to errors from part (a). 7. This was a challenging qestion for all bt the strongest candidates. Part (a) was accessible althogh some did not realise it was a circlar motion/energy qestion. Part was not recognised as an obliqe impact on most scripts, there was no attempt to resolve the before and after velocities into components and candidates sed v = e with resltant velocities instead of components perpendiclar to the wall. 8. This qestion was a very good sorce of marks with many completely correct soltions. Candidates were confident in the se of conservation of momentm and Newton s law of restittion and coped well with the reslting algebra. There were some errors in (c) with candidates sing m for both masses instead of m and km with k = ½ as printed. Several saved some calclation by realising that all the energy loss came from the components of velocity along the line of centres 9. (a) Candidates cold sally draw the plane as described bt often misplaced the angle α between the vertical and the normal to the plane. Many candidates drew the problem with the plane horizontal or sed components in ndefined directions leaving it to the examiner to work ot what was intended. Finding components along and perpendiclar to the plane was attempted by most candidates and those who sqared the components to find the resltant were sally more sccessfl than those who tried to find the angle to Edexcel Internal Review 4

25 the plane at which the ball rebonded. This part was very badly done. Most candidates did not se components of the velocity perpendiclar to the plane sing instead the initial and final speeds of the particle. Candidates who realised that components were needed often failed to take into accont the directions involved reslting in an answer of Ns rather than 6Ns. 0. Most cold score something on this nstrctred qestion bt only the best were able to get all the way throgh to a correct answer. The main problem was finding the initial angle of incidence bt weaker candidates failed to observe that velocities perpendiclar to the line of centres were nchanged by the impact.. Nearly all candidates completed part (a) sccessflly and writing ot the initial components of velocity in this way helped candidates to sort ot the directions reqired in part. Apart from a few sign errors seen in the eqations for conservation of energy and for Newton s experimental law, the majority of candidates cold find correctly the components of the velocity parallel to the line of centres after the impact. A few candidates stopped there, failing to nderstand that finding speeds reqires combining two components of velocity.. This qestion, althogh maniplatively qite straight forward, was in an nsal form and it was noticeable that many candidates needed two or three starts before recognising that this was an obliqe impact qestion and redcing the problem to a nmber of variables that cold be solved. However the great majority were able to do this and fll marks were often gained for the qestion. In part the majority sed eqations of motion with v = +as to obtain the two components of velocity after impact bt correct work-energy eqations were also sed.. Part (a) was generally well done: this was a fairly rotine problem and candidates showed confidence and good knowledge in the principles involved. Part was a slightly nsal qestion and in fact only reqired resolving the two components of the velocity fond in (a), bt the majority of candidates failed to see this: many simply left this part blank, others tried to find the velocity of the other ball. In part (c) many failed to see the connection with part and made little progress. A few managed to get throgh to the end correctly, bt the majority either gave p or made little progress. 4. The first two parts of this qestion were sally answered well. Candidates knew how to resolve the velocity of P into two components and they applied the principle of conservation of linear momentm in (a) and Newton s law of impact in along the line of centres. Sometimes the direction of the velocity component of P after the impact was not clearly explained and this sometimes led to an incorrect answer of 0.4 instead of 0.8 in part. Most candidates knew how to answer part (c) bt the last two parts were more demanding and reqired clear and carefl thinking to achieve the correct answers. A common error was to assme that the velocity of P perpendiclar to CB was the same both before and after P hit the second wall. Edexcel Internal Review 5

26 5. No Report available for this qestion. 6. No Report available for this qestion. Edexcel Internal Review 6