A level Exam-style practice

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A level Exam-style practice 1 a e 0 b Using conservation of momentum for the system : 6.5 40 10v 15 10v 15 3 v 10 v 1.

5 J d Work done by friction = loss of kinetic energy 1 F 10 1.5 F 5.

1 A level Exam-style practice 1 a e 0 b Using conservation of momentum for the system : v 15 10v 15 3 v 10 v 1.5 m s 1 c Kinetic energy lost = initial kinetic energy final kinetic energy J d Work done by friction = loss of kinetic energy 1 F F 5.65 N (1) Friction: F R R = 10g So F 10g () From equations (1) and (): 10g So g Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 1

a Power 0 kw 0 000W Power Tv 0 000 So T v Using F ma T 10 v 1400a 0 000 10 v 1400a v When v : 0 000 10 1400a 618 1400a 618 So a 0.

8 sin 6 434N The resistance to the car s motion is: 10 v 10 0 90 N The total force down the plane is greater than the resistive force, so there is a net

2 a Power 0 kw 0 000W Power Tv So T v Using F ma T 10 v 1400a v 1400a v When v : a a 618 So a 0.44 m s 1400 b Power W Power Tv T 0 So T and T 1000 N The total force down the plane is: T 1400g sin sin 6 434N The resistance to the car s motion is: 10 v N The total force down the plane is greater than the resistive force, so there is a net force down the plane. Therefore the driver will need to brake to maintain his or her original speed. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free.

c The driver places the car in neutral, so T 0 The maximum speed of the car will occur when its acceleration is 0 m s i.e. when the resultant force parallel to the plane is zero.

3 c The driver places the car in neutral, so T 0 The maximum speed of the car will occur when its acceleration is 0 m s i.e. when the resultant force parallel to the plane is zero. Therefore 10 v 1400g sin 6 3 So 1400g sin 6 10 v and v 5.6 m s 1 Since 3 tan, you have that 4 3 sin and 5 4 cos from the right-angled triangle: 5 Suppose the ball hits the plane with speed u m s 1 and rebounds with speed v m s 1 Then down the plane : usin vsin Newton s law of restitution parallel to the plane : vcos eucos (1) () Squaring equation (1) gives Squaring equation () gives u sin v sin v co s e u cos Adding these equations gives: v sin v co s u sin e u cos v sin co s u sin e u cos v u sin e u cos 3 4 v u e u 5 5 9u e u v (3) 5 5 Since the ball loses half its kinetic energy upon impact, you have mv mu mu 4 u So v (4) Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 3

4 3 continued Solving equations (3) and (4) simultaneously, you obtain: u 9u e u e e e 50 5 So e 3 7 and e a Impulse on the football = change in momentum of the football P mv mu 8 5 P P N s b Let be the angle between P and i 1. tan 0.6 So 63.4 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 5 a Consider the ball hanging in vertical equilibrium. T mg e Hooke s Law gives T l e So mg l mgl So e m 15 So PQ l e m b When the string has length 1.9 m, its extension is m So its elastic potential energy is J 1. When the string is in equilibrium, its extension is m So its elastic potential energy is 0.401J 1. Therefore the work done in stretching the string to a length of 1.9 m is J Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 5

5 c PX 1.9, so QX PX PQ 1.9 1. 0.196 0.504 m Let v be the velocity of the ball as it passes through Q. Using the conservation of energy: EPE PE KE at Q EPE at X 150.196 1 150.7 0.5 9.8 0.504 0.5 v 1.

6 5 c PX 1.9, so QX PX PQ m Let v be the velocity of the ball as it passes through Q. Using the conservation of energy: EPE PE KE at Q EPE at X v v v So v 3.6 m s d Let h be the distance travelled by the ball above point X After travelling through a distance h, the string will be slack and its velocity will be zero. Using the work-energy principle: Potential energy gained = elastic potential energy lost h 1..45h So h This is a distance of m from the ceiling. Hence, the ball will not hit the ceiling. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 6

6 a i Using conservation of momentum for the system : 33 3v 1v 1 1 9 3v 1v (1) Consider the final kinetic energy of Q : 1 1 v 3.645 v 7.9 v 7.9.7 m s -1 Substituting v into equation (1) gives 9 3v1.

7 6 a i Using conservation of momentum for the system : 33 3v 1v v 1v (1) Consider the final kinetic energy of Q : 1 1 v v 7.9 v m s -1 Substituting v into equation (1) gives 9 3v v1.1 m s -1 3 ii Newton s law of restitution: separation speed.7.1 e 0. approach speed 3 0 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 7

6 b Consider the collision between Q and R: 7 a Using conservation of momentum for the system :.7 w1 w () Newton s law of restitution: w w1 0..7 0.54 w w (3) 1 Adding equations () and (3) gives: 3.

8 6 b Consider the collision between Q and R: 7 a Using conservation of momentum for the system :.7 w1 w () Newton s law of restitution: w w w w (3) 1 Adding equations () and (3) gives: 3.4 3w So w 1.08 m s -1 Substituting w into equation (3) gives: w 1 So w m s -1 So the kinetic energy lost in this collision is given by J c P moves with speed.1 m s -1 and Q moves with speed 0.54 m s -1 Since P and Q are moving in the same direction, they will collide again. ucos60 cos 15cos 60 8 cos i ii Newton s law of restitution final speed (vertically) gives: sin 15 sin 57.8 e 0.9 approach speed (vertically) usin 60 sin 60 Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 8

7 b Conservation of momentum parallel to the line of centres: m cos mv mw cos v w (1) Newton s law of restitution: 3 w v 4 cos 45 cos 64 u w v () Adding equations (1) and () gives: 15 45 u cos u cos

9 7 b Conservation of momentum parallel to the line of centres: m cos mv mw cos v w (1) Newton s law of restitution: 3 w v 4 cos 45 cos 64 u w v () Adding equations (1) and () gives: u cos u cos w cos 64 u w 8 Substituting cos leads to u w u w u 7u Substitute w in equation (1): 7u cos v 8 7u v 15 u v 0.065u Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 7 b continued So the speed of S is given by u sin u sin sin sin tan u u v 0.065u So 85.5 Therefore S has velocity 0.795u m s 1 at 85.5 to the line of centres. T has velocity u m s 1 along the line of centres. Pearson Education Ltd 018. Copying permitted for purchasing institution only. This material is not copyright free. 10

Elastic collisions in two dimensions 5A

Elastic collisions in two dimensions 5A 1 a e= 1 3, tanα = 3 4 cosα = 4 5 and sinα = 3 5 (from Pythagoras theorem) For motion parallel to the wall: 4 vcosβ = ucosα vcos β = u (1) 5 For motion perpendicular