Thornleigh Salesian College, Bolton21.A solid iron ball of mass 890 kg is used on a demolition site. It hangs from the jib of a crane suspended by a steel rope. The distance from the point of suspension to the centre of mass of the ball is 15 m.(a)Calculate the tension in the rope when the mass hangs vertically and stationary.............................................................................................................................................................................................................................................................................................................................................................................................................(2)(b)The iron ball is pulled back by a horizontal chain so that the suspension rope makes an angle of 30 with the vertical. Calculate the new tension in the suspension rope...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(c)The ball is now released from rest and hits a brick wall just as it passes through the vertical position. It can be assumed that the ball is brought to rest by the impact with the wall in 0.2s.Calculate(i)the vertical height through which the ball falls,......................................................................................................................................................................................................................................................(ii)the speed of the ball just before impact,.................................................................................................................................................................................................................................................................................................................................................................................

(iii)the average force exerted by the ball on the wall..................................................................................................................................................................................................................................................................................................................................................................................(5)(Total 9 marks)

2.A toy locomotive of mass 0.50kg is initially at rest on a horizontal track. The locomotive is powered by a twisted rubber band which, as it unwinds, exerts a force which varies with time as shown in the table.time/s0.01.02.03.04.05.06.07.08.0force/N0.200.180.150.120.100.080.050.020.00

(a)(i)On the grid below plot a graph of force against time for the rubber band power source.

(ii)State what is given by the area between the graph and the time axis............................................................................................................................(4)(b)The rubber band is wound up and released to power the locomotive. Use your graph to show that the speed of the locomotive 8.0s after the twisted rubber band is released is l.6ms1. Ignore the effects of air resistance and energy losses due to friction.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(c)8.0s after release the locomotive collides with and couples to a toy truck, initially at rest, which has a mass of 1.50kg.(i)Calculate the speed of the coupled locomotive and truck after collision.......................................................................................................................................................................................................................................................(ii)Calculate the combined kinetic energy of the locomotive and truck immediately after collision.......................................................................................................................................................................................................................................................(iii)Show, with the aid of a calculation, whether or not the collision is elastic..................................................................................................................................................................................................................................................................................................................................................................................(5)(Total 11 marks)

3.(a)(i)Give an equation showing how the principle of conservation of momentum applies to the colliding snooker balls shown in the diagram.before collisionafter collision...........................................................................................................................(ii)State the condition under which the principle of conservation of momentum applies.......................................................................................................................................................................................................................................................(3)(b)A trolley, A, of mass 0.25 kg and a second trolley, B, of mass 0.50 kg are held in contact on a smooth horizontal surface. A compressed spring inside one of the trolleys is released and they then move apart. The speed of A is 2.2 m s1.(i)Calculate the speed of B..................................................................................................................................................................................................................................................................................................................................................................................(ii)Calculate a minimum value for the energy stored in the spring when compressed..................................................................................................................................................................................................................................................................................................................................................................................(4)

(c)The rotor blades of a helicopter sweep out a cross-sectional area, A. The motion of the blades helps the helicopter to hover by giving a downward velocity, u, to a cylinder of air, density r. The cylinder of air has the same cross-sectional area as that swept out by the rotor blades.Explaining your reasoning,(i)derive an expression for the mass of air flowing downwards per second, and.................................................................................................................................................................................................................................................................................................................................................................................(ii)derive an expression for the momentum given per second to this air.......................................................................................................................................................................................................................................................(iii)Hence show that the motion of the air results in an upward force, F, on the helicopter given byF = rAu2..................................................................................................................................................................................................................................................................................................................................................................................(5)(d)A loaded helicopter has a mass of 2500 kg. The area swept out by its rotor blades is 180 m2. If the downward flow of air supports 50% of the weight of the helicopter, what speed must be given to the air by the motion of the rotor blades when the helicopter is hovering? Take the density of air to be 1.3 kg m3................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total 15 marks)

4.A student carried out an experiment to determine the terminal speed of various ball bearings as they fell through a viscous liquid. She did this by timing their fall between two marks, P and Q, which were 850 mm apart on a vertical glass tube.

You may be awarded marks for the quality of written communication in your answer.(a)(i)Describe the motion of a ball bearing after being released from rest at the surface....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(ii)In terms of the forces acting, explain why a ball bearing reaches a terminal speed under these conditions....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(5)

(b)The students results are shown in columns A and B. Complete column C.column Acolumn Bcolumn Ccolumn Dcolumn Eradius of ball bearingr / mmtime of fall / s(through 850 mm)terminal speedn / mm s1log10(r / mm)log10(n / mm s1)1.6232.0

0.210

1.9821.4

0.297

2.2117.2

0.344

2.7311.3

0.436

3.407.2

0.531

4.124.9

0.615

(2)(c)The relationship between n and r is known to be of the formn = krn,where n and k are constants.(i)Enter the corresponding values for log10(n / mm s1) in column E of the table in part (b).

(ii)Using the grid, plot a graph of log10(n / mm s1) on the y-axis, against log10(r / mm) on the x-axis.

(4)(d)Use your graph to determine(i)the constant n,............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(ii)the constant k......................................................................................................................................................................................................................................................................................................................................................................(5)(Total 16 marks)

5.The Thrust SSC car raised the world land speed record in 1997. The mass of the car was 1.0 104kg. A 12s run by the car may be considered in two stages of constant acceleration.Stage one was from 0 to 4.0 s and stage two 4.0 s to 12s.(i)In stage one the car accelerates from rest to 44 m s-1 in 4.0s. Calculate the acceleration produced and the force required to accelerate the car.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(ii)In stage two the car continued to accelerate so that it reached 280 m s1 in a further 8.0 s. Calculate the acceleration of the car during stage two.......................................................................................................................................................................................................................................................................(iii)Calculate the distance travelled by the car from rest to reach a speed of 280 m s1.......................................................................................................................................................................................................................................................................(6)(Total 6 marks)

6.A packing case is being lifted vertically at a constant speed by a cable attached to a crane. The packing case has a mass of 640 kg.(a)With reference to one of Newtons laws of motion, explain why the tension, T, in the cablemust be equal to the weight of the packing case.You may be awarded marks for the quality of written communication in your answer...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(b)The packing case is lifted through a vertical height of 8.0 m in 4.5 s.Calculate(i)the work done on the packing case,..........................................................................................................................................................................................................................................................(ii)the power output of the crane in this situation........................................................................................................................................................................................................................................................................................................................................................................................(3)(Total 6 marks)

7.A cyclist rides along a road up an incline at a steady speed of 9.0 m s1 . The mass of the rider and bicycle is 70kg and the bicycle travels 15 m along the road for every 1.0 m gained in height. Neglect energy loss due to frictional forces.(a)(i)Calculate the component of the weight of the bicycle and the rider that acts along the incline.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

(ii)Calculate the power developed by the cyclist in riding up the slope.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(4)(b)The cyclist stops pedalling and the bicycle freewheels up the incline for a short time.(i)State the energy change taking place as the bicycle freewheels up the slope.....................................................................................................................................................................................................................................................(ii)Calculate the distance travelled along the slope from when the cyclist stops pedalling to where the bicycle comes to rest.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)(Total 7 marks)

8.A ball bearing is released into a tall cylinder of clear oil. The ball bearing initially accelerates but soon reaches terminal velocity.(a)By considering the forces acting on the ball bearing, explain its motion...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(3)

(b)How would you demonstrate that the ball bearing had reached terminal velocity?............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(2)(Total 5 marks)

9.An apple and a leaf fall from a tree at the same instant. Both apple and leaf start at the same height above the ground but the apple hits the ground first.You may be awarded marks for the quality of written communication in your answer.Use Newtons laws of motion to explain why(i)the leaf accelerates at first then reaches a terminal velocity,.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(ii)the apple hits the ground first.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(Total 5 marks)

10.The diagram shows a car travelling at a constant velocity along a horizontal road.

(a)(i)Draw and label arrows on the diagram representing the forces acting on the car.(ii)Referring to Newtons Laws of motion, explain why the car is travelling at constant velocity.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(5)(b)The car has an effective power output of l8kW and is travelling at a constant velocity of 10 m s. Show that the total resistive force acting is 1800 N..................................................................................................................................................................................................................................................................................................................................................................................(1)(c)The total resistive force consists of two components. One of these is a constant frictional force of 250 N and the other is the force of air resistance, which is proportional to the square of the cars speed.Calculate(i)the force of air resistance when the car is travelling at 10 m s1,......................................................................................................................................................................................................................................................

(ii)the force of air resistance when the car is travelling at 20 m s1,......................................................................................................................................................................................................................................................(iii)the effective output power of the car required to maintain a constant speed of20 m s1 in a horizontal road..................................................................................................................................................................................................................................................................................................................................................................................(4)(Total 10 marks)

1.(a)T = mg = 890 10 = 8900 (1) N (1)(accept alternative correct value using g = 9.81 N kg1)2(b)resolve vertically T cos 30 =mg (1)T = = 10280(1.03 104 )N (1)2(c)(i)vertical height fallen = l(1 cos q) = 15(1 0.866) = 2.0(1) m (1)(allow e.c.f if h calculated wrongly)(ii)mu2 = mgh or reference energy (1) u = = 6.34 m s1 (1)(max 1/3 if equations of motion used)(iii)F = = 2.8 104 N (1)(allow e.c.f of u and m as before)5[9]

2.(a)(i)points plotted correctly (1) (1) (deduct one for each incorrect)sensible scales chosen (1)line of best fit (1)(ii)change in momentum [or impulse] (1) (accept 0.8)max 4(b)area under graph = 0.800.05 (1) (kgms1)u = (1) = 1.6 ms1alternative:state average force = 0.10(N) (1)leading to correct derivation of 1.6ms1 (1)2(c)(i)Dmu = 0 [or statement] (1) u = 0.40ms1 (1)(ii)kinetic energy = 0.16 J (1)(iii)initial kinetic energy = 0.64 (J) (1)kinetic energy lost so inelastic (1)5[11]

3.(a)(i)equation showing momentum before = momentum after (1)correct use of sign (1)(ii)no external forces (on any system of colliding bodies) (1)3

(b)(i)(by conservation of momentum m1u1 + m2u2 = 0)0.25 2.2 = ()0.50u2 (1)u2 = ()1.1(0)ms1 (1)(ii)allow e.c.f from (i)minimum stored energy = total k.e. = 0.25 2.22 + 0.5 1.12 (1)= 0.91J (1)4(c)(i)mass of air per second = rAu (1)correct justification, incl ref to time (1)(ii)momentum per second (= Mu = u2 Ar) = u 2 Ar (1)(iii)force = rate of change of momentum (hence given result) (1)upward force on helicopter equals (from Newton third law) downward force on air (1)5(d)u2 Ar = (for 50% support) (1)u2 180 1.3 = gives u = 7.2ms1 (1) (or 7.3, g taken as 10)if not 50% of weight, max 1/3 provided all correct otherwise (gives 10.2)3[15]

4.(a)(i)initial acceleration/increase of speed (1)reaches a constant speed/velocity (1)acceleration decreases to become zero (at this speed) (1) (ii)drag/frictional forces increases with speed (1)drag equal to weight ( upthrust) (1)no resultant force at terminal speed [or balanced forces or forces cancel] (1)max 5(b)column C26.639.7four values correct (1)49.4all values correct and to 3 or 4 s.f. (1)75.2118173.52(c)(i)column E1.421.601.69 all values correct and to 3 or 4 s.f. (1)1.882.072.24(ii)axes labelled and suitable scales chosen (1)at least 5 points plotted correctly (1)acceptable line (1)4(d)(i)gradient == 2.0 (1)= n gradient (= 2) (1)(ii)intercept on y-axis = log k (1)intercept = 1.0 (1)k (= 101.0) = 10 (1)units of k: for n = 2, mm1 s1 (1)max 5[16]

5.(i)a = = 11 ms2 (1)F = ma =1.1 105 N (1)(ii)D = 236 m s1a = = 29.5 ms2 (1)(iii)sone = t = 4.0 = 88m (1)stwo = t = 8.0 (1) = 1296(m) (1)total distance = 1384 m (1)[6]

6.(a)resultant force on crate is zero (1)forces must have equal magnitudes or size (1)(but) act in opposite directions (1)correct statement of 1st or 2nd law (1)max 3QWC 1(b)(i)work done = F d = 640 9.81 8.0 (1)= 5.0(2) 104J (1)(ii)(use of P = gives) P = = 1.1(2) 104W (1)(allow C.E. for value of work done from (i))3[6]

7.(a)(i)F = W sin q = W (1) correct angle (1)= 46N (1)[2 out of 3 if no working shown](ii)P(= Fu ) = 410W (1)4

(b)(i)kinetic energy of bicycle + rider gravitational potential energy (1)(ii)initial Ek. = gain in gravitational Ep = 2.8 103 (J) (1)h = 4.1m (1)distance = 62m (1)alternative:(F = ma) a = = 0.657 (ms2) (1)v2 = u2 =2as (1)s = = 62m (1)max 3[7]

8.(a)ball bearing accelerates at first as resultant force is downwards (1)resistive force increases with speed (1)when resultant force on ball is zero, terminal velocity reached (1)3The Quality of Written Communication marks are awarded for the quality of answers to this question.(b)show ball bearing takes same time (1)to travel equal distances (1)[or measure velocity at different points (1) with appropriate method (1)]2[5]

9.(i)weight greater than air resistance[or (initially only) weight/gravity acting] (1)hence resultant force downwards or therefore acceleration (2nd law) (1)air resistance or upward force increases with speed (1)until air resistance equals weight or resultant force is zero (1)leaf moves at constant velocity (1st law) [or 1st law applied correctly] (1)(ii)air resistance depends on shape[or other correct statement about air resistance] (1)air resistance less significant (1)air resistance less, therefore greater velocity [or average velocity greater or accelerates for longer] (1)max 5QWC [2][5]

10.(a)(i)

F1 weight / mg (1)F2 reaction or normal contact force (1)F3 driving force (1)F4 friction or air resistance (1)(ii)zero acceleration (1)zero resultant force (1)max 5(b)(P = Fv gives) 18 103 = F 10 (1) (and F = 1.8 103 N)1(c)(i)1800 250 = 1.6 103 N (1) (1.55 103 N)(ii)force = 4 1.55 103 = 6.2 103 N (1)(allow e.c.f. from(i))(iii)total force = 6200 + 250(N) (1) (= 6.45 103(N))(P = Fv gives) P = 6.45 103 20 = 1.3 105 W (1) (1.29 105 W)(allow e.c.f. for value of total force)4[10]The Quality of Written Communication marks were awarded primarily for the quality of answers to (a)(ii).