IB PHYSICS HL REVIEW FORCES-MOMENTUM-ENERGY-POWER

NAME IB PHYSICS HL REVIEW FORCES-MOMENTUM-ENERGY-POWER 1. Motion of a satellite (a) Define gravitational potential....... () A satellite of mass m is in a circular orbit around the Earth at height R from the Earth s surface. The mass of the Earth may be considered to be a point mass concentrated at the Earth s centre. The Earth has mass M and radius R. orbit satellite mass m Earth mass M R R Deduce that the kinetic energy EK of the satellite when in orbit of height R is GMm EK. 4R The kinetic energy of the satellite in this orbit is 1.5 10 10 J. Calculate the total energy of the satellite. (iii) Explain how your answer to indicates that the satellite will not escape the Earth s gravitational field and state the minimum amount of energy that must be provided to this satellite so that it does escape. (Total 11 marks) 1

. This question is about the gravitational field associated with a neutron star. (a) Define gravitational field strength.......... () Neutron stars are very dense stars of small radius. They are formed as part of the evolutionary process of stars that are much more massive than the Sun. A particular neutron star has radius R of 1.6 10 4 m. The gravitational field strength at its surface is g0. The escape speed ve from the surface of the star is 3.6 10 7 m s 1. The gravitational potential V at the surface of the star is equal to g0r. Deduce, explaining your reasoning, that the escape speed from the surface of the star is given by the expression v e g 0 R. Calculate gravitational field strength at the surface of the neutron star. () (c) The period T of rotation of the neutron star is 0.0 s. Use your answer to to deduce that matter is not lost from the surface of the star as a result of its high speed of rotation................ (Total 10 marks)

3. This question is about gravitation. A spherical planet has radius R and mass M. A satellite of mass m orbits the planet with constant linear speed v at a height h above the planet s surface, as shown below (not to scale). planet mass M v R satellite mass m h (a) Outline why although the satellite is moving at constant speed, it is not in equilibrium....... () an object in the satellite appears to be weightless....... For the satellite in its orbit, state an expression, in terms of M, m, R and h, for its potential energy..... derive an expression, using the same terms as in, for its kinetic energy....... (c) The total energy of the satellite is reduced. Use your expressions in to outline what change, if any, occurs in the radius of the orbit and the speed of the satellite............. (4) 3

(d) The force of friction between the satellite and the atmospheric air increases as the speed of the satellite increases. By reference to your answer in (c), suggest why small satellites will burn up as they re-enter the Earth s atmosphere................ (4) (Total 17 marks) 4. This question is about a spacecraft. A spacecraft above Earth s atmosphere is moving away from the Earth. The diagram below shows two positions of the spacecraft. Position A and position B are well above Earth s atmosphere. Earth A B At position A, the rocket engine is switched off and the spacecraft begins coasting freely. At position A, the speed of the spacecraft is 5.37 10 3 m s 1 and at position B, 5.10 10 3 m s 1. The time to travel from position A to position B is 6.00 10 s. (a) Explain why the speed is changing between positions A and B. Calculate the average acceleration of the spacecraft between positions A and B. () (iii) Estimate the average gravitational field strength between positions A and B. Explain your working. 4

The diagram below shows the variation with distance from Earth of the kinetic energy Ek of the spacecraft. The radius of Earth is R. energy E k R 0 0 distance On the diagram above, draw the variation with distance from the surface of Earth of the gravitational potential energy Ep of the spacecraft. () (Total 8 marks) 5. This question is about gravitational potential. (a) Define gravitational potential at a point.......... () A meteorite moves towards the Moon from a long distance away. On the axes below, sketch a graph to show the variation with distance from the centre of the Moon of the gravitational potential of the meteorite as it approaches the Moon. The radius of the Moon is r. gravitational potential +ve 0 r distance from centre of Moon ve () 5

The radius r of the Moon is 1.7 10 6 m and its mass is 7.3 10 kg. Estimate the impact speed with which the meteorite hits the surface of the Moon. (iii) Suggest one factor that will make the impact speed greater than your estimate. (c) A similar meteorite moves towards the Earth from a long distance away. Suggest how the total energy of the meteorite varies with distance when the meteorite is outside the Earth s atmosphere; inside the Earth s atmosphere. (Total 10 marks) 6

6. Collisions A large metal ball is hung from a crane by means of a cable of length 5.8 m as shown below. crane cable 5.8 m wall metal ball In order to knock down a wall, the metal ball of mass 350 kg is pulled away from the wall and then released. The crane does not move. The graph below shows the variation with time t of the speed v of the ball after release. 3.0.0 v / m s 1 1.0 0.0 0.0 0. 0.4 0.6 0.8 1.0 1. 1.4 t / s The ball makes contact with the wall when the cable from the crane is vertical. (a) For the ball just before it hits the wall, state why the tension in the cable is not equal to the weight of the ball; by reference to the graph, estimate the tension in the cable. The acceleration of free fall is 9.8 m s. Use the graph to determine the distance moved by the ball after coming into contact with the wall.......... () 7

(c) For the collision between the ball and the wall, calculate the total change in momentum of the ball; () the average force exerted by the ball on the wall. () (d) State the law of conservation of momentum. () The metal ball has lost momentum. Discuss whether the law applies to this situation. () (e) During the impact of the ball with the wall, 1 of the total kinetic energy of the ball is converted into thermal energy in the ball. The metal of the ball has specific heat capacity 450 J kg 1 K 1. Determine the average rise in temperature of the ball as a result of colliding with the wall............. (4) (Total 18 marks) 8

7. This question is about momentum and energy. (a) Define impulse of a force and state the relation between impulse and momentum. definition...... relation...... () By applying Newton s laws of motion to the collision of two particles, deduce that momentum is conserved in the collision............. (5) (c) In an experiment to measure the speed of a bullet, the bullet is fired into a piece of plasticine suspended from a rigid support by a light thread. bullet speed V 4cm plasticine The speed of the bullet on impact with the plasticine is V. As a result of the impact, the bullet embeds itself in the plasticine and the plasticine is displaced vertically through a height of 4 cm. The mass of the bullet is 5. 10 3 kg and the mass of the plasticine is 0.38 kg. Ignoring the mass of the bullet, calculate the speed of the plasticine immediately after the impact......... () Deduce that the speed V with which the bullet strikes the plasticine is about 160 m s 1....... () 9

(iii) Estimate the kinetic energy lost in the impact................ (d) Another bullet is fired from a different gun into a large block of wood. The block remains stationary after impact and the bullet melts completely. The temperature rise of the block is negligible. Use the data to estimate the minimum impact speed of the bullet. mass of bullet = 5.3 10 3 kg specific heat capacity of the material of the bullet = 130 J kg 1 K 1 latent heat of fusion of the material of the bullet = 870 J kg 1 melting point of the material of the bullet = 330 C initial temperature of bullet = 30 C........................ (5) (Total 19 marks) 10

8. This question is about driving a metal bar into the ground and the engine used in the process. Large metal bars can be driven into the ground using a heavy falling object. 3 object mass =.0 10 kg bar mass = 400kg In the situation shown, the object has a mass.0 10 3 kg and the metal bar has a mass of 400 kg. The object strikes the bar at a speed of 6.0 m s 1 It comes to rest on the bar without bouncing. As a result of the collision, the bar is driven into the ground to a depth of 0.75 m. (a) Determine the speed of the bar immediately after the object strikes it................... (4) Determine the average frictional force exerted by the ground on the bar................ (c) The object is raised by a diesel engine that has a useful power output of 7. kw. In order that the falling object strikes the bar at a speed of 6.0 m s 1, it must be raised to a certain height above the bar. Assuming that there are no energy losses due to friction, calculate how long it takes the engine to raise the object to this height................ 11 (4)

9. This question is about Newton s laws of motion, the dynamics of a model helicopter and the engine that powers it. (a) Explain how Newton s third law leads to the concept of conservation of momentum in the collision between two objects in an isolated system......................... (4) The diagram illustrates a model helicopter that is hovering in a stationary position. 0.70 m 0.70 m rotating blades downward motion of air The rotating blades of the helicopter force a column of air to move downwards. Explain how this may enable the helicopter to remain stationary................... (c) The length of each blade of the helicopter in is 0.70 m. Deduce that the area that the blades sweep out as they rotate is 1.5 m. (Area of a circle = r )...... 1

(d) For the hovering helicopter in, it is assumed that all the air beneath the blades is pushed vertically downwards with the same speed of 4.0 m s 1. No other air is disturbed. The density of the air is 1. kg m 3. Calculate, for the air moved downwards by the rotating blades, the mass per second; () the rate of change of momentum. (e) State the magnitude of the force that the air beneath the blades exerts on the blades.... (f) Calculate the mass of the helicopter and its load............. () (g) In order to move forward, the helicopter blades are made to incline at an angle to the horizontal as shown schematically below. While moving forward, the helicopter does not move vertically up or down. In the space provided below draw a free body force diagram that shows the forces acting on the helicopter blades at the moment that the helicopter starts to move forward. On your diagram, label the angle. (4) 13

(h) Use your diagram in (g) to explain why a forward force F now acts on the helicopter and deduce that the initial acceleration a of the helicopter is given by a = g tan where g is the acceleration of free fall............................... (5) The helicopter is driven by an engine that has a useful power output of 9.0 10 W. The engine makes 300 revolutions per second. Deduce that the work done in one cycle is 3.0 J....... 10. This question is about mechanical power and heat engines. Mechanical power (a) Define power....... A car is travelling with constant speed v along a horizontal straight road. There is a total resistive force F acting on the car. Deduce that the power P to overcome the force F is P = Fv....... () 14

(c) A car drives up a straight incline that is 4.80 km long. The total height of the incline is 0.30 km. 4.80 km 0.30 km The car moves up the incline at a steady speed of 16 m s 1. During the climb, the average resistive force acting on the car is 5.0 10 N. The total weight of the car and the driver is 1. 10 4 N. Determine the time it takes the car to travel from the bottom to the top of the incline. () Determine the work done against the gravitational force in travelling from the bottom to the top of the incline. (iii) Using your answers to and, calculate a value for the minimum power output of the car engine needed to move the car from the bottom to the top of the incline. (4) (d) From the top of the incline, the road continues downwards in a straight-line. At the point where the incline starts to go downwards, the driver of the car in (c) stops the car to look at the view. In continuing his journey, the driver decides to save fuel. He switches off the engine and allows the car to move freely down the incline. The car descends a height of 0.30 km in a distance of 6.40 km before levelling out. 0.30 km 6.40 km The average resistive force acting on the car is 5.0 10 N. Estimate the acceleration of the car down the incline; 15 (5)

the speed of the car at the bottom of the incline. () (e) In fact, for the last few hundred metres of its journey down the incline, the car travels at constant speed. State the value of the frictional force acting on the car whilst it is moving at constant speed.... The heat engine (f) The diagram below shows the idealised pressure-volume (P-V) diagram for one cycle of the gases in an engine similar to that used in the car. pressure P B C D A volume V The changes A B and C D are adiabatic changes. Explain what is meant by an adiabatic change. () State the name given to the change B C. (g) The useful power output of the engine is 0 kw and the overall efficiency of the engine is 3. The car engine completes 50 cycles every second. Deduce that QH = 1.3 kj................ (Total 4 marks) 16

1. Motion of a satellite (a) the work done per unit mass; in bringing a small / point mass from infinity to a point (in the gravitational field); Ratio idea essential for first mark. GMm mv r equating gravitational force to centripetal ; r v GM r to get the speed ; GMm r GMm R GMm 4R 1 1 and hence EK = ; 3 potential energy in orbit is E GMm GMm r R P ; E GMm GMm GMm 4R R 4R total energy is then ; ie E = 1.5 10 10 J; 3 Award [1 max] for bare answer without explanation. (iii) at infinity the potential energy is zero and hence if satellite gets there its total energy will be E 0; but the satellite has negative energy and hence it cannot escape to infinity / the satellite is bound to the Earth forever; the minimum energy that must be supplied the value must agree with the 10 is E 1.5 10 J; or ECF from b ii candidates answer to.. Neutron star (a) the force per unit mass; exerted on a point / small mass; 3 [11] energy required to move an object of mass m from the surface of the star to infinity = mg0r; if objects KE is equal to this it will escape the gravitational influence of the star / OWTTE; therefore, mv R e mg0 ; to give v e g 0 R 3 ve g 0 ; R 14.6 10 1.6 10 4.1 10 3 10 1 N kg 4 ; v R 4 R T (c) centripetal acceleration = ; 4 40 1.6 10 9 1.6 10 ms ; 4 4 10 a comment to the effect that this is less than the gravitational field strength so that gravity will stop matter being torn away; 3 Award [1 max] for calculation of linear speed (5.0 10 6 ms 1 )and a comment that this is less than the escape speed. 17

3. (a) direction is changing and so there is an acceleration; there must be a resultant force on the satellite / force is provided by gravitational attraction; object and satellite have the same acceleration; acceleration is towards centre of planet; so no reaction force between object and satellite; 3 potential energy ( GMm) ; 1 ( R h) in orbit, mv ( GMm) ( R h) ( R h) or expressed in words; 1 use of E K mv ; 1 ( GMm) E K ; 3 ( R h) (c) (total energy = potential energy + kinetic energy) ( GMm) total energy is ; ( R h) ( GMm) as total energy is reduced, increases; ( R h) hence h decreases; Do not award if there is no reasoning or reasoning is fallacious or misleading. EK increases and v increases; 4 (d) friction reduces the total energy of the satellite; causing height to decrease and speed to increase; less height, greater frictional force; because atmosphere denser; frictional force causes heating effect; as height decreases heating effect increases / heats up more; if satellite small, sufficient heating to cause destruction; Do not allow heats up as height decreases. 4 max [17] 18

4. (a) (deceleration due to) gravitational pull of Earth; 1 v t 5100 5370 600 a ; (iii) a = 0.45ms ; ecf from : F E ; m E = a; E = 0.45Nk g 1 ; 3 Accept ms as correct units. general shape (1 / r); correct quadrant; No need to show the curve further away from the distance axis to achieve full marks. 5. (a) change in potential energy per unit mass / work done per unit mass; in moving small / point mass from infinity to the point; Do not allow from a long distance away. [8] asymptotic at large r and in negative gravitational potential region; line stops at surface line; Do not allow asymptotic to y-axis. loss of gravitational potential = 6.67 10 11 Moon ; equates loss of gravitational potential to 1 v ; m r Moon (iii) v =.4 kms 1 ; 3 meteorite may have initial speed / velocity towards Moon / contribution of Earth s gravity; 1 (c) constant; 1 decreasing; 1 [10] 19

6. Collisions (a) centripetal force is provided by the cable / the ball is moving along the arc of a circle; 1 350.6 centripetal force = ; 5.8 = 410N; tension = 410 + (350 9.8) = 3800N; 3 Award [0] if v r is not used. idea of use of area under graph / appropriate equation; distance = 1 0.15.6 (allow 0.14 0.15 s for the time) = 0.195m; (allow 0.0m, not 0.m) (c) idea of momentum as mv; total change (=.6 350) = 910Ns; p idea of average force as ; t 910 force 6100 N; 0.15 (d) for isolated / closed system; total momentum remains constant; external force acts on ball; so law does not apply to the ball; or system is ball + wall / Earth; momentum loss of ball = momentum gain of wall / Earth; (e) EK = 1 350.6 ; thermal energy = 350 450 ; idea of 0.1 EK = mc ; = 9.0 10 4 K; 4 [18] 0

7. (a) (impulse =) force time for which force acts; impulse (F t) = change in momentum ( p); The following points are needed for maximum marks. from Newton 3; when objects are in contact, the forces exerted by the objects on each other are equal and opposite; from Newton / collision time is the same; impulses are equal and opposite; therefore changes in momentum are equal and opposite / total change in momentum is zero; or Accept algebraic solution. from Newton 3; FAB = FBA from Newton ; FAB t = ma va; = mb vb; 5 (c) v = gh ; to give v =. ms 1 ; Award full marks for bald correct answer. from conservation of momentum / V 5. 10 3 = 0.38. 0.38. V = 3 5. 10 to give V = 160 m s 1 1 3 4 (iii) KE before = 5. 10 1.6 10 = 67 J; 1 KE after = 0.38 4. 8 = 0.91 J / (0.38 9.8 0.4) = 0.91 J; lost energy = 66 J 3 (d) energy to increase from 0 C to 330 C = (5. 10 3 130 300) = 00 (J) energy to melt pellet = (5. 10 3 870) = 4.5 (J); total KE = 10 J; 1 mv = 10; to give v = 80 m s 1 5 [19] 1

8. (a) momentum of object = 10 3 6.0; momentum after collision =.4 10 3 v; use conservation of momentum, 10 3 6.0 =.4 10 3 v; to get v = 5.0 m s 1 ; 4 Award [ max] for mass after collision = 400 kg and v = 30 m s 1. KE of object and bar + change in PE = 1. 10 3 5 +.4 10 3 10 0.75; use E = Fd, 4.8 10 4 = F 0.75; to give F =64 kn; Award [ max] if PE missed, F = 40 kn. or v a = ; s F mg = ma; to give F = 64 kn; 3 Award [ max] if mg missed. (c) recognize that the height is given by mgh = 1 mv ; KE = 1 mv = 1.0 10 3 36 = 3.6 10 4 J; t = P E ; 4 3.6 10 = =5.0 s; 3 7. 10 or calculation of PE = mgh using v = u + as h = 1.8 m; mgh =.0 10 3 10 1.8; t = P E ; = 3.6 10 7. 10 4 3 = 5.0 s; 4 (h) Eff = W ; Q H therefore, QH = 10 = 55 J; 0.4 []

9. (a) before and after collision there are no forces acting on the objects; from Newton 3 when the two bodies are in contact the forces that they exert on each other are equal and opposite / OWTTE; therefore, the net force on the two balls is always zero; therefore, there is no change in momentum (of the objects) / momentum is conserved; or Accept an argument based on change in momentum of each individual object. eg from Newton 3F1 = F1; (accept statement in words) p1 F1 and F t p 1 p = ; t t 1 p t ; therefore, p1 + p = 0; 4 the blades exert a force on the air and by Newton s third law the air exerts an equal and opposite force on the blades / air has change in momentum downwards giving rise to a force and from Newton 3 there will a force upwards; if this force equals the weight of the helicopter; the net vertical force on the helicopter will be zero / OWTTE; 3 (c) area = 0.7 ; = 1.5 m 1 (d) volume of air per second = 1.5 4.0(m 3 s 1 ); mass = volume density = (1. 1.5 4.0) = 7.kgs 1 ; No unit error for 7. kg. momentum per second = (7. 4.0) = 9N; 1 (e) 9 N; 1 (f) recognise that the force on the blades = Mg; to give3.0kg; (g) correct relative directions of forces; upward force length greater than weight by eye; appropriate labelling of forces; angle as shown above; 4 Award [ max] if extra force(s) drawn. 3

(h) the forward force is the horizontal component of U; resolve vertically U cos = W; horizontal component F = U sin ; divide to get = F W tan θ ; F = (W tan ) = Mg tan = Ma; to give a = g tan 5 Award [5 max] for a correctly labelled force diagram incorporating mass with a justifying statement. Award [1 max] for triangle mixing accelerations and force. 900 work done in one cycle = ; 300 = 3.0J 1 10. (a) the rate of working / work time; 1 If equation is given, then symbols must be defined. W t F d t P ; v d therefore, P = Fv; t d (c) t ; v 4800 300s; 16 W = mgh = 1. 10 4 300 = 3.6 10 6 J; 1 (iii) work done against friction = 4.8 10 3 5.0 10 ; total work done =.4 10 6 + 3.6 10 6 ; total work done = P t = 6.0 10 6 ; 6 6 10 to give P 0kW; 4 300 4

sin 0.30 6.4 (d) θ 0.047; weight down the plane = W sin = 1. 10 4 0.047 = 5.6 10 N; net force on car F = 5.6 10 5.0 10 = 60N; a F m ; 60 5.0 10 ms ; 3 1. 10 5 v = as = 5 10 6.4 10 3 ; to give v = 5 / 6ms 1 ; Give full credit for and to candidates who use energy argument to calculate v and then use this to calculate a. gain in KE = loss in PE work done against friction; 1 mv mgh Fd ; 1 6 mv 3.6 10 5.0 10 6.40; 0.6 10 3 v = 3.6 10 6 5.0 10 6.40; v = 5 / 6ms 1 ; a v ; s = 5.0 / 5.1 10 ms ; (e) 5.6 10 N; 1 (f) a compression or expansion / change in state (of the gas); in which no (thermal) energy is exchanged between the gas and the surroundings / in which the work done is equal to the change in internal energy of the gas; isobaric; 1 0 (g) (for real engine) 0. 3 P H to give PH = 63kW; time for one cycle = 0.0s; QH = PH time to give QH = 6.3 10 4 0.0; = 1.3kJ or eff W Q H ; 10 W 50 0.3 4 400 400J; to give QH = 1.3 kj; 3 QH [4] 5