Topic 2 Past Paper [175 marks]
                                 State the law of conservation of linear momentum.                                                                                             [2 marks]
                            1a.
                                    Markscheme
                                    if no external forces act / isolated system;
                                    momentum is constant / (total) momentum before=(total) momentum after;
                                 Gravel falls vertically onto a moving horizontal conveyor belt.                                                                               [7 marks]
                            1b.
                                                                                    –1
                                 (i) The gravel falls at a constant rate of 13 kg s    through a height of 1.9 m. Show that the vertical speed of the gravel as it lands on the
                                                               –1
                                 conveyor belt is about 6 m s .
                                 (ii) The gravel lands on the conveyor belt without rebounding. Calculate the rate of change of the vertical momentum of the gravel.
                                 (iii) Gravel first reaches the belt at t = 0.0 s and continues to fall. Determine the total vertical force that the gravel exerts on the conveyor belt
                                 at t = 5.0 s.
                                    Markscheme
                                    (i) use of v =    2gh;
                                                   √
                                            –1
                                    6.11ms ; (must show calculation to better than 1 sf)
                                    (ii) rate of change of vertical momentum=13×6.11;
                                    79N; (accept answers in the range of 78N to 80N)
                                    (iii) mass accrued=5.0×13=65kg;
                                                                                                –2
                                    weight of this mass (=65×9.8)=637N; (650 from g=10ms )
                                    total force (637+79=)716N; } (allow ECF from (b)(ii) and from incorrect weight)
                                                                                                            –1
                                 The conveyor belt moves with a constant horizontal speed of 1.5 m s          . As the gravel lands on the belt, it has no horizontal speed. [4 marks]
                            1c.
                                 (i) Calculate the rate of change of the kinetic energy of the gravel due to its change in horizontal speed.
                                 (ii) Determine the power required to move the conveyor belt at constant speed.
                                 (iii) Outline why the answers to (c)(i) and (ii) are different.
                                    Markscheme
                                             –1
                                    (i) 14.6Js  ;
                                                                                                        –1
                                    (ii) horizontal momentum gain per second =13×1.5( =19.5kgms );
                                    power required=29.3W;
                                    (iii) additional energy/power required to accelerate gravel (through friction at the surface of the belt) / the gravel has to slip to gain
                                    horizontal speed / OWTTE;
                                 State the difference between average speed and instantaneous speed.                                                                           [2 marks]
                            2a.
                   Markscheme
                   average speed is the speed over a period of time/distance; instantaneous speed is the speed at a particular instant in time/point in
                   space.
                  The graph shows how the acceleration a of a particle varies with time t.  [3 marks]
               2b.
                  At time t = 0 the instantaneous speed of the particle is zero.
                  (i) Calculate the instantaneous speed of the particle at t = 7.5 s.
                  (ii) Using the axes below, sketch a graph to show how the instantaneous speed v of the particle varies with t.
                   Markscheme
                   (i) speed=(area under graph =)½×7.5×3;
                               -1
                   =10 or 11 or 11.3 (ms );
                                         2
                   (ii) suitable curve approximating to v=kt ;
                  Define linear momentum.                                                   [1 mark]
               3a.
                   Markscheme
                   mass×velocity; (allow mv with symbols defined)
                                 State, in terms of momentum, Newton’s second law of motion.                                                                                    [1 mark]
                            3b.
                                    Markscheme
                                    the rate of change of momentum of a body is equal to/directly proportional to the force acting on the body;
                                    Accept 
                                           p
                                         Δ
                                    F=  only if all symbols are defined.
                                           t
                                         Δ
                                 Show, using your answer to (b), how the impulse of a force F is related to the change in momentum Δp that it produces.                         [1 mark]
                            3c.
                                    Markscheme
                                             p
                                           Δ
                                    (F= )
                                             t
                                           Δ
                                    therefore impulse FΔt =Δp ; (accept t for Δt)
                                 Show, using your answer to (b), how the impulse of a force F is related to the change in momentum Δp that it produces.                         [1 mark]
                            3d.
                                    Markscheme
                                             p
                                           Δ
                                    (F= )
                                             t
                                           Δ
                                    therefore impulse FΔt =Δp; (accept t for Δt)
                                 A railway truck on a level, straight track is initially at rest. The truck is given a quick, horizontal push by an engine so that it now     [12 marks]
                            3e.
                                 rolls along the track.
                                                                                                                                                              –1
                                 The engine is in contact with the truck for a time T = 0.54 s and the initial speed of the truck after the push is 4.3 ms      . The mass of the truck
                                           3
                                 is 2.2×10  kg.
                                 Due to the push, a force of magnitude F is exerted by the engine on the truck. The sketch shows how F varies with contact time t.
                                 (i) Determine the magnitude of the maximum force F           exerted by the engine on the truck.
                                                                                          max
                                 (ii) After contact with the engine (t = 0.54 s) the truck moves a distance 15 m along the track. After travelling this distance the speed of the
                                                 –1
                                 truck is 2.8 ms . Assuming a uniform acceleration, calculate the time it takes the truck to travel 15 m.
                                 (iii) Calculate the average rate at which the kinetic energy of the truck is dissipated as it moves along the track.
                                                                             –1                                                     3
                                 (iv) When the speed of the truck is 2.8ms  it collides with a stationary truck of mass 3.0×10 kg. The two trucks move off together with a
                                                                            –1
                                 speed V. Show that the speed V=1.2ms .
                                 (v) Outline the energy transformations that take place during the collision of the two trucks.
                                  Markscheme
                                                                               3               3
                                  (i) (impulse=) change in momentum=2.2×10 ×4.3(=9.46×10 Ns);
                                  impulse=area under graph=½F        Δt;
                                                                  max
                                                         3
                                  ½F     ×0.54=9.46×10 ;
                                     max
                                                          4
                                  F    =35k(N) or 3.5×10  (N);
                                   max
                                                                                2    2
                                                                      2  2
                                                                              4.3 −2.8
                                                                     u −v
                                                                                                   −
                                                                                                     2
                                  (ii) (magnitude of) acceleration=(       =          =)0.355ms ;
                                                                       s         30
                                                                      2
                                          u−v     1.5
                                  time=(      =       =)4.2s;
                                           a
                                                 0.355
                                  Award [1 max] if an additional 0.54 s is added to answer.
                                                  1            3
                                                                    2      2                4
                                  (iii) ΔKE = ( ×2.2×10 [4.3 −2.8 ] =)1.17×10 J;
                                                  2
                                                                     4
                                                              1.17 10
                                                                  ×
                                  rate of change of ΔKE =              =2.8kW
                                                                 4.2
                                  (mark is for division by 4.2 and correct calculation)
                                  (iv) statement of momentum conservation:
                                  e.g. momentum of the truck before collision=momentum of both trucks after collision;
                                  (allow clear symbolism instead of words)
                                                                   2.2
                                         3             3
                                  2.2×10 ×2.8=5.2×10 V or V =         ×2.8;
                                                                   5.2
                                                   -1
                                  to give V=1.2ms
                                  (v) the first truck loses kinetic energy that is transferred to internal energy in the links between the trucks (and as sound);
                                  and to kinetic energy of the stationary truck; 
                                  Award [0] for “lost as heat, light and sound”, or “in air resistance”.
                                On the diagram draw and label arrows that represent the forces acting on the block.                                                    [2 marks]
                           4a.
                                  Markscheme
                                  (normal) reaction/N/R and weight/force of gravity/gravity force/gravitational force/mg/w/W with correct directions;
                                  friction/frictional force/F/F  with arrow pointing down ramp along surface of ramp;
                                                            f
                                  Do not allow “gravity” as label. Do not allow “drag” as label for friction.
                                Calculate the magnitude of the friction force acting on the block.                                                                     [3 marks]
                           4b.
                                  Markscheme
                                  recognize that friction=T-W sin θ;
                                                  3
                                  W sin θ=3.1×10 N;
                                                 3
                                  friction=1.1×10 N;