Momentum_P2. Markscheme 1NA 2NA. 3a. [2 marks]

Size: px

Start display at page:

Download "Momentum_P2. Markscheme 1NA 2NA. 3a. [2 marks]"

Rudolph McCormick
5 years ago
Views:

1 Momentum_P2 1NA 2NA 3a. [2 marks] arrow vertically downwards labelled weight «of sledge and/ girl»/w/mg/gravitational fce/f g /F gravitational AND arrow perpendicular to the snow slope labelled reaction fce/r/nmal contact fce/n/f N friction fce/f/f acting up slope «perpendicular to reaction fce» Do not allow G/g/ gravity. Do not award MP1 if a driving fce is included. Allow components of weight if crectly labelled. Igne point of application shape of object. Igne air resistance. Igne any reference to push of feet on sledge. Do not award MP2 f fces on sledge on hizontal ground The arrows should contact the object 1

2 3b. [3 marks] gravitational fce/weight from the Earth «downwards» reaction fce from the sledge/snow/ground «upwards» no vertical acceleration/remains in contact with the ground/does not move vertically as there is no resultant vertical fce Allow naming of fces as in (a) Allow vertical fces are balanced/equal in magnitude/cancel out 3c. [2 marks] mention of conservation of momentum OR 5.5 x 4.2 = ( ) «v» 0.38 «m s 1» Allow p=p other algebraically equivalent statement Award [0] f answers based on energy 3d. [3 marks] same change in momentum/impulse the time taken «to stop» would be greater «with the snow» therefe F is smaller «with the snow» OR fce is proptional to rate of change of momentum therefe F is smaller «with the snow» 2

3 Allow reverse argument f ice 3e. [3 marks] «friction fce down slope» = μmg cos(6.5) = «5.9 N» «component of weight down slope» = mg sin(6.5) «= 6.1 N» «so a =» acceleration = = 2.2 «m s 2» Igne negative signs Allow use of g = 10 m s 2 3f. [2 marks] crect use of kinematics equation distance = «m» Alternative 2 KE lost=wk done against friction + GPE distance = «m» Allow ECF from (e)(i) Allow [1 max] f GPE missing leading to 8.2 «m» 3g. [2 marks] calculates a maximum value f the frictional fce = «μr=» 7.5 «N» sledge will not move as the maximum static friction fce is greater than the component of weight down the slope Allow crect conclusion from increct MP1 Allow 7.5 > 6.1 so will not move 3

4 4NA 5NA 6a. [4 marks] distances itemized (it must be clear through use of distance I etc) distances equated / cancel and re-arrange substitution shown / 28.1(s) seen clear written statement that the average speed of B must be the same as constant speed of I as B starts from rest the final speed must be so 28.1 (s) seen (f this alternative the method must be clearly described) attempts to compare distance travelled by I and B f 28 s I distance B distance deduces that overtake must occur about later 4

5 6b. [2 marks] use of appropriate equation of motion 1.3 (km) Award [2] f a bald crect answer. 6c. [3 marks] driver I moves at constant speed so no net (extra) fce accding to Newton 1 driver B decelerating so (extra) fce (to rear of car) (accding to Newton 1) / momentum/inertia change so (extra) fce must be present (hence) greater tension in belt B than belt I Award [0] f stating that tension is less in the decelerating car (B). 6d. [2 marks] statement that momentum is conserved Allow [2] f a bald crect answer. 6e. [2 marks] use of fce ( any variant, eg: ) } (must see matched units and value ie: without unit gains MP2, 13.2 does not) Award [2] f a bald crect answer. Allow use of 58 m s 1 from (c)(i) to give (N). 5

6 7a. [2 marks] total momentum does not change/is constant } (do not allow momentum is conserved ) provided external fce is zero / no external fces / isolated system 7b. [9 marks] (i) clear attempt to calculate area under graph initial momentum is half change in momentum Award [2 max] f calculation of total change (1.92kg ms 1 ) (ii) initial speed 15(ms 2 ) (must see negative sign a comment that this is a deceleration) average fce =12 N uses F=0.8 a 15(ms 2 ) (must see negative sign a comment that this is a deceleration) Award [3] f a bald crect answer. Other solution methods involving different kinematic equations are possible. (iii) goes through t=0.08s and from negative momentum to positive / positive momentum to negative constant sign of gradient throughout curve as shown Award marks f diagram as shown. 6

7 7c. [4 marks] impulse is the same/similar in both cases / momentum change is same impulse is fce time / fce is rate of change of momentum time to come to rest is longer f car B fce experienced by car B is less (so less likely to be damaged) 7

8 8NA 9a. [1 mark] the net (external) fce acting on the system is zero / no fce acting on system / system is isolated 9b. [5 marks] (i) no external fce/system is isolated so change in momentum is zero { (do not accept momentum is conserved/constant) fce on ball must be equal and opposite to fce on the person so ball and person/earth/pond move in opposite directions (ii) Newton s second law states that the rate of change of momentum is equal/proptional/directly proptional to the fce acting the hizontal fce acting on the ball is zero therefe the momentum must be constant/the rate of change of momentum is zero Newton s second law can be expressed as the fce acting is equal to the product of mass and acceleration the hizontal fce acting on the ball is zero therefe the acceleration is zero so velocity is constant (and therefe momentum is constant) 9c. [2 marks] 17kN 8

9 9d. [4 marks] (i) =10 v v=1.5ms 1 (ii) KE lost =18kJ 9e. [3 marks] initial KE friction =280 N use of kinematic equation to give a=0.274ms 1 use of F(=ma)= a 270/280 N 9

10 10a. [4 marks] (i) (gravitational) potential energy (of club head) goes to kinetic energy (of club head) some kinetic energy of club head goes to internal energy of club head/kinetic energy of ball (ii) equating mgh to mv 2 v=4.1(ms -1 ) Award [0] f answers using equation of motion not unifm acceleration. 10b. [2 marks] defmation prolongs the contact time increased impulse => bigger change of momentum/velocity (club head) stes (elastic) potential energy on compression this energy is passed to the ball 10c. [5 marks] (i) any value of 1.3 x 10 4 (N) (ii)-1.3 x 10-4 (N) Accept statement that fce is in the opposite direction to (c)(i). Allow the negative of any value given in (c)(i). (iii) clear use of conservation of momentum / impulse = change of momentum 21(ms -1 ) 10

11 a= v=(u+at= x =) 21(ms -1 ) Award [2] f a bald crect answer. 11a. [2 marks] if the net external fce acting on a system is zero the momentum of the system remains constant/unchanged/the same f a closed system the momentum remains constant/unchanged/the same Award [1] f momentum befe collision equals momentum after collision. Do not accept momentum is conserved. 11b. [3 marks] identifies the system as rocket + exhaust gases / total momentum of rocket and gas is equal befe and after (it must be clear that this is the system, a mention of rocket and gases is not enough) no external fces act on this system / closed system increase/change in momentum of the gases is equal and opposite to the increase/change of momentum of the rocket 11

12 11c. [4 marks] (i) attempts to use conservation of momentum, eg =52 v v=0.20(ms -1 ) Award [2] f a bald crect answer. (ii) identifies new mass as 75.3(kg) V=0.14(ms -1 ) 12a. [2 marks] if no external fces act / isolated system momentum is constant / (total) momentum befe=(total) momentum after 12b. [7 marks] (i) use of 6.11ms 1 (must show calculation to better than 1 sf) (ii) rate of change of vertical momentum= N (accept answers in the range of 78N to 80N) (iii) mass accrued=5.0 13=65kg weight of this mass (=65 9.8)=637N (650 from g=10ms 2 ) total fce (637+79=)716N } (allow ECF from (b)(ii) and from increct weight) 12

13 12c. [4 marks] (i) 14.6Js 1 (ii) hizontal momentum gain per second =13 1.5( =19.5kgms 1 ) 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 hizontal speed / OWTTE 13a. [3 marks] distance between surfaces of blocks= =0.850m relative speed between blocks = 0.36ms 1 blocks moving at same speed so meet at mid-point distance travelled by block= =0.425m Award [3] f bald crect answer. Award [2 max] if distance of 0.90 m 0.45 m used to get 2.5 s. 13

14 13b. [5 marks] (i) the collision is inelastic because kinetic energy is not conserved (although momentum is) (ii) initial final final speed =0.16ms initial E K =final E K =0.16ms -1 13c. [7 marks] (i) if object A exerts a fce on object B, then object B (simultaneously) exerts an equal and opposite fce on object A / every action has an equal and opposite reaction / OWTTE (ii) arrows of equal length (judge by eye) acting through centre of blocks crect labelling consistent with crect direction 14

15 (iii) F=0.83N 14a. [1 mark] mass velocity (allow mv with symbols defined) 14b. [1 mark] the rate of change of momentum of a body is equal to/directly proptional to the fce acting on the body Accept only if all symbols are defined. 14c. [1 mark] therefe impulse FΔt =Δp (accept t f Δt) 14d. [1 mark] 15

16 therefe impulse FΔt =Δp (accept t f Δt) 14e. [12 marks] (i) (impulse=) change in momentum= (= Ns) impulse=area under graph=½f max Δt ½F max 0.54= F max =35k(N) (N) (ii) (magnitude of) acceleration= time= Award [1 max] if an additional 0.54 s is added to answer. (iii) rate of change of (mark is f division by 4.2 and crect calculation) (iv) statement of momentum conservation: e.g. momentum of the truck befe collision=momentum of both trucks after collision (allow clear symbolism instead of wds) = V to give V=1.2ms -1 (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] f lost as heat, light and sound, in air resistance. 16

17 15NA 16NA 17a NA 17b. NA 17c. NA 17d. [2 marks] the total momentum of a system is constant provided external fce does not act the momentum of an isolated/closed system is constant Award [1] f momentum befe collision equals collision afterwards. 17e. [6 marks] (i) initial momentum final speed Watch f increct mass values in equation. (ii) initial kinetic energy of pellet clay block 17

18 use of appropriate kinematic equation with consistent sign usage e.g. 17f. [3 marks] kinetic energy of pellet is transferred to kinetic energy of clay block and internal energy of pellet and clay block clay block loses kinetic energy as thermal energy/heat 17g. [2 marks] Allow g 10 m s 2 answer 4.1 m s 2 Award [2] f bald crect answer. 18

Momentum_P2 1 NA 2NA. 3a. [2 marks] A girl on a sledge is moving down a snow slope at a uniform speed.

Momentum_P2 1 NA 2NA. 3a. [2 marks] A girl on a sledge is moving down a snow slope at a uniform speed. Momentum_P2 1 NA 2NA 3a. [2 marks] A girl on a sledge is moving down a snow slope at a uniform speed. Draw the free-body diagram for the sledge at the position shown on the snow slope. 3b. [3 marks] 1