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6 [143 marks]
1a. A small ball of mass m is moving in a horizontal circle on the inside surface of a frictionless hemispherical bowl. The normal reaction force N makes an angle θ to the horizontal. State the direction of the resultant force on the ball.
Markscheme
towards the centre « of the circle » / horizontally to the right Do not accept towards the centre of the bowl [1 mark] 1b. On the diagram, construct an arrow of the correct length to represent the weight of the ball. [1 mark] [2 marks]
Markscheme
downward vertical arrow of any length arrow of correct length Judge the length of the vertical arrow by eye. The construction lines are not required. A label is not required eg: [2 marks] 1c. Show that the magnitude of the net force^ F^ on the ball is given by the following equation.
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ALTERNATIVE 1
F = N cos θ mg = N sin θ dividing/substituting to get result ALTERNATIVE 2 right angle triangle drawn with F, N and W/mg labelled angle correctly labelled and arrows on forces in correct directions correct use of trigonometry leading to the required relationship tan θ = [3 marks]
F =
mg
tan θ
O A =^ m Fg
1d. The radius of the bowl is 8.0 m and^ θ^ = 22°. Determine the speed of the ball. [3 marks] [4 marks]
1g. The ball is now displaced through a small distance x from the bottom of the bowl and is then released from rest. The magnitude of the force on the ball towards the equilibrium position is given by where R is the radius of the bowl. Outline why the ball will perform simple harmonic oscillations about the equilibrium position.
Markscheme
the « restoring » force/acceleration is proportional to displacement Direction is not required [1 mark]
mgx
R
1h. Show that the period of oscillation of the ball is about 6 s.
Markscheme
ω = « » = « = 1.107 s » T = « = = » 5.7 « s » Allow use of or g = 9.8 or 10 Award [0] for a substitution into T = 2 π [2 marks]
√ Rg √ 9.81 8.0^ –
2 π ω 2 π
√ I g
[1 mark] [2 marks]
1i. The amplitude of oscillation is 0.12 m. On the axes, draw a graph to show the variation with time^ t^ of the velocity^ v^ of the ball during one period.
Markscheme
sine graph correct amplitude « 0.13 m s » correct period and only 1 period shown Accept ± sine for shape of the graph. Accept 5.7 s or 6.0 s for the correct period. Amplitude should be correct to ± square for MP eg: v /m s [3 marks]
[3 marks]
Markscheme
arrow vertically downwards from dot labelled weight/W/mg/gravitational force/F /F AND arrow vertically upwards from dot labelled reaction force/R/normal contact force/N/F W > R Do not allow gravity. Do not award MP1 if additional ‘centripetal’ force arrow is added. Arrows must connect to dot. Ignore any horizontal arrow labelled friction. Judge by eye for MP2. Arrows do not have to be correctly labelled or connect to dot for MP2. g gravitational N 2d. The hill at point B has a circular shape with a radius of 20 m. Determine whether the skier will lose contact with the ground at point B.
Markscheme
ALTERNATIVE 1
recognition that centripetal force is required / seen = 468 «N» W/640 N (weight) is larger than the centripetal force required, so the skier does not lose contact with the ground ALTERNATIVE 2 recognition that centripetal acceleration is required / seen a = 7.2 «ms » g is larger than the centripetal acceleration required, so the skier does not lose contact with the ground ALTERNATIVE 3 recognition that to lose contact with the ground centripetal force ≥ weight calculation that v ≥ 14 «ms » comment that 12 «ms » is less than 14 «ms » so the skier does not lose contact with the ground ALTERNATIVE 4 recognition that centripetal force is required / seen calculation that reaction force = 172 «N» reaction force > 0 so the skier does not lose contact with the ground Do not award a mark for the bald statement that the skier does not lose contact with the ground. mv^2 r v^2 r
–1 – mv^2 r 2e. The skier reaches point C with a speed of 8.2 m s^. She stops after a distance of 24 m at point D. Determine the coefficient of dynamic friction between the base of the skis and the snow. Assume that the frictional force is constant and that air resistance can be neglected.
[3 marks] [3 marks]
Markscheme
ALTERNATIVE 1
0 = 8.2 + 2 × a × 24 therefore a = «−»1.40 «m s » friction force = ma = 65 × 1.4 = 91 «N» coefficient of friction = = 0. ALTERNATIVE 2 KE = mv = 0.5 x 65 x 8.2 = 2185 «J» friction force = KE/distance = 2185/24 = 91 «N» coefficient of friction = = 0. Allow ECF from MP1. 2 − 91 65 ×9. 1 2 2 2 91 (^65) ×9. 2f. At the side of the course flexible safety nets are used. Another skier of mass 76 kg falls normally into the safety net with speed 9.6 m s . Calculate the impulse required from the net to stop the skier and state an appropriate unit for your answer.
Markscheme
«76 × 9.6»= 730
Ns OR kg ms
1
2g. Explain, with reference to change in momentum, why a flexible safety net is less likely to harm the skier than a rigid barrier.
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safety net extends stopping time F = therefore F is smaller «with safety net» OR force is proportional to rate of change of momentum therefore F is smaller «with safety net» Accept reverse argument. Δ p Δ t 3a. The gravitational potential due to the Sun at its surface is –1.9 x 10 J kg. The following data are available. Mass of Earth = 6.0 x 10 kg Distance from Earth to Sun = 1.5 x 10 m Radius of Sun = 7.0 x 10 m Outline why the gravitational potential is negative.
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potential is defined to be zero at infinity so a positive amount of work needs to be supplied for a mass to reach infinity 11 – 24 11 8 [2 marks] [2 marks] [2 marks]
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centripetal force is required and is provided by gravitational force between Earth and Sun Award [1 max] for statement that there is a “centripetal force of gravity” without further qualification. 4a. (i) Define^ gravitational field strength. (ii) State the SI unit for gravitational field strength.
Markscheme
(i) «gravitational» force per unit mass on a «small or test» mass (ii) N kg Award mark if N kg is seen, treating any further work as neutral. Do not accept bald m s
4b. A planet orbits the Sun in a circular orbit with orbital period^ T^ and orbital radius^ R. The mass of the Sun is^ M. (i) Show that. (ii) The Earth’s orbit around the Sun is almost circular with radius 1.5×10 m. Estimate the mass of the Sun.
Markscheme
i clear evidence that v in is equated to orbital speed OR clear evidence that centripetal force is equated to gravitational force OR clear evidence that a in etc is equated to g in with consistent use of symbols Minimum is a statement that is the orbital speed which is then used in Minimum is F = F ignore any signs. Minimum is g = a. substitutes and re-arranges to obtain result Allow any legitimate method not identified here. Do not allow spurious methods involving equations of shm etc ii «T = 365 × 24 × 60 × 60 = 3.15 × 10 s» 2×10 «kg» Allow use of 3.16 x 10 s for year length (quoted elsewhere in paper). Condone error in power of ten in MP1. Award [1 max] if incorrect time used (24 h is sometimes seen, leading to 2.66 x 10 kg). Units are not required, but if not given assume kg and mark POT accordingly if power wrong. Award [2] for a bald correct answer. No sf penalty here.
T = √ 4 Gπ^2 MR^3
11
v^2 = 4 π
(^2) R 2
T^2 √^
GM R
a = v R^2 g = G RM 2
√ G RM
v = 2 π TR
c g
≪ T = √ 4 π^2 R = √ ≫
( G RM 2 ) 4 π^2 R^3 GM 7
M = ≪ 4 π^2 R =≫ =
3 GT^2 4 ×3.14^2 ×(1.5× 1011 )^3 6.67× 10 −^11 ×(3.15× 107 )^2 30 7 35 [2 marks] [4 marks]
5a. The diagram shows a planet near two stars of equal mass M. Each star has mass M=2.0×10 kg. Their centres are separated by a distance of 6.8×10 m. The planet is at a distance of 6.0×10 m from each star. On the diagram above, draw two arrows to show the gravitational field strength at the position of the planet due to each of the stars.
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two arrows each along the line connecting the planet to its star AND directed towards each star arrow lines straight and of equal length Do not allow kinked, fuzzy curved lines. 30 11 11 5b. Calculate the magnitude and state the direction of the resultant gravitational field strength at the position of the planet.
Markscheme
OR 3.7×10 Nkg directed vertically down «page» OR towards midpoint between two stars OR south Allow rounding errors.
g =≪ G rM 2 = 6.67×^10 −^11 ×2.0×^1030 ≫
(6.0× 1011 )^2 -4 -
g net =≪ 2 g cos θ = 2 × 3.7 × 10 −^4 × √6.0 =≫ 6.1 × 10 −^4 Nkg−^1
(^2) −3.4 2
[2 marks] [3 marks]
Markscheme
(i) her direction is changing; hence her velocity is changing; or since her direction/velocity is changing; a resultant/unbalanced/net force must be acting on her (hence she is accelerating); (ii) arrow from Aibhe towards centre of merry-go-round; Ignore length of arrow. (iii) the force of the merry-go-round on Aibhe/her; (iv) no force is acting on the upper body towards the centre of the circle / no centripetal force acting on the upper body (to maintain circular motion); upper body (initially) continues to move in a straight line at constant speed/ velocity is tangential to circle; 6c. Euan is rotating on a merry-go-round and drags his foot along the ground to act as a brake. The merry-go-round comes to a stop after 4.0 rotations. The radius of the merry-go-round is 1.5 m. The average frictional force between his foot and the ground is 45 N. Calculate the work done.
Markscheme
distance travelled by Euan= ; ;
4.0 × 2 π × 1.5(= 37.70m)
W (= Favd = 45 × 37.70) = 1700 (J)
6d. Aibhe moves so that she is sitting at a distance of 0.75 m from the centre of the merry-go-round, as shown below. Euan pushes the merry-go-round so that he is again moving at 1.0 ms relative to the ground. (i) Determine Aibhe’s speed relative to the ground. (ii) Calculate the magnitude of Aibhe’s acceleration.
Markscheme
(i) Aibhe’s period of revolution is the same as before; from , since r is halved, v is halved; v=0.5(ms ); Award [3] for a bald correct answer. (ii) ; a=0.33(ms ); Allow ECF from (d)(i). Award [2] for a bald correct answer.
v = 2 π Tr
a (= v r^2 ) =^ 0.5 0.75^2
7a. This question is about the thermodynamics of a car engine and the dynamics of the car. A car engine consists of four cylinders. In each of the cylinders, a fuel-air mixture explodes to supply power at the appropriate moment in the cycle. The diagram models the variation of pressure P with volume V for one cycle of the gas, ABCDA, in one of the cylinders of the engine. The gas in the cylinder has a fixed mass and can be assumed to be ideal. At point A in the cycle, the fuel-air mixture is at 18 °C. During process AB, the gas is compressed to 0.046 of its original volume and the pressure increases by a factor of 40. Calculate the temperature of the gas at point B.
Markscheme
535 (K) / 262 (°C);
7b. State the nature of the change in the gas that takes place during process BC in the cycle.
Markscheme
constant volume change / isochoric / isovolumetric / OWTTE; 7c. Process CD is an adiabatic change. Discuss, with reference to the first law of thermodynamics, the change in temperature of the gas in the cylinder during process CD.
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Q/thermal energy transfer is zero; ; as work is done by gas internal energy falls; temperature falls as temperature is measure of average kinetic energy;
Δ U = − W
7d. Explain how the diagram can be used to calculate the net work done during one cycle. [1 mark] [1 mark] [3 marks] [2 marks]
Markscheme
(i) ; 2300 or 2.3k (N); Award [2] for a bald correct answer. (ii) resistive force or ; (allow ECF) so accelerating force or 1741 (N); or ; Award [2 max] for an answer of 0.49 (m s (omits 2300 N).
force = power speed
( 2300 − 580 =) 1725 (N)
a = 17251200 = 1.44 (m s−^2 ) a = 17411200 = 1.45 (m s−^2 )
7h. A driver moves a car in a horizontal circular path of radius 200 m. Each of the four tyres will not grip the road if the frictional force between a tyre and the road becomes less than 1500 N. (i) Calculate the maximum speed of the car at which it can continue to move in the circular path. Assume that the radius of the path is the same for each tyre. (ii) While the car is travelling around the circle, the people in the car have the sensation that they are being thrown outwards. Outline how Newton’s first law of motion accounts for this sensation.
Markscheme
(i) centripetal force must be ; (allow force 6000 N) ; ; Allow [3] for a bald correct answer. Allow [2 max] if 4 is omitted, giving 15.8 (m s ). (ii) statement of Newton’s first law; (hence) without car wall/restraint/friction at seat, the people in the car would move in a straight line/at a tangent to circle; (hence) seat/seat belt/door exerts centripetal force; (in frame of reference of the people) straight ahead movement is interpreted as “outwards”;
< 6000 (N)
v^2 = F × mr
31.6 (m s−^1 )
× –
8a. This question is about circular motion. The diagram shows a car moving at a constant speed over a curved bridge. At the position shown, the top surface of the bridge has a radius of curvature of 50 m. Explain why the car is accelerating even though it is moving with a constant speed. [6 marks] [2 marks]
Markscheme
direction changing; velocity changing so accelerating; 8b. On the diagram, draw and label the vertical forces acting on the car in the position shown.
Markscheme
weight/gravitational force/mg/w/F /F and reaction/normal reaction/perpendicular contact force/N/R/F /F both labelled; (do not allow “gravity” for “weight”.) weight between wheels (in box) from centre of mass and reactions at both wheels / single reaction acting along same line of action as the weight; Judge by eye. Look for reasonably vertical lines with weight force longer than (sum of) reaction(s). Extra forces (eg centripetal force) loses the second mark. w g N R 8c. Calculate the maximum speed at which the car will stay in contact with the bridge.
Markscheme
22(ms ); Allow [3] for a bald correct answer.
g = v r^2
v = √ 50 × 9.
9a.
This question is in two parts. Part 1 is about gravitational force fields. Part 2 is about properties of a gas.
Part 1 Gravitational force fields
State Newton’s universal law of gravitation.
Markscheme
the (attractive) force between two (point) masses is directly proportional to the product of the masses; and inversely proportional to the square of the distance (between their centres of mass); Use of equation is acceptable: Award [2] if all five quantities defined. Award [1] if four quantities defined. [2 marks] [3 marks] [2 marks]
10b. Deduce that the gravitational field strength^ g^ at the surface of a spherical planet of uniform density is given by where M is the mass of the planet, R is its radius and G is the gravitational constant. You can assume that spherical objects of uniform density act as point masses.
Markscheme
use of and ; evidence of substitution/manipulation; to get
g = GM
R^2
g = F m F = Gm R 2 M
g = G RM 2
10c. The gravitational field strength at the surface of Mars^ g^ is related to the gravitational field strength at the surface of the Earth^ g by g = 0.38 × g. The radius of Mars R is related to the radius of the Earth R by R = 0.53 × R. Determine the mass of Mars M in terms of the mass of the Earth M.
Markscheme
M E M E M E M E M E
= ⇒ = × [ ]
gM^2 gE MM R^2 M ME R^2 E MM ME gM gE RM RE
M M (= 0.38 × 0.53^2 M E) = 0.11 M E
[2 marks] [2 marks]
10d. (i) On the diagram below, draw lines to represent the gravitational field around the planet Mars. (ii) An object falls freely in a straight line from point A to point B in time t. The speed of the object at A is u and the speed at B is v. A student suggests using the equation v=u+g (^) Mt to calculate v. Suggest two reasons why it is not appropriate to use this equation. [3 marks]
