AQA A-Level Physics Year 2 Student Book Answers

A1 The total lateral friction on the tyres to just avoid the car going off at a tangent is equal to the required centripetal force: F = m v 2 r =690´(120´103 ¸3600)2 70 =10952=11000 N A2 The asymmetric curved shape and angle of a racing car aerofoil makes moving air flow faster below it than above, so the air pressure is lower below the aerofoil. This creates a downward force on the aerofoil pushing the car downwards. A3 At the Abbey bend, g-force = centripetal acceleration acceleration due to gravity = v 2 r g =(120´103 ¸3600)2 ¸70 9.81 =1.6 So the g-force typically experienced by a driver at the Abbey bend is 1.6g. ASSIGNMENT 2 A1 a. Estimate from the diagram of the beam length from the centre to the back of the chair  5 m. Centripetal acceleration a = r w 2 = r ´ 2p T æ è ç ö ø ÷ 2 =5´4p2 100 2 m s-2 b. The normal contact force exerted by the seat on the rider’s back provides the centripetal force = mass of rider ´ centripetal acceleration. Estimate of rider’s mass  60 kg. Normal contact force 60´2=120N A2 a. There is nothing to provide a centripetal force on them, so they tend to move tangentially until they hit the side of the drum. b. Centripetal acceleration a = r w 2 = r ´(2p f )2 =0.3´ 2p´3000 60 æ è ç ö ø ÷ 2 =30000 m s–2 c. Using the estimate of the mass of a swimming costume  50 g, the normal contact force, N, on the inside surface on the swimming costume is given by N = m a 0.05´30000=1500 N ASSIGNMENT 3 A1 2 A2 a. m g + N = m v 2 r b. N =60´8.82 5 - 60´9.81=341 N A3 N – m g = m v 2 r N =60´152 9 +(60´9.81)=2090 N A4 At the top of the loop m g + N = m v 2 r If N = 0, m g = m v 2 r , which gives v = rg = 5´9.81=7.0 m s-1 PRACTICE QUESTIONS 1a. i. 1a. ii. m g + T = m v 2 r Hence ´ = - ´ = 20.05 3 0.05 9.81 0.26N 0. ( ) 6 T 1b. i. 1b. ii. T - m g = m v 2 r 1c. The string is most likely to snap when the mass is at the bottom of its path since the tension in the string is at its greatest at this point. 2a. Although the aircraft has a constant speed, its direction is changing. Since velocity is a vector quantity, if the direction changes the velocity must also be changing. If there is a velocity change there must be some acceleration since acceleration is equal to rate of change of velocity. 2b. When the aircraft is banked, it rotates about its centre of mass so that the lift force on the wings is no longer acting in the vertical direction. The horizontal component of the lift creates a resultant force towards the centre of the circle which provides the centripetal force. (See figure 16 in the Student Book.) 3 2c. i. The banking angle  can be found from tan  = v 2 rg =(280´103 ¸3600)2 3000´9.81 =0.2056, which gives the banking angle,  = 12°. 2c. ii. centripetal acceleration a = v 2 r =(280´103 ¸3600)2 3000 =2.0 m s-2 3a. The centripetal force on the electron is given by - = = ´ 2 88.2 10mvF r N. Rearranging gives the electron’s speed - - - - - ´ = ´ ´ = ´ ´ = ´ ´ 9 8 8 6 1 31 0.053 10 8.2 10 8.2 10 2.2 10 m s 9.11 10 rv m 3b. The angular speed w - - ´ = = = ´ = ´ ´ 6 16 16 1 9 2.2 10 4.151 10 4.2 10 rad s 0.053 10 v r 3c. The time for one orbit w -p p = = = ´ ´ 16 16 2 2 1.5 10 s 4.151 10 T 4. C (Centripetal force:   w = = p = p 2 2 2 2 2 2 2 D F mr m f mD f ) 5. D (Speed v =2p r T =2´p´1.5´1011 365´24´3600 =3.0´104 ms-1) 6. A (The speed of the object is uniform so the kinetic energy æ ö ç ÷ è ø 21 2 mv must not change) 7. A (The speed of the man, v = r w= r ´2p T =15´2´p 50´60 = p 100 ms-1) 8. C (When the mass is about to slip, the centripetal force m r w2 is equal to the maximum frictional force m g 2 . Hence m r w2 = m g 2 . Rearranging gives = 2 g r ) 4 Chapter 2 REQUIRED PRACTICAL P1 Human reaction time in switching the stopwatch on and off to correspond with the mass passing the fiducial marker will probably vary and therefore is best described as a random error. The repeat measurements will help to reduce the effects of this error. P2 a. i. Range of 20T data = 0.22 s. Uncertainty in the average time for 20T is half this, 0.11 s. ii. The uncertainty in 20T determined from the range of data is less than the typical reaction time error of 0.2 s. b. Uncertainty in the average 20T value is best estimated as 0.2 s (as the value in part a.i. is based on only three repeats). This gives the uncertainty in time period T as 0.2 20 = 0.01 s. So the time period T = 0.49  0.01 s. c. i. % uncertainty in T is 0.01 0.49 ´ 100% =  2% % uncertainty in T2 = 2 ´ 2% = 4% T2 = (0.49)2 = 0.24 s2 Uncertainty in T2 = 4 100 ´ 0.24 = 0.01 s2 ii. The % uncertainty in the values of m could be 2 50 ´ 100% = 4% if 50 g masses, which is comparable to the % uncertainty in T2. If 100 g masses were used the % uncertainty in the values of m could be 2 100 ´ 100% = 2%. The student has a range of masses to choose from. If she uses masses greater than 50 g this reduces the % uncertainty in the mass, making T the greater source of uncertainty. However, if she uses masses less than 50 g this increases the % uncertainty in the mass, making m the greater source of uncertainty. iii. If 10 oscillations were counted the uncertainty in T would be 0.2 10 = ±0.02 s, giving a % uncertainty of 4% in T and hence 8% in T2. So this would be a source of error considerably greater than that of the mass values if masses of 50 g or more were to be used, but if masses of 20 g were used then % uncertainty in the values of m could be 2 20 ´ 100% = 10%, so the greater uncertainty would still be in the mass. P3 The uncertainty due to reaction time would be eliminated and the uncertainty in the time measurement itself is likely to be similar to or less than that of a digital stopwatch. If the same 50 g masses are used with an uncertainty of up to 4%, the uncertainty in the analysis and accuracy in the value of k will be improved by approximately a factor of 2 compared with the method involved with the timing of 20 oscillations. P4 The pendulum should be set in motion and after a few swings, when it is moving with small amplitude, 20 oscillations should be timed with a digital stopwatch, counting the number of times that the pendulum bob passes the fiduciary marker going in the same direction. Two repeat measurements should be made and an average time for 20 oscillations calculated. The time period T is then found by dividing this by 20. This is repeated for a range of pendulum lengths l, each time measuring the length as stated in the Technique section. A graph of T2 against l is plotted. If a best-fit line can be drawn through the points and through the origin, the proportional relationship is confirmed. A value for g can be obtained by measuring the gradient of the best-fit line, using as large a triangle as possible: gradient = 4p2 g so g = 4p2 gradient . 5 P5 Draw the line of best fit and calculate the best gradient from this. Then draw lines of maximum gradient and minimum gradient that are still a reasonable fit to the points plotted. Whichever of these two gradients is furthest from the line of best fit is called the ‘worst gradient’ and the percentage uncertainty in the best gradient is found from percentage uncertainty = best gradient- worst gradient best gradient ´ 100% The percentage uncertainty in the gradient is equal to the percentage uncertainty in the value for g. The uncertainty in g can be found from: ´ percentage uncertainty 100 g ASSIGNMENT 1 A1 [no answer required] A2 a. l m T s log l m æ è ç ö ø ÷ log T s æ è ç ö ø ÷ 0.250 1.00 −0.602 0.000 0.300 1.09 −0.523 0.037 0.350 1.18 −0.456 0.072 0.400 1.27 −0.398 0.104 0.450 1.34 −0.347 0.127 0.500 1.42 −0.301 0.152 0.550 1.48 −0.260 0.170 0.600 1.55 −0.222 0.190 6 b. From the graph, gradient = 0.21– 0 (- 0.20) - (- 0.60) = 0.21 0.40 = 0.53. This is within 6% of the theoretical prediction for the gradient, and that 6% difference could easily be accounted for by uncertainty in the measurements and by lack of precision in plotting the points and drawing the line of best fit. PRACTICE QUESTIONS 1a. The student should take at least three measurements of the time for 20 cycles and calculate an average of the three values before dividing by 20 to get the time period. 1b. The equilibrium position is the preferred location for the fiducial marker since it is easier to see the pendulum pass the equilibrium position than reach the extreme position because the amplitude of the oscillation continuously decreases during the 20 oscillations. 2a. Time period of the mass-spring system T =2p m k =2p 0.5 50 =0.628 s 2b. Percentage uncertainty in the mass: %Um = 10 500 ´100 = 2% Percentage uncertainty in the spring constant: %Uk = 2 50 ´100 = 4% Therefore the percentage uncertainty in the predicted value for the time period is 1 2 ´ (%Um + %Uk) = 1 2 ´ (2 + 4) = 3% The uncertainty in the predicted time period = 3 100 ´ 0.628 = 1.9 ´ 10–2 = 0.02 s Therefore the predicted time period T = 0.63  0.02 s 2c. i. A sample rate of 20 Hz means that a measurement is made every 1 20 s = 0.05 s = 50 ms. 7 2c. ii. The number of measurements made in 8 cycles = 8´0.628 0.05 = 100. 3a. i. The force exerted on the trolley by the spring on the left has decreased by F = k∆l = 30 ´ 0.06 = 1.8 N and the force exerted by the spring on the right has increased by 1.8 N creating a resultant force of 3.6  N to the right. 3a. ii. Acceleration a = F m =3.6 0.8 =4.5ms–2 to the right. 3b. i. For SHM, the acceleration is directly proportional to the displacement but in the opposite direction. 3b. ii. Frequency f = 1 2p 2 k m = 1 2p 2´30 0.8 =1.378Hz so time period T =1 f = 1 1.378 =0.73 s 3c. i. Vibration frequency of copper ion f = 1 2p 2 k m = 1 2p 2´200 1´10–25 =1.0´1013 Hz 3c. ii. Maximum speed vmax = w A = 2pfA = 2p ´ 1 ´ 1013 ´ 1 ´ 10–11 = 628 m s–1 3c. iii. Maximum kinetic energy Ek = 1 2 mv 2max = 1 2 ´ 1 ´ 10–25 ´ (628)2 = 2.0 ´ 10–20 J 4a. When the bob is displaced to one side it gains height so on release the energy of the bob changes from gravitational potential energy to kinetic energy and back to gravitational potential energy on reaching the other extreme. 4b. Time period T = 2p l g = 2p 0.6 9.81 = 1.55 s 4c. i. During the swing of the bob from the equilibrium position to one extreme the bob describes an arc of a circle. When the bob is first displaced, the angle between the string and the vertical is 10°, which is equal to 10 360 ´ 2p rad. The length of the arc is equal to the amplitude and is equal to rθ = 0.6 ´ 10 360 ´ 2p = 0.105 m. 4c. ii. The maximum speed is reached as the bob passes through the equilibrium position and is given by vmax = w A = 2 p T æ è ç ö ø ÷A = 2p´0.105 1.55 = 0.426 = 0.43 m s–1 4c. iii. As the bob passes through the equilibrium position, the tension exceeds the bob’s weight by an amount equal to the required centripetal force since the bob is describing the arc of a circle. The magnitude of the tension at the equilibrium position is given by mg + m v 2 r = (0.02 ´ 9.81) + 0.02´0.4262 0.6 = 0.202 N 5. D (Since the defining formula for SHM is a = – w 2 x, the graph of a versus x is a straight line with a negative slope since angular frequency, w is constant.) 6. D (The maximum speed is given by vmax = w A = 2pfA = 2p ´ 320 ´ 0.5 = 320p mm s–1) 7. B 8. A (Speed v = 2pf A 2 – x 2 = v = 2p 0.22 – 0.12 = 1 m s–1)