DHS 01 Measurements (Tutorial Solutions)

Dunman High School H2 Physics 2024/2025 1 TOPIC 1: MEASUREMENT TUTORIAL Suggested Answers 1 (a) Derived quantities Definition Defining Equation Derived units (base) Alternative unit(s) Area The extent of a two- D figure or shape in a plane. l π = × = rectangle 2 circle Aw Ar m2 - Moment Moment of a force about a point is the product of the magnitude of the force and the perpendicular distance of the force from the point. ⊥= ×moment of a force Fd kg m2 s-2 N m Pressure Force per unit area. = Fp A kg m-1 s-2 N m-2 , Pa Work done Force × Displacement in direction of force θ=W cosFd kg m2 s-2 J Power Work done per unit time. = WP t kg m2 s-3 J s-1 , W Resistance The ratio of potential difference across the conductor to the current flowing in it. I= VR kg m2 s-3 A-2 V A-1, Ω 1(b) ( ) = ⇒= = = = = = -1 2 -2 2 -1 Units of JUnits of J sUnits of s Base units of kg m s s kg m s E hf Eh f Eh f h Common mistake: Students often incorrectly equate physical quantity with units in the working, for example, by writing: h = kg m2 s−1, which is an incorrect equation. ( ) ( ) = = = = = = -2 2 -2 -1 -12 -2 2 -3 -1 1 kg m s m1 J 1 Nm1 V 1 kg m s C1 C 1 C 1 C 1 kg m s A s 1 kg m s A

Dunman High School H2 Physics 2024/2025 2 2. ( )( )( )( ) 2 2 3 2 22 2 2 base units of base units of 1base units of 2 kg m s 1 kg m m m s kg m s 1kg m s D FC Avρ − −− − − = = = = Hence, CD has no units. It is dimensionless. 3. A homogeneous equation is one in which every term has the same units. To determine whether an equation is homogeneous, find the units of each term and compare. If the units are the same, the equation is homogeneous. Unit of v = m s −1. (A) ( ) ( ) 2 2 2 -1Units of m s m m s m s vg g λ λ −− = = = = Comparing units of v and λg , we conclude that the equation λ=vg is homogeneous. (B) 2 21m sUnits of s s gv h g hm − −− = = = = Comparing units of v and g h , we conclude that the equation = gv h is not homogeneous. (C) ( )( ) ( ) 11 3 2 12 122Units of kg m m s m kg m s kg m s v gh gh ρ ρ −− − −− − = = = = Comparing units of v and ρgh , we conclude that the equation ρ=v gh is not homogeneous. (D) 12 142 212 3 m sUnits of kg m s kg m sk g m gv g ρ ρ − −−− − − = = = = Comparing units of v and ρ g , we conclude that the equation ρ= gv is not homogeneous.

Dunman High School H2 Physics 2024/2025 3 4. (i) Pressure is defined as = force areaP . Base unit of pressure P = Base unit of 2 12 2 force kg m s kg m sarea m − −−= = (ii) 12 2 2 1 3 kg m sUnits of m s m skg m Pc γ ρ −− −− −= = = = (iii) c could be velocity or speed. 5. ρρ++ = 21 2p hg v k (i) ( ) ( ) ( )( ) ( )( ) 2 2 12 3 2 12 22 3 1 12 Base units of kg m s m kg m s Base units of m kg m m s kg m s 1Base units of kg m m s kg m s2 p hg v ρ ρ − − −− − − −− − − −− = = = = = = Since all terms have the same base units, the equation is dimensionally consistent. Note: Homogenous equation may not be physically correct. (ii) The base units for k should also be 12kg m s−− . 6. (a) v x = 37 cos 28o = 33 m s−1, horizontally to the right vy = 37 sin 28o = 17 m s−1, vertically downwards (b) F x = 530 cos51o = 530 sin 39 = 334 N, downwards parallel to the incline plane Fy = 530 sin51o = 530 cos 39 = 412 N, perpendicularly towards the incline plane 39o

Dunman High School H2 Physics 2024/2025 4 7. (i) Resultant vector in the x-x direction Rx = -35 cos45o + 15 cos30o+ 0= −11.8 N Resultant vector in the y-y direction Ry = 35 sin45o + 15 sin30o - 78 = −45.8 N Vector diagram to show resultant: Magnitude of resultant force R = +=2245.8 11.8 47.3 N θ − = = 1o 45.8tan 7611.8 Resultant force R = 47.3 N in the direction 76o anticlockwise from the negative x-axis. (ii) Smallest force that must be applied such that the resultant is along x-x direction is 45.8 N in the direction 90o anticlockwise from the positive x-axis. 8. Force component along the N-S direction = 27.56 + 7.89 cos34 o + 10 sin200 – 12 sin 60o – 10 sin10o = 25.4 N Force component along the E-W direction = 0 + 7.89 sin34 o +10 cos20o + 12 cos60o -10 cos10o = 9.96 N Magnitude of resultant force = +=229.96 25.4 27.3 N θ − = = 1o 9.96tan 21.425.4 The resultant force is 27.3 N, 21.4o clockwise from the North direction. y y θ 11.8N 45.8N R 9.96 N 25.4 N θ North

Dunman High School H2 Physics 2024/2025 5 9a Taking leftward direction as positive, Change in velocity v∆  = final velocity – initial velocity = ( ) fif ivvv v = +−−   Vector diagram: Thus the change in velocity is 65 m s -1 away from racket. 9b Vector diagram: It is an equilateral triangle! Thus the change in velocity is 15 m s −1, vertically downwards. 9c Vector diagram: By Cosine Rule, ( )∆= + −××× =2 2 o -13 5 2 3 5 cos 150 7.7 m sv By Sine Rule, θ θ − = ⇒= × =  oo 1osin sin150 sin150sin 5 195 7.7 7.7 The change in velocity is 7.7 m s-1, (30o - 19o) = 11o anticlockwise from the horizontal. 60o 60o 15 m s-1 15 m s-1 30o 3 m s-1 5 m s-1 30o (3 m s-1) (5 m s-1) 150o θ 35 m s-1 30 m s-1 racket fv  iv−  fivv−  60o 60o (15 m s-1) (15 m s-1) (15 m s-1) fv  iv− 

Dunman High School H2 Physics 2024/2025 6 10. D 1.02 4 1.01 5.09Average 1.0255 C orrected diameter average diameter zero reading 1.02 ( 0.02) 1.04 mm ×+= = = = − = −− = 11. C Random errors - where repeating the measurement gives an unpredictably different result. • cannot possibly be eliminated (P2) • has varying sign and magnitude (Q2) • can be reduced by averaging repeated measurements (R1) The random errors tend to cancel each other out, and the residual error is divided by the number of readings, so it gets shared out among many readings. Increasing the number of measurements generally reduces the uncertainty in the mean value. 12. C E xperimental technique that reduces the systematic error of the quantity being investigated: Adjusting an ammeter to remove its zero error before measuring a current 13. A n object of mass 1.000kg is placed on four different balances. For each balance the reading is taken five times. The table shows the values obtained together with the means. Which balance has the smallest systematic error but is not very precise? (N2002/I/2) Balance Reading/ kg mean/kg │Mean – 1.000 kg│ 1 2 3 4 5 A 1.000 1.000 1.002 1.001 1.002 1.001 0.001 B 1.011 0.999 1.001 0.989 0.995 0.999 0.001 C 1.012 1.013 1.012 1.014 1.014 1.013 0.013 D 0.993 0.987 1.002 1.000 0.983 0.993 0.007 Smallest s ystematic error gives better accuracy , i.e. the ‘mean’ of the readings is closer to the true value of 1.000 kg. (Either balance A or B whose mean readings are 1.001 and 0.999 respectively) ‘Not precise’ means that the measurements has large random errors , i.e. the readings are ‘not close’ to each other. This can be ‘quantified’ using the ‘spread’ of the readings given by (largest reading – smallest reading). Range or ‘Spread’ for balance A = 1.002 - 1 .000 = 0.002 Range or ‘Spread’ for balance B = 1.011 - 0.989 = 0.022 Readings for Bal ance B are less precise than those for Balance A. Answer: Balance B has ‘small systematic error’ but ‘not