JPJC 2026 Quantities and Measurement Tutorial Solutions

1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS/8867 H1 PHYSICS QUANTITIES AND MEASUREMENT TUTORIAL SOLUTIONS Self-Check Questions S1 *Not in syllabus base quantity SI base unit mass kg length m time s electric current A thermodynamic temperature K amount of substance mol Luminous intensity* candela S2 The S I base units are the standard units of measurement defined by the International System of Units for the seven base quantities of what is now known as the International System of Quantities. They are a basic set from which all other SI units can be derived. Derived units are defined in terms of the base units, which are as stated in S1. S3 Incorrect constants/signs, e.g. 23s ut at=− Extra/missing terms, e.g. sv u at t= + + S4 Systematic errors result in readings taken being either always consistently larger or smaller than its true or actual value by fixed amount. Random errors result in a scatter of readings about a mean value, such that the readings have an equal chance of being larger or smaller than its true or actual value. S5 systematic errors how to eliminate forget to tare or “zero” an electronic balance gives mass measurements that always consistently larger or small than the true value by fixed amount tare or “zero” the electronic balance measuring radioactivity in the presence of background radiation before measuring the radioactivity of a specific sample, assess and record the background radiation level at the measurement location, and subtract the background radiation reading from the measurement to obtain the true radioactivity of the sample clock running consistently fast or slow regularly synchronise or calibrate the clock with timing instrument with minimal time drift

2 random errors how to reduce non-uniform variations in the diameter of wire take multiple measurements of the diameter along the length of the wire, and calculate the average value random parallax error when reading the volume of water in a measuring cylinder always position eye level at the bottom of the meniscus of the water when taking measurement on the cylinder radio frequency interference affecting electronic measurement instruments shield instruments from external interference, use proper grounding, and relocate instruments away from potential sources of interference S6 Accuracy is the degree of closeness of the average value of the measurements to the true value. It is affected by systematic error (i.e. correct calibration of instruments). Precision is the degree of agreement of repeated measurements of the same quantity. It is affected by random error. S7 A scalar quantity has a magnitude only. A vector quantity has both a magnitude and a direction. S8 Orientate the perpendicular x -y axis suitably and place the origin at the start of the 2-dimensional vector. Draw a line from the tip of the vector to each of the axis, such that a right-angle is formed on each of the axis. Apply trigonometric functions (sine or cosine) to determine the respective components. A   cos sin   = = xAA A sin cos   = = xA A A x y

3 Discussion Questions 1 23 4 ch ak= 3 1 2 4      = c akh  Units of h = Units of 3 1 2 4       c ak  3 1 2242 44 smKmW KJ         = −−− − 3 1 232 884 ssmkg smkg         = −− − 12 smkg −= Therefore, SI base units of h is 12 smkg − . 2 Units of Q 1 3 2 2 1 2 11 m (kg m s m ) kg mol m J K mol K − − − −− =   1 3 1 2 2 2 m (kg m s ) kg m (kg m s m) −− − =  SI base units of Q 1skg −= 3 SI base units of F = kg m s−2 SI base units of kAρvx = (m2)(kg m−3)(m s−1)x = kg m−1+x s−x Comparing the power for s, x = 2 4 (a) 756 37.2 0.83 2.5+ + + 796.53 797 (follow smallest d.p.) = = (b) 1.6523 0.015− 1.6373 1.637 (follow smallest d.p.) = = (c) 3.2 3.563 11.4016 11 (follow smallest s.f.) = =

4 (d) 5.6  1.7825... 1.8 (follow smallest s.f.) = = 5 3 326.15 9.7845100v =  = 10v= m s−1 (1 s.f.) Therefore, ( )10330 =v m s−1 6 V M= 3 23 4      = d M   d d M M += 3  percentage error in ρ, %100   = 1% + 3(3%) = 10% 7 ( ) 252 63.15 10 939 1.0801 10 m2.71  − − = = =    rR 0.05 3 0.012 2 0.0386 3.15 939 2.71 rR rR       = + + = + + =    ( ) 860.0386 4 10 0.04 10 m  −− = =  =   ( ) 6 1.08 0.04 10 m  − =    8 ( )1 8r Y X=− ( )1 5.0 1.08=− 0.50= cm ( )1 8r Y X =  +  ( )1 0.1 0.18=+ 0.025 0.03== cm Therefore, ( )0.50 0.03r = cm.

5 9 (a) Singapore has the shape of a rhombus, approximately 50 km east to west and 30 km north to south, as shown in diagram. Estimate area 2 1 50 302 750 km =   = (b) Consider Downtown line in Singapore MRT in which the train consists of 3 cabins, each of which has an approximate length of 20 m. Starting from rest at the station, 0u = It takes about 10 st = for the whole train to leave the station platform, 20 3 60 ms =  = Using 21 2s ut at=+ , ( ) 2 2 160 102 1.2 m s a a − = = 10 Taking leftwards as positive, Change in velocity of tennis ball ( ) ( ) 1 37 25 62 m s v v u vu −  = − = + − = − − =     Hence, change of velocity is 62 m s−1 to the left +ive 125 m su −=−  137 m sv −=  30 km 50 km

6 11 fiv v v = −  ( )fiv v v = + −  −+= 135cos2 222 ifif vvvvv ( )−+= 135cos1201602120160 222v 260v= m s−1 160 sin135sin =  v 26 = The change in velocity is 260 m s−1 at an angle of 2 6° North of East , as shown in the diagram. 12 (a) Change in velocity, 6 2 6 6 6 0 6 61 (5.0 10 ) (10 10 ) 2(5.0 10 )(10 10 )cos60 8.6603 10 8.66 10 m s v −  =  +  −   =   (b) Acceleration is rate of change of velocity. 6 7 13 13 2 6 0 06 8.6603 10 2.0 10 4.3302 10 4.33 10 m s sin 10 10 90 sin60 8.6603 10 va t   − − ==  =  =  =  The average acceleration is 13 24.33 10 m s − in the downwards direction, as shown in the diagram. 0 st = 615.0 10 m su −− =   72.0 10 st −= 6110.0 10 m sv −=  60  ( ) v v u vu  = − = + −      135° β − iv  fv  v 

7 13 Effective force along the ground 100cos30 87 N = =  Effective force normal to the ground 100sin30 50 N = =  14 Resolving F into its components, Resolving W into its components, cos// FF = sin// WW = sinFF =⊥ cosWW =⊥ 15 (a) Consider the mercury in the bore as having the same shape as a cylinder. Density of mercury, 2 2 MM V d L   ==   Diameter of bore, 2 3 2 4 4(6.884 6.610) 4.16 10 cm(13.6)(10.926 4.758)  −−= = =  − Md L 0.06449 cmd = 2      = + +d M L d M L 0.004 0.0042 0.001 0.01626.884 6.610 10.926 4.758  = + + =−− d d 0.00812d d  = 0.00812 0.06449 0.0005 cmd =  = ( )0.0645 0.0005 cmd = (b) The uncertainty of d (0.0005 cm) is much smaller as compared to the uncertainty if d is measured directly (0.004 cm). 30 o 100 N  F W ⊥W W// F// ⊥F