# Chapter 3 - Worked Solutions

### to selected extension questions from the OUP text Senior Physics - Concepts in Context by Walding, Rapkins and Rossiter

Question 30

(a) (b) (c) BC = 8 sin 18° = 2.5 ms-1

BD = 5.0 ms-1 in direction W18° S

Question 31 (a) FH = 100 sin 60° = 87 N

(b) FV = 100 cos 60° = 50N

Question 32

vPG = 250/0.5 = 500 km-1 E

vAG = 50 kmh-1 N (not S); hence vGA = 50 kmh-1 S

vPA = vPG + vGA = 500 kmh-1 E + 50 kmh-1 S vPA = 502.5 kmh-1 ; angle q = tan-1 50/500 = 5.7° ; hence direction is E 5.7° S

Question 33

dP = 4 dQ and rP = 4 rQ

(a) AP = 4p rP2

AQ = 4p (rP/4)2 = 4p rP2 /16 = AQ/16

Hence AP = AQ/16

(b) VP = 4/3 p rP3

VQ = 4/3 p rQ (1/4)3

Hence VP = VQ/64

(c) mP = r VP

mQ = r VQ

mP = mQ/64

(d) r P = r Q

Question 34

(a) V1 = 35 V2

4/3 p r13 = 35 x 4/3 p r22

(b)  Question 35

FP = FW sinQ = 1000 x 3/5 = 600N
FN = FW cosQ = 1000 x 4/5 = 800N

Question 36 PART 1: Pointing directly across
Time to cross: t = s/v = 1/6 = 0.167 hr
vBW = 1.6km/0.167hr = 9.6 km/h
Distance travelled by boat relative to ground (using Pythagoras) = √(12 + 1.62 ) = 1.887 km
Velocity of boat relative to ground
vBG = 1.887km/0.167hr = 11.3 km/hr
vBG = vBW + vWG
11.3 = 9.6 + vWG
vWG = 6 km/h (remember, this is a vector sum) PART 2: Pointing upstream
vBG = vBW + vWG
vBG = 9.6 + 6 (vectorially)
vBG = 7.5 km/h (from vector sum)
time = s/v = 1.6km/7.5km/h = 0.213 hr
sinq = 6/9.6
q = 38° upstream

Question 37

Square both sides of the equation to give: Z2 = R2 + 4p 2L2f2
Rearrange the equation to: Z2 = 4p 2L2f2 + R2
This is now in the form y = mx + c
where the slope (m) = 4p 2L2 and the intercept on the y-axis (c) = R2.
Plot the graph: Slope = (44100 - 10000)/90000 = 0.379
4p 2L2 = 0.379
L = 0.098 metres

Intercept = 10000 = R2
Therefore R = 100 ohm

Question 38