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) F_H = 100 sin 60° = 87 N

(b) F_V = 100 cos 60° = 50N

 

Question 32

v_PG = 250/0.5 = 500 km^-1 E

v_AG = 50 kmh^-1 N (not S); hence v_GA = 50 kmh^-1 S

v_PA = v_PG + v_GA = 500 kmh^-1 E + 50 kmh^-1 S

 

 

v_PA = 502.5 kmh^-1 ; angle q = tan^-1 50/500 = 5.7° ; hence direction is E 5.7° S

 

Question 33

d_P = 4 d_Q and r_P = 4 r_Q

(a) A_P = 4p r_P²

A_Q = 4p (r_P/4)² = 4p r_P² /16 = A_Q/16

Hence A_P = A_Q/16

(b) V_P = 4/3 p r_P³

V_Q = 4/3 p r_Q (1/4)³

Hence V_P = V_Q/64

(c) m_P = r V_P

m_Q = r V_Q

m_P = m_Q/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

F_P = F_W sinQ = 1000 x 3/5 = 600N
F_N = F_W cosQ = 1000 x 4/5 = 800N

Question 36

PART 1: Pointing directly across
Time to cross: t = s/v = 1/6 = 0.167 hr
v_BW = 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 v_BG = 1.887km/0.167hr = 11.3 km/hr
v_BG = v_BW + v_WG
11.3 = 9.6 + v_WG
v_WG = 6 km/h (remember, this is a vector sum)


PART 2: Pointing upstream
v_BG = v_BW + v_WG
v_BG = 9.6 + 6 (vectorially)
v_BG = 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

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

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