# Chapter 4 - Worked Solutions

### to selected Complex Reasoning questions from the OUP text Senior Physics - Knowledge, Processes and Reasoning by Walding, Rapkins and Rossiter

Question 45

(a) Up the incline: FA = mg = 0.8 x 10 = 8 N
Down the incline: FP = FW sin q + Ff = 1 x 10 x 0.57 +Ff = 5.7 +Ff
Constant speed hence FUP = FDOWN; thus 8N = 5.7 + Ff or Ff = 2.3 N

(b) FN = 1 x 10 x cos 35°
m = Ff/FN = 3.3/(1 x 10 x cos 35° ) = 0.28

Question 46
The diagram of the forces acting is as follows, where FW = weight of the bob, T = tension in the string and FH = the wind force.

Because the pendulum bob is at rest when held to one side, the sum of the three forces is zero; that is FH + FW + T = 0. The three vectors must form a closed triangle if this is true:

FW = mg = 3 x 10-3 N
Using trigonometry: the wind force FH = FW tan 35° = 2.1 x 10-3 N
Using trigonometry: the tension T = FW/cos37° = 3 x 10-3/0.8 = 3.7 x 10-3 N

Question 47
(a) FTOTAL = FW + F
3 x 104 = 2.5 x 103 + 2.5 x 103 a
5000 = 2.5 x 103 a
a = 2 ms-2 upwards
(b) s = ½ at2

Question 48
FDOWN = FW sin q - (Ff + 500 + 80 (No. people)
For one person:
F = 1060 x 10 sin 30° - (0.2 x 10600 cos30° + 580)
F = 5300 - 2416 = 2884 N
a = F/m = 2884/1060 = 2.72 ms-2

For two people:
F = 1120 x 10 sin 30° - (0.2 x 11200 cos30° + 640)
F = 5600 - 2580 = 3020 N
a = F/m = 3020/1120 = 2.69 ms-2

Therefore one person would go faster.

Question 49
s = 290 m; m = 0.60
Ff = m FN = m mg
Ff = ma
a = Ff/m = m mg/m = m g = 0.6 x 10 = 6 ms-2

v2 = u2 +2as
0 = u2 + 2 x 6 x 290
u = 59 ms-1 (212 km h-1)

Question 50
Basic relationships: FH = Ff = FAcosq ; FN = FW - FA sinq
Ff = m FN
Ff = m (FW - FAsinq )
FAcosq = m (FW - FAsinq )
FA x 0.74 = 0.1(750 - FA 0.67)
0.807 FA = 75
FA = 93 N

Question 51
u = 20 ms-1; F = -6 x 103; s = 22.0 m
a = F/m = -6 x 103/1500 = - 4.0 ms-2
v2 = u2 +2as = 202 + 2 x -4 x 22
v2 = 400 + -176
v = 15 ms-1

Question 52
Convert 60 kmh-1 to ms-1 = 16.7 ms-1
At terminal velocity FW = ½ c r A v2
½ c r A1 v12 = ½ c r A2 v22
cancelling ½ c r from both sides
A1 v12 = A2 v22
A1 v12 = 2A1 v22 (replacing A2 with 2A1)
v12 = 2 v22
16.72= 2 v22
278 = 2 v22
v2 = 11.7 ms-1

Question 53
m = 1.2 x 106 ms-1; FUP = 450 x 10-18 N; FW = 9.11 x 10-31 x 10 N
FNET = FUP - FW
FNET = 450 x 10-18 N - 9.11 x 10-31 x 10 N = 4.5 x 10-16 N
a = F/m = 4.5 x 10-16 / 9.11 x 10-31 = 4.9 x 1014 ms-2
t = s/v = 30 x 10-3 / 1.2 x 106 = 2.5 x 108 seconds
s (up) = ut + ½ at2 = 0 + ½ x 4.9 x 1014 x (2.5 x 10-8)2 = 0.153 m (15.3 cm)

Question 54

The weight of the box on the incline FW = mg = 14 x 10 = 140N
The weight of the hanging mass FW2 = m2g = 14 x 10 = 140N (also)
Ff + FW sinq = FW2
m FN + FW sinq = 140
m (140 cos 30° ) + 140 sin 30° = 140
m x 121 + 70 = 140
m = 0.58