Chapter 4 - Worked Solutions

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

Question 25

u = √2mgs
100000/3600 =  √(2 x 0.68 x 10 x s)
s = 56.7 m
 

Question 26

u = √(2 x 0.61 x 10 x 5.6 + 2 x 0.79 x 10 x 3.2) = 10.9 m/s

Question 27

mds = m - sinq
m
ds = 0.73 - sin13.8 = 0.49
u = √2mgs = √(2 x 0.49 x 10 x 17.4) = 13.1 m/s

Question 48



(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 49
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 37° = 2.26 x 10-3 N
Using trigonometry: the tension T = FW/cos37° = 3 x 10-3/0.8 = 3.7 x 10-3 N
 
Question 50
(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 = at2

 
Question 51
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 52
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 53
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 54
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 55
Convert 60 kmh-1 to ms-1 = 16.7 ms-1
At terminal velocity FW = c r A v2
c r A1 v12 = c r A2 v22
cancelling 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 56
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 + at2 = 0 + x 4.9 x 1014 x (2.5 x 10-8)2 = 0.153 m (15.3 cm)
 
Question 57


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
 

Question 58

m = m + sinq = 0.71 +sin 7° = 0.49
a = mg = 0.832 x 10 = 8.32 m/s/s
v2 = u2 + 2as
u2 = v2 - 2as = (25 x 1000/3600)2 - 2 x -8.32 x 13.7
= (6.94)- (-227.97) = 276.13
u = 16.6 m/s
 

Question 59

a1 = mg = 0.66 x 10 = -6.6 m s-2
a2 = mg = 0.76 x 10 = -7.6 m s-2

STAGE 2:
v2 = vmid2 + 2as
8.32 = vmid2 + 2 x -7.6 x 2.6
68.9 + 39.5 = vmid2
vmid = 10.4 m/s

STAGE 1:
vmid2 = u2 + 2as
10.42 = u2 + 2 x -6.6 x 9.6
u2 = 235.14
u = 15.3 m/s

Question 60

Let "T1" be the tension in the string over the top pulley.
T1 = m1g + m1a1
T1 = (m2 + m3) g - (m2 + m3) a1
m1g + m1a1 = (m2 + m3) g - (m2 + m3) a1
(m1 + m2 + m3) a1 = (m2 + m3 - m1) g
a1 = (m2 + m3 - m1) g / (m1 + m2 + m3)

a1 = g(4m2m3 - m1m2 - m1m3)/(4m2m3 + m1m2 + m1m3)

Question 61

Ff = mFN
FP = FW sinq
FN = FW cosq - FA sinq

Motion is up the incline so friction acts down the incline.
Total force down incline:
Fdown incline = FP + Ff
Fdown incline = FW sinq + mFN
Fdown incline = FW sinq + m(FW cosq - FA sinq)
Fdown incline = mg sin35 + m(mg cos35 - FA sin20)

Total force down incline:
Fup incline = FA cos20

But as velocity is constant:
Fup incline = Fdown incline
FA cos20 = mg sin35 + m(mg cos35 - FA sin20)
FA x 0.94 = 100 x 10 x sin 35 + 0.65 (100 x 10 x 0.82 - FA x 0.34)
0.94 FA = 574 + 532 - 0.22 FA
1.16 FA = 1106 N
FA = 953 N

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