PH1012
AY1819 Sem 1
Qn
Part Answer
1
ai.
𝑎=
𝑑𝑣⃗
𝑚
= 6.84 2
𝑑𝑡
𝑠
aii.
2.5
𝑥⃗ = ∫ 𝑣⃗ 𝑑𝑡 = (13.3 𝑖̂ + 17.5𝑗̂) m
0
aiii.
𝑣𝑦
𝑣𝑥
= tan 60∘ , 𝑡 = 0.32 s , 2.25 s
bi.
𝐹𝑛 − 𝑚𝑔 =
bii.
2
i.
ii.
3
i.
ii.
iii.
iv.
v.
, 𝐹𝑛 = 49.3 𝑁
1
1
𝑚𝑝 𝑔(0) + 2 𝑚𝑃 𝑣𝐵2 = 𝑚𝑃 𝑔(0.40) + 2 𝑚𝑃 𝑣𝑇2 , 𝑚𝑝 𝑢𝑝 + 0 = 𝑚𝑝 𝑣𝑝 + 𝑚𝑄 𝑣𝑄
𝑣𝑄 = 2.78
biii.
𝑚𝑣 2
𝑟
1
𝑚(0)2
2
𝑡=
𝑚
𝑠
1
− 2 𝑚𝑣𝑄2 = 𝜇𝐾 𝑚𝑔𝑥 , 𝑥 = 0.94 m
15000
1360
, 𝑥1 = 580 × 𝑡 + √150002 − 60002 = 20144 m, 𝜃 = 23.6∘
𝑣𝑖𝑦 = √2𝑔ℎ𝑚𝑎𝑥 = 600 m/s , 𝑥2 = 40500 m
𝑝𝐷 𝑉𝐷 = 𝑛𝑅𝑇𝐷 , 𝑇𝐷 = 640 K
𝛾
𝛾
𝑝𝐴 𝑉𝐴 = 𝑝𝐵 𝑉𝐵 , 𝑝𝐵 = 4.66 𝑎𝑡𝑚
5
2
𝑈𝐵𝐶 = 𝑛𝑅Δ𝑇𝐵𝐶 , 𝑈𝐵𝐶 = 118000 𝐽
7
𝐻𝑒𝑎𝑡 𝐼𝑛𝑝𝑢𝑡 𝑄 = 2 (𝑝𝐶 𝑉𝐶 − 𝑝𝐵 𝑉𝐵 ) , 𝑄 = 165200 𝐽
5
5
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = − (𝑝𝐵 𝑉𝐵 − 𝑝𝐴 𝑉𝐴 ) + (𝑝𝐶 𝑉𝐶 − 𝑝𝐵 𝑉𝐵 ) − (𝑝𝐷 𝑉𝐷 − 𝑝𝐶 𝑉𝐶 ) + 0
2
2
𝑊 = 40600 𝐽
vi.
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
vii.
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
= 0.25
𝐻𝑒𝑎𝑡 𝐼𝑛𝑝𝑢𝑡
5
5
(− 2 (𝑝𝐵 𝑉𝐵 − 𝑝𝐴 𝑉𝐴 ) + (𝑝𝐶 𝑉𝐶 − 𝑝𝐵 𝑉𝐵 ) − 2 (𝑝𝐷 𝑉𝐷 − 𝑝𝐶 𝑉𝐶 ))
7
(𝑝
)
2 𝐶 𝑉𝐶 − 𝑝𝐵 𝑉𝐵
All pressures doubled will be cancel away, so no change in efficiency
1
Finals Solution
PH1012
AY1819 Sem 1
Qn
Part Answer
4
ai.
𝑇1
Finals Solution
𝐹𝑠
𝐹𝑁
𝐹𝑁
𝑇1
𝐹𝑓
𝑃
𝑇2
𝑚𝑔
𝑚𝐴 𝑔
aii.
Block A, along slope: 𝑚𝐴 𝑔 sin 𝜃 − 𝐹𝑠 − 𝐹𝑓 + 𝑇1 = 𝑚𝐴 𝑎
Block B: 𝑚𝐵 𝑔 − 𝑇2 = 𝑚𝐵 𝑎
1
2
Pulley: (𝑇2 − 𝑇1 )𝑅 = 𝐼𝛼 = 𝑀𝑃 𝑅2 𝛼
aiii.
aiv.
b
5
ai.
𝑎 = 𝑅𝛼 , 𝑎 = 5.10
𝑚
𝑠2
1
𝑚𝐴 𝑔𝑥 sin 37∘ + 𝑚𝐵 𝑔𝑥 = 𝜇𝑘 𝑚𝐴 𝑔 cos 37∘ 𝑥 + 2 𝑘𝑥 2 , 𝑥 = 0.327 m
1
2
1
2
𝑚𝑔𝑦 = 𝑚𝑣 2 + 𝐼𝜔2 , 𝜔 = 51.9
𝑄
Φ𝐸 = 𝜖
𝑜
𝑟𝑎𝑑
𝑠
, 𝑄 = 4.03 × 10−9 C
(Positive)
aii.
Φ𝐸 = +455
b.
ci.
cii.
di.
dii.
diii.
e.
𝑘𝑞1
𝑟12
+
𝑘𝑞2
𝑟22
−
Nm2
C
𝑘𝑄
𝑟32
= 0 , 𝑟3 = −8.66 𝑐𝑚
1
1
𝑞𝑉𝐴 + 2 𝑚𝑣𝐴2 = 𝑞𝑉𝐵 + 2 𝑚𝑣𝐵2 , 𝑣𝐵 = 106000
|𝐸| =
Δ𝑉
Δ𝑟
𝑉
= 800 𝑚
downward
𝐵𝐼𝐿 − 𝑚𝑔 sin 22° = 0 , 𝐵 = 0.051 𝑇
𝐵𝑚𝑎𝑥 𝐼𝐿 − 𝑚𝑔 sin 90° = 0 , 𝐵𝑚𝑎𝑥 = 0.136 𝑇
|𝜀1 | = 𝐵𝑙𝑣1 , |𝜀2 | = 𝐵𝑙𝑣2
Using Kirchhoff’s Law: 𝐼3 = 0.029 𝐴
2
𝑚
𝑠
PH1012
AY1819 Sem 1
Qn
Part Answer
6
a.
Loop Rule:
1 − 2 + 3 = 𝑖1 1 + 𝑖1 2 + 𝑖1 3 + 𝑖1 4 − 𝑖2 5
4 + 3 = −𝑖2 5 − 𝑖3 6
Junction Rule:
𝑖1 + 𝑖2 = 𝑖3
𝑖1 = −
bi.
𝐼=
bii
13
140
𝐴, 𝑖2 = −
41
𝐴
70
, 𝑖3 = −
19
𝐴
28
15
= 1.5 × 10−3 𝐴
10 × 103
𝑉 = 𝑉𝑜 𝑒 −𝑡/𝑅𝐶
𝑡 = 0.06 s
3
Finals Solution