QUESTION 1 (5 Marks)
ANSWER
1
𝐹𝐹𝑠𝑠 = 𝐶𝐶𝐷𝐷,𝑠𝑠 𝑠𝑠 𝜌𝜌𝑉𝑉𝑠𝑠2 𝑆𝑆𝑠𝑠
2
1
𝐹𝐹𝑝𝑝 = 𝐶𝐶𝐷𝐷,𝑝𝑝 𝜌𝜌𝑉𝑉𝑠𝑠2 𝑆𝑆𝑝𝑝
2
𝐹𝐹𝑥𝑥 = 𝑚𝑚𝑚𝑚 = 𝑚𝑚 �
𝑑𝑑𝑑𝑑
� = −�𝐹𝐹𝑠𝑠 + 𝐹𝐹𝑝𝑝 �
𝑑𝑑𝑑𝑑
[1 Mark]
1
= − 𝜌𝜌𝑉𝑉𝑠𝑠2 �𝐶𝐶𝐷𝐷,𝑠𝑠 𝑆𝑆𝑠𝑠 + 𝐶𝐶𝐷𝐷,𝑝𝑝 𝑆𝑆𝑝𝑝 �
2
𝜌𝜌
Let 𝑘𝑘 = 2 �𝐶𝐶𝐷𝐷,𝑠𝑠 𝑆𝑆𝑠𝑠 + 𝐶𝐶𝐷𝐷,𝑝𝑝 𝑆𝑆𝑝𝑝 �, then,
𝑘𝑘 = �0.8 × 1000 + 1.2 × 𝜋𝜋 ×
𝑑𝑑𝑑𝑑
= 1026.19
202
1.2
�×
4
2
[1 Mark]
Integrating 𝑑𝑑𝑑𝑑 with respect to 𝑡𝑡 gives us the velocity.
𝑑𝑑𝑑𝑑
𝑘𝑘
= − � � 𝑉𝑉 2
𝑑𝑑𝑑𝑑
𝑚𝑚
𝑉𝑉
𝑑𝑑𝑑𝑑
𝑘𝑘 𝑡𝑡
� 2 = − � 𝑑𝑑𝑑𝑑
𝑚𝑚 0
0 𝑉𝑉
𝑉𝑉𝑜𝑜
𝑉𝑉 =
𝑘𝑘
1 + �𝑚𝑚� 𝑉𝑉0 𝑡𝑡
[1 Mark]
For the displacement of the spaceshuttle 𝑥𝑥,
𝑑𝑑𝑑𝑑 = 𝑉𝑉𝑉𝑉𝑉𝑉
𝑚𝑚
𝑘𝑘
𝑥𝑥 = ln �1 + � � 𝑉𝑉0 𝑡𝑡�
𝑘𝑘
𝑚𝑚
The final answer is obtained by substituting the known values into the equation for
𝑥𝑥. The second part of the question is answered by doubling the mass while retaining
the other known quantities.
[2 Marks]
QUESTION 2 (5 Marks)
ANSWER
We first compute the Reynolds number of the real object (hornet) moving in the
original environment.
𝑅𝑅𝑒𝑒𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝑅𝑅𝑒𝑒𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
5 × 0.05
=
= 1785
14 × 10−6
This indicates that the flow is laminar.
[1 Mark]
[1 Mark]
Then, to keep the same Reynolds number under different external conditions for the
model,
𝑉𝑉1 𝑥𝑥1 𝑉𝑉2 𝑥𝑥2
=
𝜈𝜈1
𝜈𝜈2
1785 × 110 × 10−6
𝑉𝑉2 =
0.25
= 7.854 m/s
[3 Marks]