PHY1505 MAY JUNE 2016 TO OCT/NOV 2018 EXAM PACK with detailed step by step calculation, sketches and explanations perfect for exam preparation and revision. Feel free to contact me if you need further explanations on the solutions. All the best in your studies! Contact details are in the exam pa...
0 Time (t) s
The movement starts at “u” m/s and has a linear acceleration am/s2.
The straight line graph
∆𝑦 ∆𝑉
Y=mx+c …………………. (1) but c=u; and 𝑚 = ∆𝑥 = ∆𝑡
=𝑎
Thus equation (1) becomes:
𝑉 = 𝑢 + 𝑎𝑡 ….. Equation of motion
Final velocity = initial velocity + acceleration(time)
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
∆𝑉
Where 𝑎 = 𝑡
and ∆𝑉 = 𝑉 − 𝑢
2𝑎𝑠 = 𝑣 2 − 𝑢2 𝑜𝑟 𝑉 2 = 𝑢2 + 2𝑎𝑠 … 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛
Question 2:
Newton’s third law of motion
a. For every action there is an equal and opposite reaction.
b.
ℎ = 600 − 260
ℎ = 440𝑐𝑚
Therefore 𝑇 = 0.1(10 − 9.81) = 0.023𝑁
Question 3
For representative element
∆𝑚 = 𝑃∆𝑉𝑥 … 𝑤𝑖𝑡ℎ 𝑃 𝑡ℎ𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
∆𝑚 =
Question 4:
When objects collide in the absence of:
a. External forces, the total momentum of the system is conserved. If collision happens
to be elastic, then kinetic energy will be conserved.
b. The conservation of momentum gives:
i. 𝑃𝑓 = 𝑃𝑜
𝑀𝐴 𝑉𝑓1 + 𝑀𝑏 𝑉𝑓2 = 𝑀𝑎 𝑉𝑜1 + 𝑀𝑏 𝑉𝑜2
𝑀𝐴 𝑉𝑓1 + 𝑀𝑏 𝑉𝑓2 = 𝑀𝑏 𝑉
𝑀𝐴 𝑉𝑓1 = 𝑀𝑏 𝑉 − 𝑀𝑏 𝑉𝑓2
𝑀𝐴 𝑉𝑓1 = 𝑀𝑏 (𝑉 − 𝑉𝑓2 )
𝑀
𝑉𝑓1 = 𝑀𝑏 (𝑉 − 𝑉𝑓2 ) … (1)
𝑎
𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝐴 𝑎𝑓𝑡𝑒𝑟 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 𝑖𝑠:
𝑀
𝑉𝑓1 = 𝑀𝑏 (𝑉 − 𝑉𝑓2 )
𝑎
ii. To solve Vf2, we use the conservation of kinetic equation:
1 1 1 1
(2)𝑀𝑏 𝑉𝑓1 2 + (2) 𝑚𝑎 𝑉𝑓2 2 = 2
(𝑀𝑏 𝑉01 2 ) + 2 (𝑀𝐴 𝑉02 2 )
Note: Vo1 = 0 and Vo2 = v m/s
𝑀𝐵 𝑉𝑓1 2 + 𝑀𝐴 𝑉𝑓2 2 = 𝑀𝐵 (𝑂)2 + 𝑀𝑎 (𝑉)2
𝑀𝐵 𝑉𝑓1 2 + 𝑀𝐴 𝑉𝑓2 2 = 𝑀𝑎 (𝑉)2
𝑀𝐵 𝑉𝑓1 2 = 𝑀𝑎 (𝑉)2 − 𝑀𝐴 𝑉𝑓2 2
𝑀
𝑉𝑓1 2 = 𝑀𝑎 (𝑉 2 − 𝑉𝑓2 2 )
𝐵
𝑀
𝑉𝑓1 = √𝑀𝑎 (𝑉 2 − 𝑉𝑓2 2 )
𝐵
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