PHY EXAM MEMOS

PHY1505 EXAM PACK
2016-2018 SOLUTIONS




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, Exam Pack: PHY 1505
OCT/NOV 2015
Question 1:
Velocity (V) m/s 𝑉 = 𝑢 + 𝑎𝑡


∆𝑉


U ∆𝑡
U


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

∆𝑉 = 𝑎𝑡
Also: Area under the velocity/time graph is equal to displacement.
𝐴𝑟𝑒𝑎 = 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 + 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒
1
𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 2 ∗ 𝑡 ∗ ∆𝑉 + 𝑢𝑡 … 𝑤ℎ𝑒𝑟𝑒 ∆𝑉 = 𝑎𝑡
1
𝑆 = 2 𝑡(𝑎𝑡) + 𝑢𝑡
1
𝑆 = 2 𝑎𝑡 2 + 𝑢𝑡
1
𝑆 = 𝑢𝑡 + 2 𝑎𝑡 2 … 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛

,Also the relationship between V, u, a and s is as follows:
𝑉 = 𝑢 + 𝑎𝑡 … (1);
𝑣−𝑢
𝑡= 𝑎
… (2);
𝑢+𝑉
𝑠=( 2
) 𝑡 … (3)

Substitute for t
1 𝑉−𝑢 2 (𝑣−𝑢)
𝑆= ( ) +𝑢
2 𝑎 𝑎

1 (𝑣 2 −2𝑉𝑢+𝑢2 )𝑎
𝑆= + 𝑢𝑉 − 𝑢2
2 𝑎2
1
𝑎𝑠 = 2 (𝑣 2 − 2𝑉𝑢 + 𝑢2 )
(𝑢+𝑣)(𝑣−𝑢)
𝑠= 2𝑎

𝑣 2 −𝑢2
𝑠=
2𝑎

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




600cm 60cm
B (fixed point)
mg


Floor
Where 𝑚 = 100𝑔, 𝑠 = 600 − 260 = 440𝑐𝑚
𝐾𝐸 = 𝑚𝑔ℎ 𝑇 = 𝑚𝑎 − 𝑚𝑔
𝐾𝐸 = 0.1 ∗ 9.81 ∗ 4 𝑇 = 𝑚(𝑎 − 𝑔)

𝐾𝐸 = 3.92 𝐽𝑜𝑢𝑙𝑒𝑠

, 1 1
𝑚𝑣 2 = 𝑚𝑔ℎ 𝑠 = 𝑢𝑡 + 𝑎𝑡 2 … (𝑖)
2 2

𝑣 2 = 2𝑔ℎ 𝑣 2 = 𝑢2 + 2𝑎𝑠 … (𝑖𝑖)

𝑣 2 = 2 ∗ 981 ∗ 4.4 = 88.32 𝑢𝑠𝑖𝑛𝑔 (𝑖𝑖) … 𝑢 = 0

𝑣 = √88.32 𝑣 2 − 𝑢2 = 2𝑎𝑠
88.32
𝑣 = 9.397𝑚/𝑠 𝑎 = 2(4.4)

9.4𝑚 10𝑚
𝑣= 𝑠
𝑎= 𝑠2

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