1. 4.1 Vector Addition & Subtraction

2. Question 1 (a) What is the net force on the object below Fnet 5.0N 
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Fnet
5.0N 1 
12N
Find 
𝑡𝑎𝑛𝜃= 𝑜𝑝𝑝 𝑎𝑑𝑗
𝑡𝑎𝑛𝜃= 5 12
𝜃 = 𝑡𝑎𝑛 −
𝜃 = o
𝜃  23o
So Fnet is
Find Fnet
h2 = a2 + b2
Fnet2 =
Fnet2 = 169
Fnet = 13N
13N
23o

3. Question 1 (b) What is the net force on the object 2Acos 𝜽 𝟐
= 2  8  cos 20o =
 15N
20o
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4. Question 1 (c) What is the net force on the object 2Acos 𝜽 𝟐 40N
= 2  40  cos 25o =
 73N
40N
25o
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5. Question 4 (d) What is the net force on the object 2Acos 𝜽 𝟐
= 2  10  cos 30o =
60o
30o
Fnet =
Fnet =
Fnet  17N
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6. Question 5 (e) What is the net force on the object
Symmetry in this situation means that Fnet = 0N
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7. Question 2
Work out the change in velocity for the situations below (show your answer in diagram form) (a) (b) (c)
4 ms -1
Intial
6 ms -1
Final
2 ms -1 v
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8ms -1
Intial
3 ms -1
Final
5 ms -1 v
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3 ms -1
Intial
2 ms -1
Intial
5 ms -1 v
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8. Question 2
Work out the change in velocity for the situations below (show your answer in diagram form) (d) (e)
4 ms -1
Intial
6 ms -1
Final
10 ms -1 v
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12ms -1
Intial
5 ms -1
Final
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9. Question 2
Work out the change in velocity for the situations below (show your answer in diagram form) (d) (e)
4 ms -1
Intial
6 ms -1
Final
10 ms -1 v
12ms -1
Intial
5 ms -1
Final v 
5.0N
12N
Find 
𝑡𝑎𝑛𝜃= 𝑜𝑝𝑝 𝑎𝑑𝑗
𝑡𝑎𝑛𝜃= 12 5
𝜃 = 𝑡𝑎𝑛 −
𝜃 = o
𝜃  67o
Find v
h2 = a2 + b2
v2 =
v2 = 169
v = 13N
So v is
23o
67o
13ms-1

10. Question 2
Work out the change in velocity for the situations below (show your answer in diagram form) (f) v
6 ms -1 vf
30o
v = 2Acos 𝜽 𝟐
v = 2  6  cos 60o
v = 6.0ms-1
6 ms -1
–vi
30o
120o
60o
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