[<< wikibooks] A-level Physics (Advancing Physics)/Vectors/Worked Solutions
1. Which of the following are vectors?

20 cm is scalar, not a vector - it does not have a directional component.
9.81 ms−2 towards the centre of the earth is a vector.
5 km south-east is a vector.
500 ms−1 on a bearing of 285.3° is a vector.2. A displacement vector a is the resultant vector of two other vectors, 5 m north and 10 m south-east. What does a equal, as a displacement and a bearing?

So, for the resultant vector:

Horizontal displacement

=
0
+

10

2

≈
7.07

m

{\displaystyle {\mbox{Horizontal displacement }}=0+{\frac {10}{\sqrt {2}}}\approx 7.07{\mbox{ m}}}

Vertical displacement

=
5
−

10

2

≈
−
2.07

m

{\displaystyle {\mbox{Vertical displacement }}=5-{\frac {10}{\sqrt {2}}}\approx -2.07{\mbox{ m}}}

This gives us a right-angled triangle. So:

θ
=
arctan
⁡

2.07
7.07

≈
16.3

{\displaystyle \theta =\arctan {\frac {2.07}{7.07}}\approx 16.3}
°, so the bearing equals 90° + 16.3° = 106.3°.

Resultant displacement

=

7.07

2

+
(
−
2.07

)

2

≈
7.4

m

{\displaystyle {\mbox{Resultant displacement}}={\sqrt {7.07^{2}+(-2.07)^{2}}}\approx 7.4{\mbox{ m}}}

3. If I travel at a velocity of 10 ms−1 on a bearing of 030°, at what velocity am I travelling north and east?

Vertical velocity (North)

=
10
×
cos
⁡

30

=
5

3

ms

−
1

≈
8.66

ms

−
1

{\displaystyle {\mbox{Vertical velocity (North)}}=10\times \cos {30}=5{\sqrt {3}}{\mbox{ ms}}^{-1}\approx 8.66{\mbox{ ms}}^{-1}}

Horizontal velocity (East)

=
10
×
sin
⁡

30

=
5

ms

−
1

{\displaystyle {\mbox{Horizontal velocity (East)}}=10\times \sin {30}=5{\mbox{ ms}}^{-1}}

4. An alternative method of writing vectors is in a column, as follows:

a

=

(

x
y

)

{\displaystyle \mathbf {a} ={\binom {x}{y}}}
,
where x and y are the vertical and horizontal components of the vector respectively. Express |a| and the angle between a and

(

1
0

)

{\displaystyle {\tbinom {1}{0}}}
in terms of x and y.
By Pythagoras' theorem:

(

|

a

|

)

2

=

x

2

+

y

2

{\displaystyle (|\mathbf {a} |)^{2}=x^{2}+y^{2}}

|

a

|

=

x

2

+

y

2

{\displaystyle |\mathbf {a} |={\sqrt {x^{2}+y^{2}}}}

Let θ be the angle between a and

(

1
0

)

{\displaystyle {\tbinom {1}{0}}}
.

tan
⁡

θ

=

y
x

{\displaystyle \tan {\theta }={\frac {y}{x}}}

θ
=
arctan
⁡

y
x

{\displaystyle \theta =\arctan {\frac {y}{x}}}

This angle θ is known as the argument of a.
5. A more accurate method of modelling the trajectory of a ball is to include air resistance as a constant force F. How would this be achieved?
Once the arrow representing the acceleration due to gravity has been added on, add an horizontal arrow pushing against the motion of the ball. Since F = ma, the magnitude of this acceleration is F divided by m.
Note that this model is still not perfect. In fact, F is not constant - it depends on the horizontal component of the ball's velocity.