An aircraft is travelling in wind. Fig. 1.2 shows the velocities for the aircraft in still air and for the wind.

Question 8

(a) State what is meant by a scalar quantity and by a vector quantity. [2]

(b) Complete Fig. 1.1 to indicate whether each of the quantities is a vector or a scalar.

quantity          vector or scalar

power

temperature

momentum

Fig. 1.1

[2]

(c) An aircraft is travelling in wind. Fig. 1.2 shows the velocities for the aircraft in still air and for the wind.

Fig. 1.2

The velocity of the aircraft in still air is 95 m s^-1 to the west.

The velocity of the wind is 28 m s^-1 from 65° south of east.

(i) On Fig. 1.2, draw an arrow, labelled R, in the direction of the resultant velocity of the

aircraft. [1]

(ii) Determine the magnitude of the resultant velocity of the aircraft. [2]

[Total: 7]

Reference: Past Exam Paper – June 2018 Paper 21 Q1

Solution:

(a)

A scalar quantity is one that has magnitude only while a vector has both magnitude and direction.

(b)

power: scalar

temperature: scalar

momentum: vector

(c)

(i)
arrow labelled R in a direction from 5° to 20° north of west

{To find the resultant on a vector diagram:

[You may check solution 703 at Doubt Page 142] 

Step 1: Lay down the first vector: here it is the ‘aircraft velocity’

Step 2: Then, place the tail of the second vector at the head of the first one: this is the ‘wind velocity’ reproduced in blue.

Step 3: The resultant starts from the tail of the first vector (aircraft velocity) to the head of the second vector (wind velocity) as placed in the second step.}

(ii)

{Let the resultant (R) be v.

Consider the triangle formed. We know the 2 sides (28 m s^-1 and 95 m s^-1) and the angle opposite to v (= 115°).}

{cosine rule:    c² = a² + b² – 2bc cos(C)}

v² = 28² + 95² – (2 × 28 × 95 × cos 115°)

OR

{We can alternatively resolve the horizontal (which add to the aircraft velocity) and vertical components of the wind velocity, and then find the resultant using Pythagoras’ theorem.}

v² = [(95 + 28 cos 65°)² + (28 sin 65°)²]

v = 110 ms^-1 (109.8 ms^-1)