Distinguish between a vector and scalar - Leaving Cert Physics - Question 14 - 2022

The diagram should illustrate two force vectors acting at an angle, a common setup for determining the resultant vector. This involves using a table or a board to represent the vectors, with labeled arrows showing their directions and magnitudes. Label the vectors as A and B, and indicate the resultant vector R, which is formed by connecting the tail of vector A to the head of vector B.

Given the initial velocity of 150 m s^-1 at an angle of 20°:

• The horizontal component (v_x) can be calculated using the cosine function: $v_x = 150 \cos(20^\circ) \approx 141.0 \text{ m s}^{-1}$

• The vertical component (v_y) can be calculated using the sine function: $v_y = 150 \sin(20^\circ) \approx 51.3 \text{ m s}^{-1}$

Using the equations of motion, particularly the vertical component:

1. The vertical velocity after 8 seconds can be calculated as: $v_y = v_{y0} - gt$ where ( g = 9.8 , \text{m s}^{-2} ) is the acceleration due to gravity. Thus, $v_y = 51.3 - (9.8 \times 8) = -27.1 \, \text{m s}^{-1}$ \ (below the horizontal).

2. The magnitude of the resultant velocity can be calculated using the Pythagorean theorem: $|v| = \sqrt{v_x^2 + v_y^2} = \sqrt{(141.0)^2 + (-27.1)^2} \approx 143.5 \, \text{m s}^{-1}$

3. The direction can be evaluated using the tangent function: $\theta = \tan^{-1}\left( \frac{v_y}{v_x} \right) = \tan^{-1}\left( \frac{-27.1}{141.0} \right) \approx -10.9^\circ \ (below\ the\ horizontal)$.