Sure! Let's address each part of the problem step-by-step.
### Part (a): Finding the Speed of the Puck
The velocity of the puck is given as a complex number: [tex]\( (141 + 22j) \)[/tex] m/s.
Here, 141 is the real component (velocity in the x-direction) and 22 is the imaginary component (velocity in the y-direction).
The speed of the puck is the magnitude of this velocity vector. We can find the magnitude using the Pythagorean theorem:
[tex]\[
\text{Speed} = \sqrt{ \text{Re}(v)^2 + \text{Im}(v)^2 }
\][/tex]
Where:
- [tex]\( \text{Re}(v) = 141 \)[/tex]
- [tex]\( \text{Im}(v) = 22 \)[/tex]
[tex]\[
\text{Speed} = \sqrt{ 141^2 + 22^2 }
\][/tex]
Calculating these squares:
[tex]\[
141^2 = 19881
\][/tex]
[tex]\[
22^2 = 484
\][/tex]
Adding them together:
[tex]\[
19881 + 484 = 20365
\][/tex]
Taking the square root:
[tex]\[
\text{Speed} = \sqrt{20365} \approx 142.7
\][/tex]
So, the speed of the puck is approximately [tex]\( 142.7 \)[/tex] m/s.
### Part (b): Finding the Angle with the Unit Vector [tex]\( j \)[/tex]
We need to find the angle θ that the velocity vector makes with the y-axis (the unit vector [tex]\( j \)[/tex]). This angle can be found using the arctangent function, where:
[tex]\[
\theta = \arctan \left(\frac{\text{Im}(v)}{\text{Re}(v)}\right)
\][/tex]
Using the values:
[tex]\[
\theta = \arctan \left(\frac{22}{141}\right)
\][/tex]
[tex]\[
\theta \approx \arctan (0.1560)
\][/tex]
To find this angle in radians:
[tex]\[
\theta \approx 0.154 \text{ radians}
\][/tex]
To convert this into degrees, we multiply by [tex]\(\frac{180}{\pi}\)[/tex]:
[tex]\[
\theta \approx 0.154 \times \frac{180}{\pi} \approx 8.82 \text{ degrees}
\][/tex]
So, the angle of direction of the puck's motion with the unit vector [tex]\( j \)[/tex] is approximately [tex]\( 0.154 \)[/tex] radians or [tex]\( 8.82 \)[/tex] degrees.
### Part (c): Effect of Modelling the Ice as a Smooth Surface
If we model the ice as a smooth surface, it implies that there is no friction acting on the puck. This means that:
- The puck would continue to travel indefinitely in a straight line at a constant speed unless acted upon by another force.
This follows Newton's First Law of Motion, indicating that an object in motion will remain in motion unless acted upon by an external force.
### Part (d): Converting Density from [tex]\( \text{g/cm}^3 \)[/tex] to [tex]\( \text{kg/m}^3 \)[/tex]
The density of the hockey puck is given as [tex]\( 1.4 \ \text{g/cm}^3 \)[/tex].
To convert this into [tex]\( \text{kg/m}^3 \)[/tex], we use the fact that [tex]\( 1 \ \text{g/cm}^3 = 1000 \ \text{kg/m}^3 \)[/tex]:
[tex]\[
1.4 \ \text{g/cm}^3 \times 1000 \ = 1400 \ \text{kg/m}^3
\][/tex]
So, the density of the hockey puck is [tex]\( 1400 \ \text{kg/m}^3 \)[/tex].
This concludes the detailed step-by-step solution for each part of the question.
