Cards In This Deck

• $\text{First step to find the bearing of a velocity vector }\mathbf{v}=-3\mathbf{i}+4\mathbf{j}\text{?}$
  $\text{Draw a sketch of the vector starting from the origin to identify the correct quadrant.}$
• $\text{How do you find the angle between a vector }\mathbf{a}=x\mathbf{i}+y\mathbf{j}\text{ and the positive x-axis (}\mathbf{i}\text{)?}$
  $$\begin{gathered} \text{Use trigonometry on the components:} \\ \tan (\theta )=\frac{y}{x} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If a particle moves with velocity }\mathbf{v}=-3\mathbf{i}-4\mathbf{j}\text{ m/s, its speed is }-5\text{ m/s.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: }5\text{ m/s.} \\ \text{Speed is a scalar quantity and is the magnitude of the velocity vector. It must always be positive.} \\ |\mathbf{v}|=\sqrt{(-3{)}^2+(-4{)}^2}=5\text{.} \end{gathered}$$
• $\text{If a position vector }\mathbf{r}=(4+2\lambda )\mathbf{i}+(3-6\lambda )\mathbf{j}\text{ is parallel to the vector }\mathbf{j}\text{ (due North), how do you find }\lambda \text{?}$
  $$\begin{gathered} \text{Set the }\mathbf{i}\text{ component equal to zero.} \\ 4+2\lambda =0 \end{gathered}$$
• $\text{You are given the mass }m\text{ and the acceleration vector }\mathbf{a}=x\mathbf{i}+y\mathbf{j}\text{. How do you calculate the magnitude of the force }|\mathbf{F}|\text{?}$
  $$\begin{gathered} \text{First find the force vector }\mathbf{F}=m\mathbf{a}\text{, then calculate the magnitude using Pythagoras.} \\ |\mathbf{F}|=|m(x\mathbf{i}+y\mathbf{j})|=m\sqrt{x^2+y^2} \end{gathered}$$
• $\text{When resolving a force }F\text{ given an angle }\theta \text{, how do you decide which component takes }\cos \theta \text{ and which takes }\sin \theta \text{?}$
  $$\begin{gathered} \text{The component adjacent (touching) the angle }\theta \text{ is always }F\cos \theta \text{.} \\ \text{The component opposite the angle (or perpendicular to the adjacent side) is }F\sin \theta \text{.} \end{gathered}$$
• $\text{If particle A is due North of particle B, what equation can you form regarding their position vectors }{\mathbf{r}}_A\text{ and }{\mathbf{r}}_B\text{?}$
  $$\begin{gathered} \text{Their }\mathbf{i}\text{ (East-West) components must be equal.} \\ {x}_A=x_B \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{A particle travels from point A to B and then back to A. Its total displacement is zero, therefore the total distance travelled is zero.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: Displacement is zero, but distance is the total length of the path (}AB+BA\text{).} \end{gathered}$$
• $\text{What is the formula for the position vector }\mathbf{r}\text{ of a particle at time }t\text{, given its initial position }{\mathbf{r}}_0\text{ and constant velocity }\mathbf{v}\text{?}$
  $\mathbf{r}={\mathbf{r}}_0+\mathbf{v}t$
• $\text{What is the mathematical condition for two particles }P\text{ and }Q\text{ to collide at time }t\text{?}$
  $$\begin{gathered} \text{Their position vectors must be equal at the same time }t\text{.} \\ {\mathbf{r}}_P={\mathbf{r}}_Q \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{To find the speed of a particle given its position vector }\mathbf{r}=3t\mathbf{i}+4t\mathbf{j}\text{, you calculate }|\mathbf{r}|\text{ and then divide by }t\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False (mostly)} \\ \text{✅ Correct Method: Differentiate }\mathbf{r}\text{ (or identify coefficients of }t\text{ if linear) to find velocity }\mathbf{v}\text{, then find the magnitude }|\mathbf{v}|\text{.} \\ \text{In this specific linear case it works, but conceptually speed is }|\mathbf{v}|\text{, not }|\mathbf{r}|/t\text{ (which is average speed from origin).} \end{gathered}$$
• $\text{When finding the resultant of two forces given as vectors (e.g., in a diagram with angles), when should you use the Triangle Law (Tip-to-Tail) vs. Resolving Components?}$
  $$\begin{gathered} \text{Use Resolving Components when you need to calculate specific acceleration or force values in coordinate directions (standard M1 approach).} \\ \text{Use Triangle Law (Sine/Cosine Rule) if you only need the magnitude/direction of the resultant and are dealing with non-right-angled geometry.} \end{gathered}$$
• $\text{How do you calculate the resultant velocity of an object moving with velocity }{\mathbf{v}}_1\text{ subject to a wind/current velocity }{\mathbf{v}}_2\text{?}$
  $$\begin{gathered} \text{Add the vectors together:} \\ {\mathbf{v}}_{resultant}={\mathbf{v}}_1+{\mathbf{v}}_2 \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If a car is moving in the negative direction (velocity is negative) and is slowing down (decelerating), its acceleration }a\text{ must be negative.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: The acceleration }a\text{ is positive.} \\ \text{Deceleration means acceleration acts in the opposite direction to motion. If }v\text{ is negative (}-\text{), }a\text{ must be positive (}+\text{) to oppose it.} \end{gathered}$$
• $\text{You are solving a statics problem with exactly three non-parallel coplanar forces acting on a particle. Which method is often faster than resolving forces?}$
  $$\begin{gathered} \text{Triangle of Forces (or Lami’s Theorem).} \\ \text{Sketch a closed vector triangle. You can then use the Sine Rule or Cosine Rule to find unknown magnitudes or angles without resolving into }x\text{ and }y\text{ components.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{When drawing a Triangle of Forces for a particle in equilibrium, the arrows representing the forces must all point away from the same vertex.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Method: The arrows must follow a cyclic order (nose-to-tail) around the triangle. } \\ \text{If you start at a point and follow the arrows, you should arrive back at the start (representing a resultant force of zero).} \end{gathered}$$
• $\text{What is the standard first step when resolving forces for a particle on an inclined plane?}$
  $$\begin{gathered} \text{Resolve forces in two perpendicular directions:} \\ \text{1. Parallel to the line of greatest slope (}\nearrow \text{ or }\swarrow \text{).} \\ \text{2. Perpendicular to the plane (}\nwarrow \text{ or }\searrow \text{).} \end{gathered}$$