Example 1.
Gonick-Huffman: Cartoon guide to Physics
An 80-kg football player going north at 3 m/s tackled by a 100-kg player going east at 2 m/s. After the impact, which way are they going and how fast?[br]
Solution 1.
The information about players enables us to draw a vector format of the situation:
Eastward momentum is[br][br]  [math]\Large p_x=m_1v_x=200\frac{kg\cdot m}{s},[/math]  [br][br]northward momentum is[br][br]   [math]\Large p_y=m_2v_y=240\frac{kg\cdot m}{s},[/math] [br][br]and the mass after the collision ("they continue together") on m[sub]t[/sub] = 180 kg. [br][br]After the collision the momentum of the system stays constant. Thus, the speeds can be obtained the combined mass:
[math]\Large\begin{eqnarray}[br]v_x=\frac{p_x}{m_t}=\frac{200}{180} &(\approx 1.11\;\frac m s)\\[br]v_y=\frac{p_y}{m_t}=\frac{240}{180} &(\approx 1.33\;\frac m s)[br]\end{eqnarray}[/math][br] [br][br]The final speed after the collision is obtained with Pythagoras:[br]  [br]   [math]\Large v_t=\sqrt{\left (\frac{200}{180}\right )^2+\left (\frac{240}{180}\right )^2}\approx 1.7 \;\frac m s.[/math] [br][br]The system continues at the angle of 50° from east to north -direction, as[br]  [br]   [math]\Large \alpha = \arctan\left (\frac{240/180}{200/180}\right )\approx 50.2.\degree[/math]
Example 2.
There are three points A, B and C. Let us define [math]\Large \overrightarrow{CA}=\vec a[/math] and [math]\Large \overrightarrow{CB}=\vec b.[/math]. There is a point P on the line segment AB so that it divides the line segment in a ratio 3:4. Define [math]\Large \overrightarrow{CP}[/math] with known [math]\Large \vec a[/math] and [math]\Large \vec b[/math].
Solution 2.
  [br]If ratio is thought as a distance, it could mean, for example, that the distance between A and P is 3 cm and between B and P 4 cm. Thus, the total distance between A and B is 7 cm and from A to P is only 3/7 of AB:[br][br][math]\Large \begin{eqnarray}[br]\overrightarrow{CP}&=&\overrightarrow{CA}+\overrightarrow{AP}\\[br]&=& \overrightarrow{CA}+\frac 3 7 \overrightarrow{AB}\\[br]&=&\vec a +\frac 3 7(\vec b - \vec a)\\[br]&=&\frac{4\vec a+3\vec b}{7}.[br]\end{eqnarray}[/math][br]
Example 3
Let [math]\Large \overrightarrow{OA}[/math] and [math]\Large \overrightarrow{OB}[/math] be known as vectors. There is a point on the extension of line segment [i]AB[/i] so that [math]\Large PA:PB = 3:1.[/math] Define a vector [math]\Large \overrightarrow{OP}.[/math]  [br]  [br]
Solution 3.
Because the point [i]P[/i] is on the extension of the line segment [i]AB[/i], it cannot situate between the points. The ratio tells us that the point [i]P [/i] is closer to the point [i]B[/i] (in this case on the right side). Now, the line segment [i]AB[/i] is of the line segment [i]AP[/i], so [i]BP[/i] is half of the line segment [i]AB[/i].[br][br]  [math]\Large \begin{eqnarray}[br]\overrightarrow{OP}&=&\overrightarrow{OB}+\overrightarrow{BP}\\[br]&=&\overrightarrow{OB}+\frac 1 2 \overrightarrow{AB}\\[br]&=&\overrightarrow{OB}+\frac 1 2 \left ( \overrightarrow{OB}-\overrightarrow{OA}\right )\\[br]&=&-\frac 1 2 \overrightarrow{OA}+\frac 3 2 \overrightarrow{OB}[br]\end{eqnarray}[br][/math][br]   

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