physicsvectors
An object moving 12m/s passes north and hits an object. Due to the wind from a west direction, it is pushed sideways at 5m/s. Find the resultant velocity.
I don't know where to start with this one, I can do the other ones just fine. It involves vector addition and subtraction. Would anyone know? I have the answers (13m/s north at 22.6 degrees).
Any help is much needed!
The resulting vector is:
$$ \vec{v} = \begin{bmatrix} 30\cos(45°) \\ -30\sin(45°)\end{bmatrix} + \begin{bmatrix} 200\cos(30°) \\200\sin(30°)\end{bmatrix} = \begin{bmatrix} 194,42 \\ 78,79 \end{bmatrix} $$
$||\vec{v}||$ = 209,8 m/s under an angle of 22.05 degrees
If the object moves with velocity $\vec v_0$ the ball with velocity $\vec v$ and the normal at the impact point is $\vec n$ then we have:
$$ \vec v = \vec v_{\vec n}+\vec v_{\Pi}\\ \vec v_{\Pi} = \vec v - \vec v_{\vec n_1} \\ \vec v_r = (\vec v-\vec v_0)_{\Pi}-(\vec v-\vec v_0)_{\vec n}\\ \vec v_r = (\vec v-\vec v_0) -2((\vec v-\vec v_0).\vec v_{\vec n})\vec v_{\vec n} $$
with
$$ \vec v_{\vec n} = \left(\vec v\cdot\left(\frac{\vec n}{||\vec n||}\right)\right)\frac{\vec n}{||\vec n||} $$
where $\vec v_r$ represents the reflected ball velocity after collision
NOTE
Here $\Pi$ represents the plane passing by the impact point with normal $\vec n$
Attached three cases. Here
$$ \begin{cases} \vec v \ \ \mbox{red}\\ \vec v_0 \ \ \mbox{green}\\ \vec n \ \ \mbox{black}\\ \Pi \ \ \ \mbox{dashed cyan}\\ \vec v_r \ \ \mbox{blue} \end{cases} $$
Best Answer
Fairly simple. I hope you can imagine it: The resultant works along the diagonal of a rectangle, each of the component velocities work along the sides of it. The sides are proportional to $5$ and $12$. So, the length of the diagonal is $\sqrt{5^2+12^2}=13$.
To find the angle with the side of length $12$, note that the angle $\theta$ is such that $\tan \theta = \frac{5}{12} \implies \theta = \tan^{-1} \frac{5}{12} = 22.618864^\circ$.
For a better understanding, you may like the image: