- #1
Christoffelsymbol100
- 19
- 1
Homework Statement
I am currently solving a problem and I am not sure if it is correct.
There are two particles A and B. A has a constant velocity with [tex] |\vec{v}| = 3 [/tex] and starts from y = 30
B has constant acceleration with [tex] |\vec{a}| = 0,4 [/tex]
The goal is to find the angle between the y-Axis and the path of particle B under which collision with particle A happens.
All I want to know if my result is right, since I am not sure, and if there is another way to solve this
Homework Equations
[tex]|\vec{v}| = 3, y = 30[/tex]
[tex]|\vec{a}| = 0,4 [/tex]
The Attempt at a Solution
For particle A and I can say that: [tex]\vec{v} = (3,0) [/tex]
Integrating this with respect to time we get: [tex]\vec{x_A} = (3t,0) + (0,30)[/tex] where the second term represents the starting position
For particle B I can say that: [tex]\vec{a} = (sin(\theta),cos(\theta))\cdot 0,4 [/tex]
Integrating this two times with respect to time and setting the starting velocity and starting position as 0 we get: [tex]\vec{x_B} = (sin(\theta),cos(\theta))\cdot 0,2 \cdot t^2 [/tex]
Now defining [tex] t_C [/tex] as the time at which the collision happens, the x- and y-components of both [tex] x_A [/tex] and [tex] x_B [/tex] must be the same so:
[tex] 3\cdot t = sin(\theta) \cdot 0,2 t_C^2[/tex] and [tex] 30= cos(\theta) \cdot 0,2 t_C^2[/tex]
From the 1st equation it follows: [tex]t_C = \frac{15}{sin(\theta)}[/tex]
Plugging this into the second equation we will get:
[tex]\frac{2}{3} = \frac{cos(\theta)}{sin^2(\theta)}[/tex]
Solving this for [tex]\theta[/tex] we get: [tex]\theta = \frac{\pi}{3}[/tex] or [tex]\theta[/tex] equals 60 degrees.