Collision Angle for Two Particles with Constant Velocity and Acceleration

[Christoffelsymbol100]

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 $$|\vec{v}| = 3$$ and starts from y = 30

B has constant acceleration with $$|\vec{a}| = 0,4$$

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

$$|\vec{v}| = 3, y = 30$$
$$|\vec{a}| = 0,4$$

The Attempt at a Solution

For particle A and I can say that: $$\vec{v} = (3,0)$$

Integrating this with respect to time we get: $$\vec{x_A} = (3t,0) + (0,30)$$ where the second term represents the starting position

For particle B I can say that: $$\vec{a} = (sin(\theta),cos(\theta))\cdot 0,4$$

Integrating this two times with respect to time and setting the starting velocity and starting position as 0 we get: $$\vec{x_B} = (sin(\theta),cos(\theta))\cdot 0,2 \cdot t^2$$

Now defining $$t_C$$ as the time at which the collision happens, the x- and y-components of both $$x_A$$ and $$x_B$$ must be the same so:

$$3\cdot t = sin(\theta) \cdot 0,2 t_C^2$$ and $$30= cos(\theta) \cdot 0,2 t_C^2$$

From the 1st equation it follows: $$t_C = \frac{15}{sin(\theta)}$$

Plugging this into the second equation we will get:

$$\frac{2}{3} = \frac{cos(\theta)}{sin^2(\theta)}$$

Solving this for $$\theta$$ we get: $$\theta = \frac{\pi}{3}$$ or $$\theta$$ equals 60 degrees.

 

[DoItForYourself]

Hi,

It seems correct. Good job!

 

[RedDelicious]

You need to rethink your equations for B.

In the x direction, B has no acceleration. Ok. When doing the integrations, you set its initial velocity in both directions to zero as well. Without any acceleration or initial velocity in the x-direction, how do you have it moving in the x-direction?

 

[Christoffelsymbol100]

You need to rethink your equations for B.

In the x direction, B has no acceleration. Ok. When doing the integrations, you set its initial velocity in both directions to zero as well. Without any acceleration or initial velocity in the x-direction, how do you have it moving in the x-direction?

But it has acceleration in the x- and y- direction namely $$a_x = sin(\theta)*0,4$$ and $$a_x = cos(\theta)*0,4$$. Particle B starts with zero velocity and starts the origin so $$\vec{x_0} = 0 , \vec{v_0} = 0$$