# Collision of Particles

1. Oct 10, 2005

### shawpeez

Could someone point me in the right direction, im totally lost

particle A moves along the line y=30m with a constant velocity v of magnitude 3.0m/s and parallel to the x axis. At the instant particle A leaves the y axis, particle B leaves the origin with zero initial speed and constant acceleration a of magnitude 0.40m/s^2. What angle between a and the positive direction of the y axis would result in a collision?

2. Oct 10, 2005

### mezarashi

A math exercise once you can visualize what's going on. You have a particle moving along y=30m. Now you are at the origin with zero velocity. You want to crash into the dude moving at y=30. The only thing is that you don't have a steering wheel. Meaning you have to calculate the angle you want to start off with knowing that you can accelerate at 0.40 m/s/s.

Kinematics equations. You have two equations, one in the x and the other in the y.
What are the conditions for collision?

1. The displacement in y is equal to 30m.
2. The displacement in x is equal to the displacement of particle A at collision time (which we don't yet know).

With those two conditions, we write out the variables we know and are looking for in the x and y (I gave you some hints, fill in the rest):

x-axis
a: asin(angle)
vinitial:
vfinal: ?
displacement:
time: ?

y-axis:
a: acos(angle)
vinitial:
vfinal: ?
displacement:
time: ?

3. Oct 10, 2005

### shawpeez

Wouldn't x - axis be a: acos(angle) and y -axis be a: asin(angle)

4. Oct 10, 2005

### shawpeez

Ok this is what i've tried so far
Particle B
Y-axis
displacement= 30m

30 = V0(t) + 1/2a(t^2)
t= 12.25s

v = v0 + at
v = (.40)12.25
v= 4.9m/s

In 12.25s particle A travels 3.0m/s * 12.25s = 36.75m

X-axis
displacement = 36.75m
36.75 = vo + 1/2a(t^2)
36.75 =(.2)(t^2)
t= 13.56s

v= vo + at
= (.4)(13.56)
= 5.42m/s

I don't know if im on the right track or not but this is as far as i can get

5. Oct 10, 2005

### shawpeez

still pretty much stuck, could anyone help

6. Oct 11, 2005

### shawpeez

from this i took the position r=(36.75m)i + (30m)j
then i found the angle relative to the x axis

Tan(theta)= 30/36.75
theta = arctan .82
= 39.4

90 - 39.4 = 50.6 which would be the angle from the y-axis

can someone tell me if im doing this right i would appreciate it.

7. Oct 11, 2005

### mezarashi

Normally yes, except the question asks

so you've got to be more flexible and truly understand the geometry problem instead of thinking that cosine is always an x-axis "thing".

How did you get the value for t? The value for a is acos(angle). I thought we were just discussing how we don't know the angle and that's what we're finding? Same goes for your x-axis analysis. The conditions we DO know are:

$$d_b_y = \frac{1}{2}a_1t^2 = 30m$$
$$d_b_x = \frac{1}{2}a_2t^2 = d_a_x = vt$$

where a1 and a2 are acos(angle) and asin(angle). How many unknowns, how many variables do we have?