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?