What Angle Causes Particles A and B to Collide?

[BeyondBelief96]

Homework Statement

A particle A moves along the line y = d (30 m) with a constant velocity

(v= 3.0 m/s) directed parallel to the positive x-axis (Fig. 4-40). A second particle B starts at the origin with zero speed and constant acceleration

(a = 0.40 m/s2) at the same instant that particle A passes the y axis. What angle θ between

and the positive y-axis would result in a collision between these two particles?[/B]

Homework Equations

Kinematic Equations

The Attempt at a Solution

So I listed out all knowns and unknowns for both particle A & B as follows:

Particle A: Xi = 0, Xf = ?, Vix = 3 m/s, Vif = 3m/s, a = 0 t = ? and y = 30

Particle B: Xi = 0, Xf = ?, Vix = 0 m/s, Vif = ?, ax = asin(theta), t = ?
Yi = 0, Yf = 30, Viy = 0 m/s, Vif = ?, ay = acos(theta), t = ?

From there I used the kinematic equation d = vit + 1/2at^2 for particle A and B and got the following:

Particle A: x = 3t, y =30
Particle B: x = 1/2asin(theta)t^2, y = 1/2acos(theta)t^2

I set each of the equations equal to each other and tried to solve the system of equations for theta as follows:

3t = 1/2asin(theta)t^2 (Eq1)

30 = 1/2acos(theta)t^2 (Eq2)

3 = 1/2asin(theta)t so t = 6/asin(theta)

So I plugged t into Eq2:

30 = 1/2acos(theta)[6/asin(theta)]^2

from here I am at a loss on how to solve for theta. To be honest, my algebra solving skills arent super spectacular, which is a bit sad haha. I was able to simplify this as:

30 = 1/2acos(theta)[36/a^2sin^2(theta)] == 5/3 = cot(theta)/asin(theta) ...aaaaaand this is where I am lost.[/B]

 

[tnich]

Homework Statement

A particle A moves along the line y = d (30 m) with a constant velocity View attachment 229683 (v= 3.0 m/s) directed parallel to the positive x-axis (Fig. 4-40). A second particle B starts at the origin with zero speed and constant acceleration View attachment 229681 (a = 0.40 m/s2) at the same instant that particle A passes the y axis. What angle θ between View attachment 229682 and the positive y-axis would result in a collision between these two particles?[/B]

Homework Equations

Kinematic Equations

The Attempt at a Solution

So I listed out all knowns and unknowns for both particle A & B as follows:

Particle A: Xi = 0, Xf = ?, Vix = 3 m/s, Vif = 3m/s, a = 0 t = ? and y = 30

Particle B: Xi = 0, Xf = ?, Vix = 0 m/s, Vif = ?, ax = asin(theta), t = ?
Yi = 0, Yf = 30, Viy = 0 m/s, Vif = ?, ay = acos(theta), t = ?

From there I used the kinematic equation d = vit + 1/2at^2 for particle A and B and got the following:

Particle A: x = 3t, y =30
Particle B: x = 1/2asin(theta)t^2, y = 1/2acos(theta)t^2

I set each of the equations equal to each other and tried to solve the system of equations for theta as follows:

3t = 1/2asin(theta)t^2 (Eq1)

30 = 1/2acos(theta)t^2 (Eq2)

3 = 1/2asin(theta)t so t = 6/asin(theta)

So I plugged t into Eq2:

30 = 1/2acos(theta)[6/asin(theta)]^2

from here I am at a loss on how to solve for theta. To be honest, my algebra solving skills arent super spectacular, which is a bit sad haha. I was able to simplify this as:

30 = 1/2acos(theta)[36/a^2sin^2(theta)] == 5/3 = cot(theta)/asin(theta) ...aaaaaand this is where I am lost.[/B]

Here is a technique that you can use on a lot problems like this to eliminate angle $\theta$.

$3t = \frac 1 2 a\,t^2 \sin(\theta)$
$30 = \frac 1 2 a\,t^2 \cos(\theta)$

$9t ^2= \frac 1 4 a^2t^4 \sin^2(\theta)$
$900 = \frac 1 4 a^2t^4 \cos^2(\theta)$
_____________________
$9t ^2 + 900 = \frac 1 4 a^2t^4$

In the last step, I just added the two equations. Now you have a quadratic equation to solve for $t^2$.

 

[BeyondBelief96]

Here is a technique that you can use on a lot problems like this to eliminate angle $\theta$.

$3t = \frac 1 2 a\,t^2 \sin(\theta)$
$30 = \frac 1 2 a\,t^2 \cos(\theta)$

$9t ^2= \frac 1 4 a^2t^4 \sin^2(\theta)$
$900 = \frac 1 4 a^2t^4 \cos^2(\theta)$
_____________________
$9t ^2 + 900 = \frac 1 4 a^2t^4$

In the last step, I just added the two equations. Now you have a quadratic equation to solve for $t^2$.

NEW
oh okay, that makes sense! So once I've solved for t^2 just square root it for t, and then plug it into one of my original equations then solve for theta?

 

[tnich]

NEW
oh okay, that makes sense! So once I've solved for t^2 just square root it for t, and then plug it into one of my original equations then solve for theta?

Right.