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Collision of Two Particles

  • #1

Homework Statement



A particle A moves along the line y = d (30 m) with a constant velocity
upload_2018-8-22_16-24-23.png
(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
upload_2018-8-22_16-24-23.png
(a = 0.40 m/s2) at the same instant that particle A passes the y axis. What angle θ between
upload_2018-8-22_16-24-23.png
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 im 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 im lost.
[/B]
 

Attachments

Answers and Replies

  • #2
tnich
Homework Helper
1,048
336

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 im 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 im 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##.
 
  • #3
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?
 
  • #4
tnich
Homework Helper
1,048
336
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.
 

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