# What Angle Causes Particles A and B to Collide?

• BeyondBelief96
In summary: Or you could just solve for ##\sin(\theta)## or ##\cos(\theta)##, then use an inverse trig function on your calculator. The advantage of solving for ##t^2## is that it eliminates the trig function entirely.In summary, to find the angle θ between particle A and the positive y-axis that would result in a collision with particle B, we can use the kinematic equations to set up a system of equations and eliminate the trig function by adding them together. This will result in a quadratic equation that can be solved for t^2, which can then be used to find the value of theta using an inverse trig function on a calculator.
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]

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BrandonBerisford said:

## 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##.

tnich said:
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?

BrandonBerisford said:
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.

## 1. What is a collision of two particles?

A collision of two particles is when two individual particles come into contact with each other, resulting in a transfer of energy and possibly a change in direction or velocity.

## 2. What factors can affect the outcome of a collision between two particles?

The outcome of a collision between two particles can be affected by factors such as the mass, velocity, and type of particles involved, as well as the angle and force of the collision.

## 3. How is the collision between two particles analyzed and calculated?

The collision between two particles can be analyzed and calculated using principles of conservation of momentum and energy. This involves considering the initial and final momentum and energy of the particles before and after the collision.

## 4. Can the collision between two particles be perfectly elastic?

Yes, the collision between two particles can be perfectly elastic, meaning that there is no loss of kinetic energy during the collision. This is possible when the particles involved are of the same mass and there is no external force acting on them.

## 5. How is the collision between two particles relevant in real-world applications?

The collision between two particles is relevant in various real-world applications, such as in particle accelerators, where particles are collided at high speeds to study their properties. It is also relevant in fields like astrophysics, where the collision of particles in space can lead to the formation of celestial bodies.

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