# When do these two moving particles come together?

• Yigitu17
In summary: What the conditions are for when the two particles collide?In which case the two particles collide?What the result of the collision is?I actually understand the situations by sketching, and I can give some mathematical statements about them. But for example for the situation where they don't collide, the factor time remains in the inequality and I have no idea how to eliminate it.For the second statement I can say that X1(t)=X2(t), v1+a1*t>0, v2+a2*t<0. But again I don't know how to combine those statements. Do you have any tips for me?Can you show some of
Yigitu17
Homework Statement
Hello everyone, I have recently joined a physics summer school, for which I have to do homework every day. But I couldn't manage to solve today's problem, so I'm seeking help from you guys.

The question is about two particles moving on a one dimensional path. First one is at x=0 when t=0, and moves with a velocity of V1 and acceleration a1. The other one is at x=L>0 when t=0 and moves with a velocity of V2 and acceleration a2.
Relevant Equations
x(t)=x0+v*t+1/2*a*t^2
Hello everyone, I have recently joined a physics summer school, for which I have to do homework every day. But I couldn't manage to solve today's problem, so I'm seeking help from you guys.

The question is about two particles moving on a one dimensional path. First one is at x=0 when t=0, and moves with a velocity of V1 and acceleration a1. The other one is at x=L>0 when t=0 and moves with a velocity of V2 and acceleration a2. Questions:

In which cases do the objects:
a) not crash
b)crash one time as they go opposite directions
c)crash one time as they go in the same directions
d)crash twice in opposite directions
e)crash twice in same directions
f)crash twice, first traveling in the same direction and then in the opposite direction
g)crash twice, first traveling in the opposite direction and then in the same direction.

Now I understand that it must be x2(t)>x1(t) if they don't crash. I tried solving this inequality but I couldn't eliminate t and my math wasn't good enough for this inequity. I assume I have to find a way to eliminate t to solve further questions as well. So if you can help me I would really appreciate that. Thank you in advance for you help.

Delta2
Welcome to PF.

You are using the right Relevant Equation, so can you say a few things about the initial velocities and accelerations to result in them crashing going opposite directions versus going the same direction?

Also, in problems like this it is helpful to draw a 2-d graph of the motion of each particle. You can put x on the horizontal axis and time on the vertical axis. Then you can sketch different scenarios with mixes of initial velocities and accelerations to see when you get the 2 particles coming together...

(Also, there is a LaTeX Guide at the bottom of the page -- learning to post equations in LaTeX makes them much more readable)

Delta2
berkeman said:
Welcome to PF.

You are using the right Relevant Equation, so can you say a few things about the initial velocities and accelerations to result in them crashing going opposite directions versus going the same direction?

Also, in problems like this it is helpful to draw a 2-d graph of the motion of each particle. You can put x on the horizontal axis and time on the vertical axis. Then you can sketch different scenarios with mixes of initial velocities and accelerations to see when you get the 2 particles coming together...

(Also, there is a LaTeX Guide at the bottom of the page -- learning to post equations in LaTeX makes them much more readable)
Thank you for your answer, forgive my way of writing the equation, I am new to the site. I actually understand the situations by sketching, and I can give some mathematical statements about them. But for example for the situation where they don't collide, the factor time remains in the inequality and I have no idea how to eliminate it. For the second statement I can say that X1(t)=X2(t), v1+a1*t>0, v2+a2*t<0. But again I don't know how to combine those statements. Do you have any tips for me?

Delta2
For a) you don't have to solve an inequality. Instead find the condition such that the equation $$X_1(t)=X_2(t)\iff X_1(t)-X_2(t)=0$$ has no real solutions. It would be a quadratic equation, you know when it doesn't have real solutions when the discriminant is less than zero.

Yigitu17 said:
I actually understand the situations by sketching
Can you show some of your sketches for each of the situations in the question? If you scan them or take good quality pics of them, you can use the "Attach files" link below the Edit window to upload PDF or JPEG copies of them.

The different scenarios are kind of fun to think about and visualize, actually. I'm not sure yet if they are all possible or not, but most seem just to be related to the signs and magnitudes of the initial velocities and the accelerations. FYI, here are your equations in LaTeX:
$$x_1(t) = x_{1i} + v_{1}t + \frac{1}{2}a_1t^2$$
$$x_2(t) = x_{2i} + v_{2}t + \frac{1}{2}a_2t^2$$

Generally all the sub questions are about when the "proper" quadratic equation has none, one positive (and one negative) or two positive solutions (solutions for the time t when the collision happens). Remember , time can be only positive.

berkeman
It seems to me that you can answer the question quite easily and without equations if you consider the situation from the point of view of an observer sitting at rest on particle 1. He sees particle 2 having (relative) acceleration ##a_{rel}=a_2-a_1## and (relative) velocity ##v_{rel}=v_2-v_1##. There are four possibilities
1. ##a_{rel} >0## and ##v_{rel} >0##
2. ##a_{rel} >0## and ##v_{rel} <0##
3. ##a_{rel} <0## and ##v_{rel} >0##
4. ##a_{rel} <0## and ##v_{rel} <0##
For each of these possibilities, can you figure out the motion of particle 2 as seen by the observer on particle 1? The number of crashes will be the same as seen by this observer and an observer who sees both particles accelerate.

Delta2
Yigitu17 said:
Homework Statement:: Hello everyone, I have recently joined a physics summer school, for which I have to do homework every day. But I couldn't manage to solve today's problem, so I'm seeking help from you guys.

The question is about two particles moving on a one dimensional path. First one is at x=0 when t=0, and moves with a velocity of V1 and acceleration a1. The other one is at x=L>0 when t=0 and moves with a velocity of V2 and acceleration a2.
Relevant Equations:: x(t)=x0+v*t+1/2*a*t^2

couldn't eliminate t
It is not a matter of eliminating t. They crash if ##x_1(t)=x_2(t)## for some t. For what relationships between ##v_1, v_2, a_1, a_2, L## does such a t exist?

## 1. What is the formula for calculating when two moving particles will come together?

The formula for calculating when two moving particles will come together is t = (d2 - d1) / (v1 - v2), where t is the time, d1 and d2 are the initial distances of the particles, and v1 and v2 are the velocities of the particles.

## 2. Can the time for two particles to come together be negative?

No, the time for two particles to come together cannot be negative. If the calculated time is negative, it means that the particles have already come together at a previous time.

## 3. How does the mass of the particles affect the time for them to come together?

The mass of the particles does not affect the time for them to come together. The time is only dependent on the initial distances and velocities of the particles.

## 4. What happens if the velocities of the particles are the same?

If the velocities of the particles are the same, the time for them to come together will be undefined. This is because the denominator in the formula will be zero, making the calculation impossible.

## 5. Is there a way to predict the exact moment when two particles will come together?

No, there is no way to predict the exact moment when two particles will come together. The formula only gives an estimate of the time, and factors such as external forces or changes in velocity can affect the actual time for the particles to come together.

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