# Analysis of collision of particles from graph

• coldblood
The slopes are different for the two particles, but they are the same for each particle before and after the collision. This tells us that the masses are equal. In summary, the conversation discusses an issue with analyzing the collision of particles from a graph. The question asks for clarification on certain points in the graph, including the prediction of the instant of collision, the direction of the velocities, and the change in slopes. The conversation concludes with the confirmation that the collision occurs over a period of time and that the slopes of the graphs indicate equal masses for the particles.

#### coldblood

Hi friends,
I have an issue in Analysis of collision of particles from graph.

The problem is as:

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-ash4/q71/s720x720/1381762_1432382976988808_995987085_n.jpg

Attempt:

https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/1378626_1432380616989044_74050265_n.jpg

Please try to help me in this.
I will appreciate the help.

In your answer marked "(I)" you say that "(I) is correct" - what does this mean?
... is it not possible for two velocities on the same axis to be in opposite directions?

In your answer marked "(II)" you say that "it is clear from the graph" but you don't say how. What is it about the graph that tells you the velocities are the same at the instant of the collision?
Note: if both objects are traveling with the same velocity (i.e. same speed and direction) then how can they collide?

In your answer marked "(III)" you say there is no change in the slopes of the curves - yet don't both curves change slope twice? i.e. at t=1s and at t=3s.

Try describing, in words, the motion of S and R as represented by the v-t graphs.

You are right on I, though the wording is a little unclear. What exactly do you mean by being "on the same axis"?
In II you refer to the instant of collision. In the graph, collision occurs over a time period, and the question asks about the middle of that period. Where is that on the graph?
For III, compare the slopes during collision. Are they the same? What does that tell you about the masses?

Simon Bridge said:
In your answer marked "(I)" you say that "(I) is correct" - what does this mean?
... is it not possible for two velocities on the same axis to be in opposite directions?

In your answer marked "(II)" you say that "it is clear from the graph" but you don't say how. What is it about the graph that tells you the velocities are the same at the instant of the collision?
Note: if both objects are traveling with the same velocity (i.e. same speed and direction) then how can they collide?

In your answer marked "(III)" you say there is no change in the slopes of the curves - yet don't both curves change slope twice? i.e. at t=1s and at t=3s.

Try describing, in words, the motion of S and R as represented by the v-t graphs.

Well,
I though that the collision takes place at t = 2 sec. Please tell me how can we predict the instant of collision or the time limit. So far I am getting that wherever the slope of velocity - time graph is changing, these are the instants of collision. For ex. in the case of R, at points t= 1 and t = 3 the slopes are changing hence these are the instants of collision. Is it correct?

Well, about the direction I am saying that, the velocity time graphs for the both particles lies on the same quadrant. That's why I was saying the the velocities are in the same direction for both the particles.

Still have a big confusion in this problem.

haruspex said:
You are right on I, though the wording is a little unclear. What exactly do you mean by being "on the same axis"?
In II you refer to the instant of collision. In the graph, collision occurs over a time period, and the question asks about the middle of that period. Where is that on the graph?
For III, compare the slopes during collision. Are they the same? What does that tell you about the masses?

Simon Bridge said:
In your answer marked "(I)" you say that "(I) is correct" - what does this mean?
... is it not possible for two velocities on the same axis to be in opposite directions?

In your answer marked "(II)" you say that "it is clear from the graph" but you don't say how. What is it about the graph that tells you the velocities are the same at the instant of the collision?
Note: if both objects are traveling with the same velocity (i.e. same speed and direction) then how can they collide?

In your answer marked "(III)" you say there is no change in the slopes of the curves - yet don't both curves change slope twice? i.e. at t=1s and at t=3s.

Try describing, in words, the motion of S and R as represented by the v-t graphs.

Well,
I though that the collision takes place at t = 2 sec. Please tell me how can we predict the instant of collision or the time limit. So far I am getting that wherever the slope of velocity - time graph is changing, these are the instants of collision. For ex. in the case of R, at points t= 1 and t = 3 the slopes are changing hence these are the instants of collision. Is it correct?

Well, about the direction I am saying that, the velocity time graphs for the both particles lies on the same quadrant. That's why I was saying the the velocities are in the same direction for both the particles.

Still have a big confusion in this problem.

coldblood said:
Well,
I though that the collision takes place at t = 2 sec. Please tell me how can we predict the instant of collision or the time limit. So far I am getting that wherever the slope of velocity - time graph is changing, these are the instants of collision. For ex. in the case of R, at points t= 1 and t = 3 the slopes are changing hence these are the instants of collision. Is it correct?

Well, about the direction I am saying that, the velocity time graphs for the both particles lies on the same quadrant. That's why I was saying the the velocities are in the same direction for both the particles.

Still have a big confusion in this problem.
There is not an instant of collision here, the collision occurs over a period of time from t=1 to t=3.
Yes, same quadrant makes more sense than same axis. Better still, just say they're both positive.
Now concentrate on the period of collision. Each v-t graph has a slope there, one positive one negative. What does the slope represent? If the force between the particles at some instant is F, what equations would you write for their accelerations?

The collision is not an instantaneous process. See video.

When two balls collide they stay together for some time interval, and both deforming during that impact process.
The velocity of the CM-s change with respect to the initial velocity after getting into contact. At an instant, they move together, and after that, the elastic force between them starts to accelerate both balls away from each other, till the instant when they get separated again.

ehild

Last edited by a moderator:
coldblood said:
Well,
I though that the collision takes place at t = 2 sec.
That is only an instant of time.
A collision does not take an instant.
That is the time that the velocities are equal - the velocities start changing before then.

Please tell me how can we predict the instant of collision or the time limit.
You read it off the graph.

So far I am getting that wherever the slope of velocity - time graph is changing, these are the instants of collision. For ex. in the case of R, at points t= 1 and t = 3 the slopes are changing hence these are the instants of collision. Is it correct?
That is correct - the collision takes place over the time period from t=1 to t=2 seconds. During that time the objects experience a force which causes the acceleration.

Note: in a real life collision, the v-t graph won't be that simple.

Well, about the direction I am saying that, the velocity time graphs for the both particles lies on the same quadrant. That's why I was saying the the velocities are in the same direction for both the particles.
You mean to say that the velocities are both positive.

Look at the graph, and try to picture what is going on: take positive distances to the left:
- R travels to the left at +0.8m/s for 1μs, then decelerates to +0.2m/s over 2 micro-seconds.
- S is stationary for the first second, then it starts accelerating to the left. It reaches a velocity of +1.0m/s in 2μs and continues at a constant speed.

So: what causes S to start moving?

Simon Bridge said:
the collision takes place over the time period from t=1 to t=2 seconds.
t=1 to t=3.

haruspex said:
Each v-t graph has a slope there, one positive one negative. What does the slope represent? QUOTE]

Slope of S represents acceleration and slope of R represents de-acceleration.

haruspex said:
If the force between the particles at some instant is F, what equations would you write for their accelerations?

For R -

a = 0; 0< t ≤1
a = -0.3; 1≤ t ≤ 3
a = 0 ; t≥3

For S:

a = 0.5; 1≤ t ≤ 3
a = 0 ; t≥3

haruspex said:
t=1 to t=3.
I'm getting tired - should take a break <makes another coffee>

• 1 person
coldblood said:
For R -

a = 0; 0< t ≤1
a = -0.3; 1≤ t ≤ 3
a = 0 ; t≥3

For S:

a = 0.5; 1≤ t ≤ 3
a = 0 ; t≥3
I was thinking in terms of F = ma. If you apply that to each ball at some arbitray instant during collision, can you see a relationship between their respective accelerations and their respective masses?

• 1 person
haruspex said:
I was thinking in terms of F = ma. If you apply that to each ball at some arbitray instant during collision, can you see a relationship between their respective accelerations and their respective masses?

haruspex said:
I was thinking in terms of F = ma. If you apply that to each ball at some arbitray instant during collision, can you see a relationship between their respective accelerations and their respective masses?

Taking equation for the interval t= 1 to t = 3

For R -

F = -0.3m1; 1≤ t ≤ 3

For S:

F = 0.5m2; 1≤ t ≤ 3

During collision "F" will be same for both the objects. so, mass will be inversionally proportional to magnitude of acceleration. Hence mass of R will be grater than mass of S.

All the doubts of this assignment are cleared with your and Simon's help. Thanks a lot to you guys. I appreciate the help.

No worries ;)