# Thread: What is velocity of the 2 kg block immediately before the collision?

1. A block with a mass of 2kg has an elastic collision with a block of mass 1 kg that is at rest. After the collision velocity of 1kg block is 1.4 m/s. What is velocity of the 2 kg block immediately before the collision? There is no friction. Please show the steps.

2. Originally Posted by R Morrison
A block with a mass of 2kg has an elastic collision with a block of mass 1 kg that is at rest. After the collision velocity of 1kg block is 1.4 m/s. What is velocity of the 2 kg block immediately before the collision? There is no friction. Please show the steps.
Show the steps. What did you try?

3. Originally Posted by R Morrison
A block with a mass of 2kg has an elastic collision with a block of mass 1 kg that is at rest. After the collision velocity of 1kg block is 1.4 m/s. What is velocity of the 2 kg block immediately before the collision? There is no friction. Please show the steps.
This is either a homework problem or looks like one. In either case you're required to post it in the Homework Help sub forum and follow the guidelines given there for asking for help. See READ FIRST BEFORE POSTING - Homework Help Guidelines

4. Write down the equations for conservation of energy and conservation of momentum. (The problem is one dimensional). The two unknowns are the before and after velocities of the 2 kg. block. The algebra is an elementary quadratic equation.

5. A sketch of the blocks before and after the collision will help make it clear what the equations will be. The whole thing is sign sensitive, be carefull with the directions of movement.

6. Originally Posted by pikpobedy
A sketch of the blocks before and after the collision will help make it clear what the equations will be. The whole thing is sign sensitive, be carefull with the directions of movement.
No, the correct speed, with the correct sign falls out no matter how you write the equations. You only have to make sure that all entities have positive sign in the equation of energy conservation, you can put any signs you want in the equation of momentum conservation, things "take care of themselves". Magic :-)

7. Originally Posted by mathman
The algebra is an elementary quadratic equation.
You are absolutely correct but the solution is given by an equation degree 1 because the term in cancels out: where , , .

8. XOX
You are muddying up this thread.
I do not need incorrect corrections or any algebra lessons what so ever from you. I did not ask for any and none were warranted.
Do that we will be OK.

9. Originally Posted by pikpobedy
XOX
You are muddying up this thread.
Why? It is correct physics. Try writing the equations and you will find out exactly what I told you.

I do not need incorrect corrections or any algebra lessons what so ever from you.
It was not a correction, it was just an observation. If you want to take it personally, it is your problem. The observation is correct, it is meant to help the OP solve the problem.

10. x0x

I checked your posts. You are some gotcha kind of girl. You post on physicsforums.com. So why don't you leave me alone. Is that fair with you. All comments about my posts by you will be flagged for stalking. Starting with ones above.

Your answer of 0.7 is completely wrong. NOt only that it is fundamentally impossible to be 0.7. Just by looking at it I can tell you have no clue what an elastic collision is and how to create the momentum vector equations and the KE scalar equations. You do not know how to calculate this stuff at all.

The fact you criticised Mathman furthermore proves you have no clue. There are quadratic terms. The velocity squared term.

I think I've written enough. I think your three strikes are up here.

PS
And stop starting your "gotcha" posts with "no" or "actually". Especially when you are off track and off target. It makes you look like a pedantic school marm, and makes you ridiculous when you are wrong.
Oh. Don't take this personally.

11. Originally Posted by pikpobedy

The fact you criticised Mathman furthermore proves you have no clue. There are quadratic terms. The velocity squared term.
Yes, I dropped one term in the final equation, the equation is indeed degree 2, mathman is right. The comment to you about the signs in the conservation of momentum equation stand.

12. Took x0x till 10:23am to correct** her mistake of method. This early morning it took me 5 minutes to calculate it two different ways. But I did not need to calculate to see that x0x had an impossible answer... it stood out like a sore thumb. I hope she does not set her sights too high because the Pik Pobedy is more than 7.4 km high..

Look, I am not snide here, but to be played around by a snide amateur does amuse me... some what, sometimes.

**not "correct" but erase is what I should have written, it is 6:03pm edt,

13. Originally Posted by pikpobedy
Took x0x till 10:23am to correct her mistake of method.
There is no error in the method, there was a dropped free term in the final equation. I admitted to the error.

This early morning it took me 5 minutes to calculate it two different ways. But I did not need to calculate to see that x0x had an impossible answer... it stood out like a sore thumb. I hope she does not set her sights too high because the Pik Pobedy is more than 7.4 km high..

Look, I am not snide here, but to be played around by a snide amateur does amuse me... some what, sometimes.
I wasn't playing you, I simply pointed out that one can use arbitrary signs in the equation of momentum conservation.

14. No answers for homework questions. Wait till the OP pipes up.
If he dares after seeing the crap you wrote, rewrote, and rewrote again.

Oh. The big block's magnitude change in velocity was -0.7 m/s and it continued in its same direction. I put thenegative sign in rather than stating "decreased" . Just to wash the floor some more. Who knows maybe I'm wrong? It has happened, just ask my wife. I kid, I kid.

15. Originally Posted by pikpobedy
Most likely the answer is in the book, the method is what is interesting. So , what is your answer? Mine is

And the equation giving the solution is indeed degree 1, not 2.

Who knows maybe I'm wrong?
Yes, you are. .

16. X0X, it is much better if you can help the OP arrive at the method and solution themselves. I think it is OK to confirm a solution if they offer it, or offer other suggestions if it is wrong. It is certainly not good form to hijack someone else's attempts to help the OP.

17. Originally Posted by Jilan
X0X, it is much better if you can help the OP arrive at the method and solution themselves.
This is exactly what I did.

I think it is OK to confirm a solution if they offer it, or offer other suggestions if it is wrong. It is certainly not good form to hijack someone else's attempts to help the OP.
Well, pickpobedy is not helping since his directions are wrong, and so is his answer, so I am not hijacking anything. Perhaps you could weigh in with your solution? What answer do you get? Last time you claimed my answer is wrong turned out that it was your answer that was wrong.

18. Go for it Gotcha Girl, you impress no one. LOL!

19. Originally Posted by Jilan
Go for it Gotcha Girl, you impress no one. LOL!
So, what is the speed before collision?
What is the speed after collision?

20. To xox. You made a mistake in your 11:26 pm post. m1 = 2, not 1.

The 2 kg block starts out at .7 and ends at 0.

21. Originally Posted by mathman
To xox. You made a mistake in your 11:26 pm post. m1 = 2, not 1.
Yes, (it was a typo, has no bearing on the answer), . This is why the answer to the OP (the initial speed of the 2kg block) is 1.05 as I said in my post to you. , do you agree?

The 2 kg block starts out at .7 and ends at 0.
I do not think that the above is correct. According to the correct equations, it starts at 1.05 and it ends at .

22. Here's another clue. 1.4 was chosen because it is close to the square root of two, thus creating a 1 x 2/2 situation for the small mass's KE.
This is the stuff taught to 16 to 18 year olds.

23. Originally Posted by pikpobedy
Here's another clue. 1.4 was chosen because it is close to the square root of two, thus creating a 1 x 2/2 situation fot the small mass's KE.
This is the stuff taught to 16 to 18 year olds.
So what is your answer? The OP exercise has a very nice symbolic solution that falls out the simple equation degree 1. The beauty of symbolic solutions is that they work with any numbers.

24. Mathman
Your PM function is disabled. I have a question for you.

25. x0x How did you know I blew in here with yesterday's road dust? That I fell with yesterday's fly ash? That I was washed in with yesteredays gutter runoff?

26. Originally Posted by pikpobedy
x0x How did you know I blew in here with yesterday's road dust?
What does all the above have to do with the simple exercise we were discussing?

That I fell with yesterday's fly ash? That I was washed in with yesteredays gutter runoff?

27. What is that you didn't understand that this a homework question. Thus I am falling forum rules. Whether you or I agree with them is irrelevent.

Write the KE equation for me. Humour me.

28. Moderator note: Regardless of who is right or wrong here, can we please stay on topic and act with some decorum?

29. Originally Posted by pikpobedy
What is that you didn't understand that this a homework question. Thus I am falling forum rules. Whether you or I agree with them is irrelevent.

Write the KE equation for me. Humour me.

The division by 2 can be omitted. So, now that I answered that, what is your (symbolic and numeric) answer for the exercise question. .

30. You are correct. This is becoming a bit too much. Problem is I need to PM someone who has shut off his PM.

I come the cigaraficionado site and am quite adept at dealing with Obama, gun control, global warming and evolution discussions. They basically allow us to use brass knuckles, razors and clubs. But the members are a hairy crusty bunch of %\$??&?.

31. x0x
Okay.
Now put in the given mass values for the masses, the given velocity for the small mass and the before and after velocities for the large mass.

32. Originally Posted by pikpobedy
x0x
Okay.
Now put in the given mass values for the masses, the given velocity for the small mass and the before and after velocities for the large mass.
I already solved this problem , I gave out BOTH the symbolic AND the numeric solution. I am now asking you : do you agree with my solution. Yes or no?

For completness, here is the set of conservation equations:

You can see that the large mass did not start at 0.7 and finish at zero.
Sure that is one solution for the momentum equation but it is not a solution for the KE equation.
Now I have known that since 2 or 3 this morning. I even got up out of bed so that all this would be cleaned up by morning.

34. Originally Posted by pikpobedy

You can see that the large mass did not start at 0.7 and finish at zero.
I never claimed that, mathman did, and I corrected him. He never replied after that, never acknowledged his error.

35. Originally Posted by pikpobedy
I come the cigaraficionado site and am quite adept at dealing with Obama, gun control, global warming and evolution discussions. They basically allow us to use brass knuckles, razors and clubs. But the members are a hairy crusty bunch of %\$??&?.
Well here we don't allow any brawling. The only weapon you should be using is logic.

36. x0x
You did state 0.7 for V before and then erased it later.

Because of the moderators intervention I will give you that the big mass is at V before = 1.06 and at V after = 0.36.

And momentum, being a vector is sign sensitive. Sometimes when the interaction are complicated we have no idea what direcion forces, flows, currents, objects are going. We define a direction and when we get a negative answer it means the opposite direction of the arrow, or the compression is really a tension.,

37. Originally Posted by pikpobedy
You did state 0.7 for V before and then erased it later.

Because of the moderators intervention I will give you that the big mass is at V before = 1.06 and at V after = 0.36.
You mean 1.05 and 0.35. Exactly. I am glad that you agree with me.

And momentum, being a vector is sign sensitive.
Yet, one can write the momentum conservation equation with ANY combinations of signs and still get the correct result. Like so:

Sometimes when the interaction are complicated we have no idea what direcion forces, flows, currents, objects are going.
You don't need to know, as I explained to you early in the thread.

38. What ever.

39. Originally Posted by pikpobedy
You will remember what happened in this thread all your life and you will chuckle.
Momentum equations must absolutely have the correct signs.
No, they do not, the momentum equations take care of themselves, the speed always comes up with the correct sign, as I explained to you at post 6. This is the beauty of the approach, you do not need to guess the sense of the vector. Even if you guess wrong it comes out right. Chuckle.

40. Can't understand normal thinking.

41. Originally Posted by pikpobedy
Can't understand normal thinking.
For chuckles and grins let's invert such that the conservation of momentum looks like:

If you do the calculations (correctly) you will get:

exactly as before. There is a good reason why this is true, can you figure it out?

So, no need to guess the sign of any of the variables.

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