Need help with a Collisions Question

[physics_geek]

I need help with this question plz:

A 25.0g object moving to the right at 20.0cm/s overtakes and collides elastically with a 10.0g object moving in the same direction at 15.0cm/s. Find the velocity of each object after the collision.

The answer is: 17.1cm/s (25.0g obect), 22.1cm/s (10.0g object)

Can someone please explain how to get these answers..im soo lost

[ranger]

Before you receive help, you will have to attempt the problem. What formulas do you think need to be applied?

[physics_geek]

i can only think of
mv1(initial) + mv2(initial)= mv1(final) + mv2(final)

i dont know what to do next

[physics_geek]

umm yea is anyone going to help me..??

[Ki Man]

This is posted in the wrong section. somebody move this please.

You came up with an equation. have you plugged in the numbers yet?

[physics_geek]

Yea..and it doesnt work!!!...i guess nobody knows how to do this

[ranger]

i can only think of
mv1(initial) + mv2(initial)= mv1(final) + mv2(final)

i dont know what to do next

That step that you did was important. You've written down the momentum conservation for the system. And if we also know that total kinetic energy remains the same before and after the collusion (write it down), can you solve for v1 final and v2 final?

[cristo]

umm yea is anyone going to help me..??

Firstly, please note that we have homework forums, where these questions should be posted. Posting there will enable your question to be answered quicker. Secondly, the people helping you on this site are volunteers, and you cannot expect an answer to your question within ten minutes! Please be patient!

i can only think of
mv1(initial) + mv2(initial)= mv1(final) + mv2(final)

i dont know what to do next

With regard to your question; you have the right equation here. Which variables do you know? Can you plug these into the equation? Do you know any other quantity that is conserved here?

[physics_geek]

lol..alrite..sorry..guess im a little pushy..anyways ill try the homework forums..thanks

[cristo]

That's ok. You should continue this question here now though, since you've started a thread. (This will probably be moved by a mentor at a later time anyway)

Have you thought about my questions in the last post?

[physics_geek]

1. The problem statement, all variables and given/known data
A 25.0g object moving to the right at 20.0cm/s overtakes and collides elastically with a 10.0g object moving in the same direction at 15.0cm/s. Find the velocity of each object after the collision.


2. Relevant equations
mv1(initial) + mv2(initial)= mv1(final) + mv2(final)


3. The attempt at a solution
umm i dont even know where to begin or if the above equation is even the right one..please help

[physics_geek]

umm yea and i plugged in the information i know but im not getting the right answer..

[turdferguson]

Energy is also conserved, and you need a second equation to find the two unknowns

[cristo]

Well, post what you get when you plug the numbers into the equation (and not just the answer.. show your working) then I'll be able to see where you're going wrong!

[chemhelper]

You need another equation since you have 1 equation with two unknowns. That's why cristo was asking if there was another quantity conserved.

[physics_geek]

hmmm ummm uhhh....alrite im a blithering idiot 'cuz i cant come up with any equations..

[ranger]

hmmm ummm uhhh....alrite im a blithering idiot 'cuz i cant come up with any equations..

did you read my second reply to see what else is conserved?

[cristo]

Ok, you have this equation m1v1(initial) + m2v2(initial)= m1v1(final) + m2v2(final). Now, you know the masses, and the initial speeds, so plug them in!

Also, since we know the collision is elastic, kinetic energy is conserved. Do you know the equation for kinetic energy?

[physics_geek]

umm ok how about:
.5mv1(initial)^2 + .5mv2(initial)^2 = .5mv1(final)^2 + .5mv2(final)^2

what now?

[physics_geek]

yea i know..energy is also conserved..but how is that going to help me?

[cristo]

Here's what you wrote in the other thread:

umm ok how about:
.5mv1(initial)^2 + .5mv2(initial)^2 = .5mv1(final)^2 + .5mv2(final)^2

what now?

Now you have two equations. Plug the values you know into each of these equations, and you will have two equations, with two unknowns. Post what you get.

(edit: Thanks to the mentor that locked the other thread; it was getting rather confusing!)

[berkeman]

I merged the two threads on this topic by the OP.

[physics_geek]

umm ok

(25g)(20cm/s) + (10g)(15cm/s) = (25g + 10g)V(final)

.5(25g)(20cm/s)^2 + .5(10g)(15cm/s)^2 = .5(25g + 10g)V(final)^2 = 18.7

now wat?

[cristo]

umm ok

(25g)(20cm/s) + (10g)(15cm/s) = (25g)V1 + (10g)V2

.5(25g)(20cm/s)^2 + .5(10g)(15cm/s)^2 = .5(25g)V1^2+.5(10g)V2^2

now wat?

Since this is an elastic collision, the objects will still be separate after the collision. See the red amendments to your equation above.

Now, you solve the equations for the two variables you have.

[physics_geek]

umm i still dont understand...how do u solve for the two variables?

[cristo]

Well, the left hand side of both equations are just numbers, so firstly you should work those out. Then, you will have the first equation in the form of A=25v1+10v2 (where A is the number on the left hand side- I'm not doing the arithmetic for you!) Rearrange this to get v1 on its own, and the substitute this into the second equation. The second equation is then an equation for only v2 which you can solve.

Then, substitute the value obtained for v2 into the first equation, to give you the value of v1

[physics_geek]

ok well it sounds sooo easy...but for whatever reason i just cant solve it..

[physics_geek]

umm help someone..plz

[cristo]

ok well it sounds sooo easy...but for whatever reason i just cant solve it..

Why can't you do it? Have you simplified the equations into the form I showed you? If so, post them so I can see what you've done.

[HallsofIvy]

So far, you have not even tried anything. You have been told that, in addition to "conservation of momentum", since the collision is elastic, you also have "conservation of "energy". You have two equations:
(25g)(20cm/s) + (10g)(15cm/s) = (25g)V1 + (10g)V2 and
.5(25g)(20cm/s)^2 + .5(10g)(15cm/s)^2 = .5(25g)V1^2+.5(10g)V2^2

You ask "now what?" Well, how about actually multiplying out the numbers you have above? (25g)(20 cm/s)= 500 gcm/s, etc.
You will get
25V1+ 10V2= 650 and 12.5 V1^2+ 5V2^2= 6125

Try solving the first equation for V2 and putting that formula into the second so you get a quadratic equation for V1.

[AngeloG]

Draw a picture of what's happening and then after that it should become more clear on what equations you need to apply =).