physics anyone?

[cecile]

hm yeah, I'm pretty bad at physics I'm not sure how to set up this problem. care to help? I know the answer is 4.51 m/s....

A 5.00x10^2 kg log collides inelastically with a second with the same mass. These combined logs then collide with a third log with a mass of 5.00x10^2 kg. The final speed of the three combined logs is 3.67 m/s. If the speed of the third log before collision was 3.00 m/s, and the speed of the second log before collision was 3.50 m/s, what was the speed of the first log before collision?

[Brad Barker]

Originally Posted by cecile

hm yeah, I'm pretty bad at physics I'm not sure how to set up this problem. care to help? I know the answer is 4.51 m/s....

A 5.00x10^2 kg log collides inelastically with a second with the same mass. These combined logs then collide with a third log with a mass of 5.00x10^2 kg. The final speed of the three combined logs is 3.67 m/s. If the speed of the third log before collision was 3.00 m/s, and the speed of the second log before collision was 3.50 m/s, what was the speed of the first log before collision?

it's a conservation of momentum problem, but with two parts. there's one collision, where you have to use conservation of momentum and the definition of an inelastic collision to find the speed of the logs after the collision. and then you use THAT information as the initial speed heading into the second collision.

write down everything they give you, draw pictures before and after each collision, and write down the speeds and masses on your diagrams. helps out quite a bit.


oh, and be wary of signs. the direction of motion of one log wrt another may be antiparallel, which would require a negative value for the momentum.

[cecile]

I know the formula for conservation of momentum is: (M1V1,i)+(M2V2,i)= (M1V1,f)+(M2V2,f)

and the formula for inelastic collisions is:
(M1V1,i)+(M2V2,i)= (M1+M2)Vf


i just don't understand what I'm suppose to plug where because there are 3 logs and it gives you the speed before the collision for the other 2.

[Geoff St. Germaine]

It's easiest if you break the problem down and treat the two collisions separately. You know the final speed of the three logs after the second collision and the speed of the third log just before that collision so from that you can find the speed of the two logs before the second collision (which is their speed after the first collision). This speed can then be used in the same way for the first collision to get the speed of the first log since you know the speed of the second log prior to the first collision.

[Jared Lash]

It is a pair of inelastic collisions where the first two logs collide inelastically and then hit the third in another inelastic collision.

The only tricky part of this question is that you have to work the problem backwards. They give you the final speed of all three as well as the speed of the third one. So you can find the speed of the first two logs before the second collision.

After that you have the speed of logs one and two after their collision as well as the speed of log two. That's enough info to solve for the initial speed of the first log.

EDIT: Apparently I don't type fast enough as Geoff beat me to the punch.

[middy]

Thank God we have you physicists around to calculate log speeds!

[TrooperFarva]

Conservation of Momentum

(1500*3.67) = (500*3) + (500*3.5) + (500*X)

X=4.51

It's probably easier to treat it as 2 separate collisions, and calculate the momentums separately, but it's not necessary

[Brad Barker]

Originally Posted by middy

Thank God we have you physicists around to calculate log speeds!

i'm not sure that the logarithm of speed is a physically relevant quantity.

[Ericman197]

those are some pretty heavy logs i tell you hwhat

[*smb]

Originally Posted by TrooperFarva

Conservation of Momentum

(1500*3.67) = (500*3) + (500*3.5) + (500*X)

X=4.51

It's probably easier to treat it as 2 separate collisions, and calculate the momentums separately, but it's not necessary

As this guy said - the final momentum is just the sum of the individual momenta of each object.

[Jared Lash]

Originally Posted by Ericman197

those are some pretty heavy logs i tell you hwhat

I said virtually the same thing this afternoon, albeit in a very different context. Damn you Chipotle. . .

[Sir Edward V]

Originally Posted by TrooperFarva

Conservation of Momentum

(1500*3.67) = (500*3) + (500*3.5) + (500*X)

X=4.51

It's probably easier to treat it as 2 separate collisions, and calculate the momentums separately, but it's not necessary

+1

Funny thing is I didn't remember any of this on my own! I have a horrible memory... I figured out that I absorb the material by paying attention in class, but it stays dormant and unorganized in my brain until I need it for a test or something. Then I figure out how everything relates to each other just in time for the test, then let it go back to how it was after it is over... The funny thing is that I decided with the help of my Data Structures teacher that my method is more efficient time-wise than people who study all the time! Cause I still get A's and B's, I just don't know anything unless I NEED to!

[hbarcat]

Originally Posted by Brad Barker

i'm not sure that the logarithm of speed is a physically relevant quantity.

That's nerd humor if I ever saw it.


Then I must be a nerd because I laughed.

[Brad Barker]

Originally Posted by hbarcat

That's nerd humor if I ever saw it.


Then I must be a nerd because I laughed.

i made a facebook group called "particle physics gives me a hadron." it's up to 803 members.

[CrazyArcher]

Originally Posted by hbarcat

That's nerd humor if I ever saw it.

Then I must be a nerd because I laughed.

LOL I had a similar thought looking through the topic. "Log speeds? ***?"