Question

A 0.150 kg air-track glider moving at 1.75 m/s bumps into a 0.350 kg glider at rest.

Part 1: Find the total kinetic energy after collision if the collision is elastic.

Part 2: Find the total kinetic energy after collision if the collision is completely inelastic.

***How can I solve this and show the work for it? :/

Answer 1

Okay hmm, this is taking me back to physics 1...argh 3 year old memory haha

well basically since it would be elastic (part a) there would be no loss of kinetic energy..so we have

\[\large K.E_{initial} = K.E_{final}\]

so what the the total kinetic energy in the beginning?

\[\large K.E_{air glider 1} + K.E_{air glider 2}\]
\[\large \frac{1}{2}mv^2 + \frac{1}{2}mv^2\]...

Answer 2

hahaha :P

and oaky! so from there, we know it is elastic, so we have to find the total kinetic energy? what would we plug in? :O

Answer 3

It is elastic so we know whatever we get from the initial total kinetic energy is our answer because it is the same as AFTER the collision

So before we have

For air glider 1

\[\large K.E = \frac{1}{2}mv^2 = \frac{1}{2}(0.150kg)(1.75m/s)^2 = ?\]

And for air glider 2

\[\large K.E = \frac{1}{2}mv^2 = \frac{1}{2}(0.350kg)(0m/s)^2 = 0\]

Right? and the total kinetic energy is the sum of those...so what is

\[\large \frac{1}{2}(0.150kg)(1.75m/s)^2 + 0 = ?\]

Answer 4

ahh okay!!

so it would equal 0.13125?

not sure if i calculated that correctly?

Answer 5

Argh you're gonna make me calculate it? >.< haha no jk hang on...

Answer 6

Nope not quite what I get...

Answer 7

You didn't square the velocity...

Answer 8

haha im not sure... a lot of times i tend to do something wrong lol

like i forget to change something in my calc :/ i get confused

ohhh darn :/ okay, let me try again :)

Answer 9

ohh so it would be 0.2296875? :/

Answer 10

Indeed....but I think just rounding to 0.23 is fine

Answer 11

ahh okay yay!! :) so that is the total kinetic energy for part 1? 0.23? also, i forgot, what are the units again? :/

Answer 12

Well here we have

kg * (m/s)^2 would be

kg m^2 / s^2 *kilogram- meters squared per seconds squared

Answer 13

And a hint in case you hate all that

1 joule ( J ) is equal to 1 kg m^2 / s^2

So just 0.23 J would be good

Answer 14

ohh okay! :) awesome!! i will try to remember that :P

so onto part 2? :O

Answer 15

Sure....so part 2 we have an INELASTIC collision...meaning the 2 objects stick together after they collide right?

Answer 16

yes! :)

Answer 17

So...we start off with the normal kinetic energy and also the original momentum

1/2 mv^2 and mv

then they collide (boom!) lol

now they are stuck together

Answer 18

ok:) haha boom, got it! :P

Answer 19

Thanks to the conservation of momentum...the momentum before and after the collision are the same...

\[\large m_1 v_1 = (m_1 + m_2) v_2 \]

why is it (m1 + m2) over on the right? because they are now stuck together...so the total mass would be the mass of both

good so far?

Answer 20

ahh okay:) yes, good :)

Answer 21

And the kinetic energy at the end is going to be basically the same

\[\large K.E = \frac{1}{2}(m_1 + m_2)v_2^2 \]

Since this is AFTER the collision...we use the final velocity (v2) and we still have the 2 masses because they are stuck together....still good?

Answer 22

yes:)

Answer 23

So we have 2 equations

The momentum: \(\large m_1 v_1 = (m_1 + m_2)v_2\)

and the kinetic energy \(\large \frac{1}{2}(m_1 + m_2)v_2 ^2\)

So what we need, is to know the final velocity....looking at the momentum equation...how would we solve for the v2?

Answer 24

isolate it?

Answer 25

Right...so we would get

\[\large v_2 = \frac{m_1 v_1}{m_1 + m_2}\]

right?

Answer 26

right :)

Answer 27

Now let's just plug that into our kinetic energy equation and we will finally be ready to solve

\[\large K.E = \frac{1}{2}(m_1 + m_2)(\frac{m_1v_1}{m_1 + m_2})^2\]

man that's a long one...but we have everything we need...we know both the masses...and we know the original velocity...so just plug everything in and we will have the answer

Answer 28

and yes I'll do it too to check :P

Answer 29

haha okie :)

so we will get this? hopefully i am plugging in correctly lol

KE=1/2(0.150+0.350)((0.150*1.75)/(0.150+0.350))^2 ?

and that equals 7686.905625 ? :/

ooh something looks weird :( not sure what i did weirdly and incorrectly though :/

Answer 30

Oh boy no lol

You did indeed plug everything in right though

\[\large \frac{1}{2}(0.150 + 0.350)(\frac{0.150\times 1.75}{0.150 + 0.350})^2\]
\[\large \frac{1}{2}(0.5)(\frac{0.2625}{0.5})^2\]
\[\large \frac{1}{2}(0.5)(0.275625)\]
\[\large 0.0689\]

Answer 31

oh yay at least i plugged in right! haha oops i was really off :(
i'm not very good with the calculator!! :(

so it would be 0.0689 J ? that is the final total kinetic energy for part b? :O

Answer 32

Correct :)

Answer 33

yay!! thank you so much! :)

Answer 34

Not a problem...and now...I must bid you farewell D: lol

Answer 35

:)