Question

Can someone please check my work?

Answer 1

page 2:

Answer 2

page 3:

Answer 3

page 4:

Answer 4

@Sam

Answer 5

i don't think 5b is correct. Can you please check my work?

Answer 6

Only one spring right? Hold on

Answer 7

yeah there is only one spring and we assume the carts are moving to the right when colliding with one another. So right is positive x-direction

Answer 8

We can say that the spring is at maximum compression when the speeds of both blocks are equal

Answer 9

I am not sure if you've done your work correctly, I haven't checked

Answer 10

can you please check my work? :)

Answer 11

the velocity of both carts are given. The momentum of cart A is bigger than cart B. so i think the net velocity of both carts should be right (positive). One thing to note is that the moment both carts collide the spring is compressed and as cart A moves cart B to the right because again due to bigger momentum, the spring is still compressed as they move to the right/east.

Answer 12

Yes, what 5b means when the 1.2kg cart "after" the collision? Is it when the carts touch each other or when the second block changes its direction?

Answer 13

i think they want to know the velocity of cart B after the collision.

Answer 14

Yes that can mean anything

Answer 15

or maybe they want to know the velocity of cart B right after the kinetic energy transfers from cart A to cart B.

Answer 16

So you're stating after the collision cart B is still moving west before it stops and changes direction

Answer 17

i'm saying that cart A is heading east and cart B is heading west for a head-on collision. Once they collide, cart A and Cart B move to the right/east together but...since the momentum of cart A is greater than cart B. After the collision...cart A will stop cart B and change cart b's direction from west to east minus the energy wasted in the collision.

Answer 18

Right, I found PE= 3.09J when the spring is at its maximum collision, close to yours.

Answer 19

Give me a min

Answer 20

yes sir :)

Answer 21

Almost done

Answer 22

i dont understand what you fdid on page 3 :\

Answer 23

I gotta type everything again, ugh

Answer 24

Wait

Answer 25

so after the collision, the conservation of energy and conservation of momentum stays the same. so i took vf1 and energy equation plugged it into the second equation and solved for vfB

Answer 26

so from equation 1, i isolated the "v1i" and solve for that and then i took that value and plugged it into the second equation and solved for "v2f"

Answer 27

Hmm there's something wrong with mathjax

Answer 28

yeah i can't copy previous drawing either... :/

Answer 29

yea^

Answer 30

*i'm confused because everything is working fine for me*

Answer 31

ok ...thank gawd!! :)

Answer 32

OK so, the total energy before the collision is given by \(\sum E= E_1 + E_2 + E_S\) but since the spring isn’t compressed, \(\sum E= E_1 + E_2 + \cancel{E_S}\).
\[\sum E_{TOT_1}= E_1 + E_2 + E_S \\ \\ \sum E_{TOT_1}= \frac{1}{2}m_1v_{1i}^2+ \frac{1}{2}m_2v_{2i}^2 \\ \\ \sum E_{TOT_1}= 34.2J \]
Now, when the spring is compressed there is potential energy.
\[\sum E_{TOT_2}= \frac{1}{2}m_1v_{1}^2+ \frac{1}{2}m_2v_{2}^2 + PE \\ \\ \sum E_{TOT_2}= 31.11 +PE \]
Because energy is conserved, both the total energies before and after the spring is compressed are equal. So
\[\sum E_{TOT_1}= E_1 = \sum E_{TOT_2} \]
So \(34.2=31.11+PE, PE=3.09J\)
The total energy after collision is
\[ \sum E_{TOT_3}= E_{1f}+E_{2f} + \cancel{E_Sf} \\ \\ =\frac{1}{2}m_1 v_{1f}^2+ \frac{1}{2} m_2 v_{2f}^2\]
We know that \(\sum E_{TOT_1}= \sum E_{TOT_2} = \sum E_{TOT_3}\).
So we'll have simultaneous equations with both unknowns
\[34.2= \frac{1}{2}(1.6) v_{1f}^2+ \frac{1}{2}(1.2)v_{2f}^2 ...(1) \\ \\ 31.11= \frac{1}{2}(1.6)v_{1f}^2+ \frac{1}{2}(1.2)v_{2f}^2 ...(2)\]

Answer 33

so from here i jst have to solve for v_2^2f? on both equations and thats it?

Answer 34

Yeah

Answer 35

I got 5.816m/s for v_2f and 4.6485m/s for v_1f

Answer 36

btw can i use those two equations that i was using in the pictures?

Answer 37

You can, you got pretty close to what I've got for total energy during collision

Answer 38

ok Thanks papi. I was struggling on this problem for days.,.... thanks you are a hero for me <3 god bless!! :)

Answer 39

Hahah I've spent an hour on this as well, god with that server hiccup I had to type again xD, good luck anyways

Answer 40

God bless india and its people. you are a lifesaver!! :)

Answer 41

every effort was appreciated :)

Answer 42

I'm not from india though, but anyways yw

Answer 43

and thanks random i love you both to pieces. <3

Answer 44

oh my bad ^~^

Answer 45

sorry i jus lost my internet connection for a bit

Answer 46

it's ok hunny!! :) thanks

Answer 47

haha thanks :)

Answer 48

and @Sam thats some nice work m8

Answer 49

Lots of love :)

Answer 50

Yeah tis annoying question anyway xD I hate doing momentum problems

Answer 51

XD

Answer 52

lawl, these were some of my fav questions back in HS

Answer 53

i hated questions related to Optics :\

Answer 54

gawd XD really? XD

Answer 55

ye

Answer 56

Thsi is how i solve it btw

Answer 57

but sam's method looks more efficient

Answer 58

Optics isn't my favourite either, I love Newton's law problems, Conservation of energy, fluid mechanics, E&M and Quantum mechanics

Answer 59

You've got different answers than mine @random231

Answer 60

9.793 and 1.94

Answer 61

lol....i love newton's law problems too... XD but i think the conservation of energy and momentum problems are starting to grow me on so mehh... im getting better

Answer 62

what?!?!?! so who's right? XD

Answer 63

Conservation of energy problems are easy, \(E_i =E_f\) always

Answer 64

Damn i might have made some calculation error

Answer 65

i dont have a calculator w me :\

Answer 66

Mine, I did the safest and the most secure way possible

Answer 67

^

Answer 68

ok i guess i'll try both ways and see if your answers match.... thanks guys..

Answer 69

just sold a chemistry exam to someone online XD