Question

A 1.6 Kg cart moving at 6.0 m/s [E] collides with a 1.2 Kg cart moving at 3.0 m/s [W]. The head-on collision is completely elastic and is cushioned by a spring.

a) Find the velocity of the two carts when the spring is at maximum compression. (i've already done this part)

b) Find the velocity of the 1.2 Kg cart after the collision. (i need help with this question).

Answer 1

This is how i did part A. let's hope my pen works...uhg

Answer 2

@random231

Answer 3

How do i do the second question? what formula do i use plz?

Answer 4

it's in the formula sheet idk tbh

Answer 5

is it 2m? i cant understand what it is. :P

Answer 6

yeah it's 2m

Answer 7

mhmm idk where to use that formula.

Answer 8

well what do you think how should we approach this problem?

Answer 9

Anyways what i understand from the question is: after the collision takes place the spring is compressed to the maximum. then it expands again fully. We need to find the velocity when the spring has fully expanded. what do you think?

Answer 10

this is what we have for variables..

Answer 11

no if you look at the question again. the momentum of cart A with a mass of 1.6kg is more than cart B with a mass of 1.2 kg. When they both collide with eachother head-on. the net velocity will be due west. after the collision.

Answer 12

.

Answer 13

east.

Answer 14

i dont

Answer 15

^right?

Answer 16

the arrow sign

Answer 17

what about it? :D

Answer 18

look what happens is: the two bodies collide--->spring compresses fully and stores potential energy-->both bodies start moving in same direction with same velocity-->then spring relaxes releasing the potential energy-->changes to kinetic energy changing the velocities of the bodies

Answer 19

^this is what i think. otherwise the question doesnt make sense.

Answer 20

ok :)

Answer 21

the n the answer should be 4.7 for both a and b

Answer 22

what do you think?

Answer 23

i do agree with your response and i think it does sound logical but for part be if the spring is compressed, that means there is some stored energy in the spring that will work against the momentum of the cart after the collision. But lets just focus on the main part of question b, which is the velocity of cart 1.6kg. i think...

Answer 24

B*

Answer 25

i think the stored energy in the sprign will slow down the cart A with 1.6 kg moving east. what do u say?

Answer 26

after the collision.....that is.^^^

Answer 27

well we could just solve the problem and find out ourselves. :)

Answer 28

so lets start from when the spring is fully compressed.

Answer 29

sounds awesome.

Answer 30

so before proceeding further we need to find how much energy the spring holds

Answer 31

yeah. :/

Answer 32

W = change in energy.

Answer 33

so can you find it? use energy conservation

Answer 34

let me draww the equation...

Answer 35

is this it?

Answer 36

wait cant we use the following equation?

Answer 37

no no this is momentum conservation. theres no way to find potential energy stored in spring from this

Answer 38

plug in the values and find PE(spring)

Answer 39

omg i have that equation written above the conservation of momentum list...uhg im hopeless :/ one sec

Answer 40

i'll get back to you once i solve. :) <3

Answer 41

or should i just leave that for now?

Answer 42

just keep that as PE and find its value.

Answer 43

yes sir

Answer 44

im geting my answer in meters per second

Answer 45

let me send you the full solution.

Answer 46

shet i cant view it. its not loading

Answer 47

try again plz :)

Answer 48

yea now i can

Answer 49

swee

Answer 50

you made a couple of mistakes

Answer 51

btw the value wont be 6.85. you forgot to square 4.7 on the right side

Answer 52

how can the unit on the left side be joules and the right side be kg m/s ??

Answer 53

because i forgot to square on the right side... sorry let me try again. :)

Answer 54

i get 3.27 Joules of work.

Answer 55

ok

Answer 56

how does that sound? logical?

Answer 57

ig

Answer 58

yeah i mean its only 3.3 joules. on 1.6 kg cart...yeah i think its fine

Answer 59

so is that it?

Answer 60

nope, wait 3.3J is the spring PE right?

Answer 61

yeah i mean.....

Answer 62

its realistic

Answer 63

yea ofc :D

Answer 64

so now you have the PE of the spring. we'll start from here

Answer 65

yeah it's the PE of spring cuz thats what we were solving for. there's else except the PE of spring

Answer 66

ok :)

Answer 67

btw this question is only worth 3 marks.... isn't it getting too long for just 3 marks?

Answer 68

now here the momentum and energy remains the same throughout the process of spring relaxation

Answer 69

well tbh its probably because im explaining it in detail. this sum cud be done in 5 lines

Answer 70

wow lol

Answer 71

plug in the values in eqn 1

Answer 72

so you have the value of vf1 in terms of vf2 and a constant

Answer 73

now plug this in place of vf1 in eq2

Answer 74

now you just have 1 unnown value in the equation

Answer 75

ok

Answer 76

its a quadratic equation. i hpe you do know how to solve it?

Answer 77

yup i do :)

Answer 78

ok i'll get back to you once i'm done solving it :)

Answer 79

awesome. solve for vf2

Answer 80

:D

Answer 81

heres some ice-cream for being such a good student :P

Answer 82

wow i was never taught this in my course. so why am i being tested on it?

Answer 83

nom nom nom nom :D

Answer 84

anyway i'll figure it out.

Answer 85

:D yeah youll figure it out. youre smart. just dont make silly mistakes :P

Answer 86

i'll try my best XD