Question

Can someone help me with this Physics problem?! (The kinetic energy is 1.1x10^9) Im stuck on how to actually find the answer! Is that what you're supposed to do, and if yes then how do I put it into the calculator?

Answer 1

you are heading on the right lines, but you should cancel the 1/2 and the mass from the right hand side as you have moved them across. Putting them into your calculator correctly takes practise and you must understand some basic algebraic rules. from where you are, divide the KE by 0.5. Divide THAT answer by the mass. then you must square root that answer.
Alternatively, rearrange the KE formula for the thing you want:\[v = \sqrt{\frac{ 2 KE }{ m }}\] With the new calculators, you can probably insert the numbers into this arrangement straight away. post your answer if you want it checked

Answer 2

1.1 x 10^9 =1/2 . (8 x 10^4) v^2
v^2 = 1.1 x 10^9/(40000)
v^2 = 27500
v = sqrt(27500)

Answer 3

@furnessj but it doesnt give us the mass

Answer 4

@gerryliyana thank you!

Answer 5

@furnessj thank you to you too!

Answer 6

ur welcome :)

Answer 7

The mass is 8x10^4 - make sure you know your units! it has kg after it so must be the mass. Like i have told so many pupils, learn the units, especially as exams often give you the equations, you just have to know which numbers are which things.

Answer 8

just kidding it does @furnessj

Answer 9

thank you, sorry i was looking at another problem!

Answer 10

no problem :)

Answer 11

are you a physics teacher? cause i have another question!

Answer 12

yup, go ahead

Answer 13

p.s. - you can tell i'm a physics teacher, I have a bucket on my head in my picture...! (we are all very normal :-/ )

Answer 14

okay so we're learning about energy so im confused on what formulas to use when! and how to use algebra to change up the formula to meet the correct equation!

Answer 15

haha true that!

Answer 16

Here's an example. I have NO idea how to start the problem! I don't understand when to use the formulas and whenever i use algebra to change the formulas something always go wrong.

Answer 17

Right, with rearranging formulas, you have to learn the technique. some people find it easier, but you'll get a eureka moment at some point. At your level, the thing to start doing is moving letters from one side of the equals to the other, and making what is does opposite. You are already doing that a bit. For example, if there is a ' v= d /t' you can move the t across and make it a times. leaving v x t = d.

you'll only need these opposites:
multiply / divide
positive / negative
square / square root.

Finally, for which equation, look at what the question gives you. Is is moving? If so its kinetic energy. Is there a height, then it is gravitational potential energy. those are the only two i think you'll need. are there others?
And remember - all energies are measured in Joules, whatever type it is - you have to learn the units of things!

Answer 18

That one is many times harder than the previous one! First of all - which kind of energy is the egg starting with..? (read my previous answer and the question!)

Answer 19

there are other equations such as the m1v1 + m2v2 = m1v1 + m2v2
and m1v1 + m2v2 = m3v3
i dont know when to use which of these

Answer 20

btw thank you so much for your help, i really appreciate it.

Answer 21

it is starting with kinetic because it is dropped, right?

Answer 22

and then potential because it lands.

Answer 23

so i would use 1/2mv^2 ?

Answer 24

idk what it is asking for though?

Answer 25

m1v1 + m2v2 = m1v1 + m2v2
is momentum conservation law

you must know what are they condition, it is elastic? inelastic? or??

for elastic e =1, for inelastic e =0

where e = -(v1'-v2')/(v1-v2)

and

m1v1 + m2v2 = m3v3

if e = 0, so m3 = m1+m2,

Answer 26

how do you know if it is elastic or inelastic?

Answer 27

right! the m1v1... is momentum, you will be given velocities (m/s) and masses (g or kg) in those questions. so it is not that. Here, it starts still, at a height. So it is not kinetic into potential, it is the opposite!
\[mgh = \frac{ 1 }{ 2 } m v ^{2}\] is what you need. then this gets tough if you don't like rearranging. [are you in uk or us? - the difference in your first question and this one is a whole one,maybe two years of learning)

Answer 28

(us) and why do we equal mgh to KE?

Answer 29

if the question say "perfect bouncing" or "perfect collision" so that's elastic, but if in the question say " after collision, they are moving in together" so that's inelastic..,

Answer 30

because the potential energy TURNS INTO kinetic energy at the bottom of the fall (assuming no loss of energy through air resistance)

Answer 31

because base on the law of conservation of energy

EP1 + EP2 = EK1 +EK2

Answer 32

so what if it was turning kinetic to potential?

Answer 33

well, that is what the question is telling you to ignore in the parentheses at the bottom. this would happen if you threw a ball upwards. it STARTS with speed (kinetic) then gains height (potential). so then it is the other way around, but you would still be able to make them equal to each other (conservation of energy)

Answer 34

okay, so are we solving for the height?

Answer 35

yes of course.,

hope this helpful for u, check this out!
http://www.physicsclassroom.com/Class/momentum/u4l2b.cfm

Answer 36

This is not a momentum question! You are solving for v, natasha (then we are halfway...!)

Answer 37

m1v1 + m2v2 = m1v1' + m2v2'

is The Law of Momentum Conservation

Answer 38

how do we know that we are solving for v? (im sorry that im so bad at this!)

Answer 39

v = sqrt(2 g h)

Answer 40

well, that is the stupidly hard bit. hopefully after you will see. it is the only thing we can find with what we are given and the physics that is happening. we will then use it in ANOTHER equation.
p.s gerryliyana has put the correct formula for getting v for you

Answer 41

mgh = 1/2 mv^2
gh = 1/2 v^2
v^2 = 2gh
v= sqrt(2gh)

Answer 42

okay so whenever it is asking for how much it is compressed by, it is asking for v?

Answer 43

yay!! i think i get the algebra part!!! :)

Answer 44

congrats! @natasha.aries

Answer 45

what unit would you measure compression in?

Answer 46

thanks! and what do you mean @furnessj ?

Answer 47

is the answer 15. 344? then we dont do anything to its mass?

Answer 48

yes it dont do anything to its mass

the unit in m/s

Answer 49

that is correct for v, well done! I mean, if you want to measure how much something is compressed, what units would you measure it in? (what would you use to measure it?)
p.s. by the beauty of physics, we didn't need the mass

Answer 50

mgh = 1.2 mv^2 (divded all by m)

so, We have been eliminating m

nice

Answer 51

thanks! and grams? idk :/

Answer 52

doesnt it have to do with mass?

Answer 53

yes its doesn't!

Answer 54

just kidding so meters?

Answer 55

m/s, or meters per seconds

Answer 56

oh okay, and if you are solving for height would be 1/2v^2/g ?

Answer 57

yes of course!

Answer 58

or 0.5v^2/g

Answer 59

compression is a distance! so we need two last (ugly) formulas:
First:
acceleration = change in velocity / time

The velocity we just worked out IS the change, as it goes from that to zero. It takes 6.25 x10^-3 seconds so you can divide by that to get the acceleration.

Finally, \[v ^{2} = u ^{2} + 2 a s \]
s is the compression distance, u is the starting speed (v from before) and v = 0 now, so it is gone from the formula.
then:
-u^2 = 2as
-u^2/2a = s
try that...?!

Answer 60

okay!! cool! but on another questions i have this equation but they dpnt give a velocity

Answer 61

great for u @natasha.aries

Answer 62

That question is an AP question really - is that what you are taking?

Answer 63

yes, and im confused on why we are doing all next?

Answer 64

because the question asks for the compression, not the velocity! It is a really long problem solving question, requiring several steps and lots of techniques and foresight - as I said, its only just below college level.

Answer 65

okay so to answer this question i would find the change in velocity, then the acceleration, and speed?

Answer 66

yes, use the velocity to determine the acceleration (notice it gives you a time it takes to stop, and mentions deceleration), then the long formula of motion i put up to use the things you know to get the distance compressed. (you know start speed, finish speed, and the deceleration, and can work out distance).

Answer 67

so would u be 0 and how can i figure out speed?

Answer 68

Ok, at this point, forget what we have done so far with the falling part, EXCEPT, that the egg is currently going at 15.3m/s as it starts to hit the pad. from now, this is the starting speed. It will then decelerate as the pad pushes upward on it, until it reaches a speed of zero. Using this change in velocity, and the time it took, we can work out the (de)-acceleration. Finally, we put
v = 0
u = 15.3
a = .... (the velocity difference divided by the time to do it)
then solve for s.

Answer 69

for the acceleration, i got 2455 :/

Answer 70

ohh nevermind! 4.073x10^-4

Answer 71

No, you were right!

Answer 72

oh so would i just make that into a negative since it was a positive number?

Answer 73

technically yes, because we have been treating down as positive.

Answer 74

so when i work it all i got 274.48 for the compression

Answer 75

but the answer says its .048 m

Answer 76

not quite, you should get 0.048 (which is metres). I.e. 4.8cm - it doesn't use up the whole 5cm (why it didn't break!)

Answer 77

look upwards, i explained this last tiny bit:
-u^2 = 2as
so square u first. then divide by 2. then divide by 2455. What you get?

Answer 78

i did (15.54)(2)(2455) and then i square rooted the answer

Answer 79

not quite. 15.54 squared first. then divide by 2 and again by 2455..

Answer 80

ohh i see!! in the equation above it was split up!

Answer 81

yes i think i did one step at a time

Answer 82

have you got it yet?! :)

Answer 83

i did but only when you told me to do this and that first, in -u^2 = 2as wouldnt you square root 15.54 and then to get s by itself, divide that by 2a ?

Answer 84

it just says to square the u, i will write it as a proper equation below. then with your new found algebra skills you are correct, divide that by 2a!
\[\frac{u ^{2}} { 2a} = s \]

Answer 85

but when i do that, i get 296430!

Answer 86

do it one step at a time.
(15.34)x(15.34) = ANS
ANS / (2 x a) = s

Answer 87

oh i got it, thank you soooo much!!!!

Answer 88

(i also feel i should mention - because you did part of it anyway, that the acceleration is negative, and so should the u^2 value. but then the negatives on each side cancel)

Answer 89

okay, thank you!

Answer 90

Fantastic! Well done @natasha.aries , as a teacher I enjoy helping people, its great when someone puts in some hard work to understand it. I'll be on this site quite a bit so see you around, don't be afraid to ask stuff again.

Answer 91

THANK YOU SO MUCH! I will! Do you think you will be on tomorrow? I have 2 more questions which I attempted but didnt get the correct answer :( But I don't want to bother you anymore than I already have today!

Answer 92

Yes I should be. And no problem, overdoing it is no good anyway, plus its late here now!

Answer 93

Okay! :)