Kinetic energy and potential energy

[look416]

1. The problem statement, all variables and given/known data
http://i923.photobucket.com/albums/ad73/look416/IMG_3785.jpg
the diagram shows two bodies X and Y connected by a light cord passing over a light, free-running pulley. X starts from rest and moves on a smooth plane inclined at 30 degrees to the horizontal.
What will be the total kinetic energy of the system when X has travelled 2.0m along the plane?(g=9.8m/s2)


2. Relevant equations
E=mgh


3. The attempt at a solution
my teacher ask me to find the perpendicular distance of y and x, then use formula E=mgh to find the differences between potential energy between X and Y, and the differences will equal to the total kinetic energy since energy is conserved during the whole motion
BUT, i cant even find the perpendicular distance of X and Y




1. The problem statement, all variables and given/known data
A stationary thoron nucleus (A=220, Z=90) emits an alpha particle with kinetic energy Ea. What is the kinetic energy of the recoiling nucleus?


2. Relevant equations



3. The attempt at a solution
i totally have no idea about it.


1. The problem statement, all variables and given/known data
Particles X (of mass 4 units) and Y ( of mass 9 units) move directly towards each other, collide and then separate. If \lambdaVx is the change of the velocity of X and \lambdaVy is the change in velocity of Y, the magnitude of ratio \frac{\lambdaVx}{\lambdaVy}


2. Relevant equations



3. The attempt at a solution
well the answer state is \frac{9}{4}
but i get \frac{2}{3}, just by rationing both magnitude right?

[tiny-tim]

…i cant even find the perpendicular distance of X and Y …

Hi look416! :smile:

When Y moves down vertically 1 m, how much higher does X get? :wink:

[look416]

but in this case does we assume that Y is moving down 1m?
ops, i think i have found the height of y,
but i still couldnt find the perpendicular height of x

[look416]

just to push to thread up

[tiny-tim]

but in this case does we assume that Y is moving down 1m?
ops, i think i have found the height of y,
but i still couldnt find the perpendicular height of x

(just got up … :zzz:)

Do you mean "vertical"? (But height is vertical anyway)

The height of x is the length along the slope, multiplied by the sine of the angle, isn't it?

What's worrying you about that?

[look416]

(just got up … :zzz:)

Do you mean "vertical"? (But height is vertical anyway)

The height of x is the length along the slope, multiplied by the sine of the angle, isn't it?

What's worrying you about that?

lolz you just reminded me of that
btw it just afternoon for here
ok i will try it ltr as im in cc now

[look416]

tiny-tim
for your information i have just added another 2 questions,
so could you teach me how to do it?

[look416]

push push

[tiny-tim]

A stationary thoron nucleus (A=220, Z=90) emits an alpha particle with kinetic energy Ea. What is the kinetic energy of the recoiling nucleus?
…
3. The attempt at a solution
i totally have no idea about it.

(It would better to start a new thread)

An alpha particle has 4 nucleons (ie protons and neutrons, with the same mass), and the thoron nucleus has 220 nucleons, ie 55 times as massive.
Particles X (of mass 4 units) and Y ( of mass 9 units) move directly towards each other, collide and then separate. If \lambdaVx is the change of the velocity of X and \lambdaVy is the change in velocity of Y, the magnitude of ratio \frac{\lambdaVx}{\lambdaVy}
…
well the answer state is \frac{9}{4}
but i get \frac{2}{3}, just by rationing both magnitude right?

?? :confused: Show us your calculations.

[look416]

(It would better to start a new thread)

An alpha particle has 4 nucleons (ie protons and neutrons, with the same mass), and the thoron nucleus has 220 nucleons, ie 55 times as massive.



Still cant get it very well,
does that mean to calculate the kinetic energy of recoiling nucleus?
Btw, could you explain what is recoiling nucleus?
which is the one it refers to?
the thoron nucleus or the alpha particle?


For another question, here is how i did it
by here i assume that after they collide particle X and particle Y move in the direction that oppose the direction before they collide (elastic collision) and considering the direction to the right is positive.
therefore for paricle X
mxu1-mxv1=0

and for particle y
my-u2+myv2=0

after that i extracted it out
mx(u1-v1)

my(-u2+v2)

the ratio between velocity of X and Y
= \frac{{mx(u1-v1)}}{{my(-u2+v2}}

but they want the magnitude of ratio of change in speed so by putting the value of mass X and Y into the equation and pull it out
\frac{4}{9}
then simplify it then i get \frac{2}{3}

[tiny-tim]

Hi look416! :smile:

(in LaTeX, mx is m_x, and mx1 is m_{x1} :wink:)
Still cant get it very well,
does that mean to calculate the kinetic energy of recoiling nucleus?
Btw, could you explain what is recoiling nucleus?
which is the one it refers to?
the thoron nucleus or the alpha particle?

the thoron nucleus … it ejects part of itself, so it recoils, just like a gun recoils when it ejects a bullet :wink:
… (elastic collision) …

You're only using conservation of momentum, and that applies to all collisions …

you don't need it to be elastic

(elastic is only necessary for conservation of energy)
the ratio between velocity of X and Y
= \frac{{mx(u1-v1)}}{{my(-u2+v2}}

but they want the magnitude of ratio of change in speed so by putting the value of mass X and Y into the equation and pull it out
\frac{4}{9}
then simplify it then i get \frac{2}{3}

I don't understand … how exactly did you get 2/3 ? :confused:

[look416]

Hi look416! :smile:

(in LaTeX, mx is m_x, and mx1 is m_{x1} :wink:)


the thoron nucleus … it ejects part of itself, so it recoils, just like a gun recoils when it ejects a bullet :wink:


You're only using conservation of momentum, and that applies to all collisions …

you don't need it to be elastic

(elastic is only necessary for conservation of energy)


I don't understand … how exactly did you get 2/3 ? :confused:

well i just take the \frac {m_{x}}{m_{y}} out and simplify it
thats all

[tiny-tim]

well i just take the \frac {mx}{my} out and simplify it
thats all

take it out of what? simplify how? :confused:

[look416]

take it out of what? simplify how? :confused:

\frac {m_x}{m_y} is the coefficient of the ratio change between Vxand vy, since mx values 4 units and my values 9 units then it turns out to be \frac {4}{9}, so i just simplify it

[look416]

about the second question
i still cant figure it out sorry for im stupid:tongue2:
which equation are we using
kinetic equation which resembles
\frac{1}{2}mu2+\frac{1}{2}mu12=\frac{1}{2}mv2+\fra c{1}{2}mv12

[tiny-tim]

\frac {m_x}{m_y} is the coefficient of the ratio change between Vxand vy, since mx values 4 units and my values 9 units then it turns out to be \frac {4}{9}, so i just simplify it

oh I see … but 6/9 is 2/3.

get some sleep! :zzz:
about the second question
i still cant figure it out sorry for im stupid:tongue2:
which equation are we using
kinetic equation …

(btw, the element with Z = 90 is thorium … thoron was an old name for one of the isotopes of radon :wink:)

Although energy is conserved, you don't know the exact mass of the decay product (A = 216, Z = 88), and a very small error in the m makes a big difference in mc2, which is what you need in the conservation of energy equation.

So instead, use conservation of momentum (which of course applies to all collisions), to find the momentum, and then use m = 216 to find the kinetic energy (the small error in m won't make much difference here).

[look416]

oh I see … but 6/9 is 2/3.

get some sleep! :zzz:


(btw, the element with Z = 90 is thorium … thoron was an old name for one of the isotopes of radon :wink:)

Although energy is conserved, you don't know the exact mass of the decay product (A = 216, Z = 88), and a very small error in the m makes a big difference in mc2, which is what you need in the conservation of energy equation.

So instead, use conservation of momentum (which of course applies to all collisions), to find the momentum, and then use m = 216 to find the kinetic energy (the small error in m won't make much difference here).

lolz its pretty obvious im lack of sleep haha
for the question regarding thoron, thx ya for your little info
and i have got the answer hurray

what abt the question regarding the two particles hoho?
i am still circling inside it

[tiny-tim]

what abt the question regarding the two particles hoho?
i am still circling inside it

I suspect you're keeping the equation in your head instead of bothering to write it out properly:

if ∆x and ∆y are the changes in the velocities of x and y,

then mx∆x = -my∆y,

so |mx/my| = … ? :smile:

[look416]

oic now i got it
yea
so happy to find out the answer