Kinetic energy and potential energy

[look416]

Homework Statement

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 traveled 2.0m along the plane?(g=9.8m/s²)

Homework Equations

E=mgh

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 can't even find the perpendicular distance of X and Y

Homework Statement

A stationary thoron nucleus (A=220, Z=90) emits an alpha particle with kinetic energy E_a. What is the kinetic energy of the recoiling nucleus?

Homework Equations

The Attempt at a Solution

i totally have no idea about it.

Homework Statement

Particles X (of mass 4 units) and Y ( of mass 9 units) move directly towards each other, collide and then separate. If $$\lambda$$V_x is the change of the velocity of X and $$\lambda$$V_y is the change in velocity of Y, the magnitude of ratio $$\frac{\lambdaV_x}{\lambdaV_y}$$

Homework Equations

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 can't even find the perpendicular distance of X and Y …

Hi look416!

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

 

[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 couldn't 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 couldn't 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 I am 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 E_a. What is the kinetic energy of the recoiling nucleus?
…

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 $$\lambda$$V_x is the change of the velocity of X and $$\lambda$$V_y is the change in velocity of Y, the magnitude of ratio $$\frac{\lambdaV_x}{\lambdaV_y}$$
…
well the answer state is $$\frac{9}{4}$$
but i get $$\frac{2}{3}$$, just by rationing both magnitude right?

?? 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 can't 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
m_xu₁-m_xv₁=0

and for particle y
m_y-u₂+m_yv₂=0

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

m_y(-u₂+v₂)

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

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!

(in LaTeX, m_x is m_x, and m_x1 is m_{x1} )

Still can't 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

… (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
= $$\frac{{m_x(u₁-v₁)}}{{m_y(-u₂+v₂}}$$

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 ?

 

[look416]

Hi look416!

(in LaTeX, m_x is m_x, and m_x1 is m_{x1} )the thoron nucleus … it ejects part of itself, so it recoils, just like a gun recoils when it ejects a bullet 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 ?

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?

 

[look416]

take it out of what? simplify how?

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

 

[look416]

about the second question
i still can't figure it out sorry for I am stupid
which equation are we using
kinetic equation which resembles
$$\frac{1}{2}$$mu²+$$\frac{1}{2}$$mu₁²=$$\frac{1}{2}$$mv²+$$\frac{1}{2}$$mv₁²

 

[tiny-tim]

$$\frac {m_x}{m_y}$$ is the coefficient of the ratio change between V_xand v_y, since m_x values 4 units and m_y 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 can't figure it out sorry for I am stupid
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 )

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 )

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 I am lack of sleep haha
for the question regarding thoron, thanks you 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 m_x∆_x = -m_y∆_y,

so |m_x/m_y| = … ?

 

[look416]

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