Kinetic energy and potential energy

1. The problem statement, all variables and given/known data

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 E_a. 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 [tex]\lambda[/tex]V_x is the change of the velocity of X and [tex]\lambda[/tex]V_y is the change in velocity of Y, the magnitude of ratio [tex]\frac{\lambdaV_x}{\lambdaV_y}[/tex]


2. Relevant equations



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

 

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 to push to thread up

 

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

 

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

 

push push

 

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
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
= [tex]\frac{{m_x(u₁-v₁)}}{{m_y(-u₂+v₂}}[/tex]

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
[tex]\frac{4}{9}[/tex]
then simplify it then i get [tex]\frac{2}{3}[/tex]

 

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

 

[tex]\frac {m_x}{m_y}[/tex] 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 [tex]\frac {4}{9}[/tex], so i just simplify it

 

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

 

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

 

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