Question

Please help me... @Catch.me
An alpha particle travelling at a velocity u makes an oblique elastic collision with a helium atom moving with a velocity which negligible compared to u.After collision , the alpha particle moves in the direction inclined at 30 degree to its initial direction of motion.What is the angle between the directions of motion of the helium atom and the initial direction of motion of alpha particle?

Answer 1

Do you know dimension collision
then just follow this instructions here
http://www.physicsforums.com/showthread.php?t=415537

Answer 2

Not really understand dimension collision...

Answer 3

momentum is a vector
apply conservation of momentum for x direction
and y direction.

Answer 4

get their component and you will get the unknown theta.
x direction
mu = mv1cos 30 + Mv2cos theta .
y direction
0 =- Mv2sin theta +mv1sin 30 .

Answer 5

And the mass of alpha particle and helium is known

Answer 6

I try first...

Answer 7

I cant solve...

Answer 8

@Catch.me

Answer 9

Wait

Answer 10

oh uh....alpha particle and helium has the same mass right.

Answer 11

Then the angle angles after collision should be 90 degree
thus the angle is 60 degree.

Answer 12

@MichelleShum you don't know that right??

Answer 13

yes...i don't know...

Answer 14

@MichelleShum OK anyway the answer is 60
NOW we are going to prove that the are 90 degree.

Answer 15

k...

Answer 16

we got that
conservation of kinetic energy,
\[ m u = m V1 \cos 30 + mv2 \cos \theta.\]
and \[0 = m v1 \sin 30 - m v2 \sin \theta. \]
cancel masses in these equations and make mv2 in separated equation .
\[V2 \cos \theta= u - v1 \cos 30 . \]
\[V2 \sin \theta = v1 \cos 30.\]
square both equations then add them
sin ^2 + cos ^2 = 1
\[V _{2} ^{2} = u ^{2} - 2 u V _{1} \cos 30 + V _{1}^{2}.\]
From kinetic equation
\[0.5 u ^{2} = 0.5 V _{1}^{2} + 0.5 V _{2}^{2}\]
now use v2^2 to substitute in kinetic equation.
\[0 = 2 V _{1}^{2} - 2 V _{1} u \cos 30 \]
\[V _{1} = u \cos 30 . \]
\[u = \frac{ V _{1} }{ \cos 30 }\]Put that in your head we will use it later eq 1 .
Now we will divide these equations we got them previous.
\[V2 \sin \theta = v1 \cos 30.\]
\[V2 \cos \theta= u - v1 \cos 30 . \]
equal
\[\tan \theta = \frac{ V _{1} \sin 30 }{ u - V _{1 \cos30} }\]
use equation 1 instead of u
\[\tan \theta = \frac{ \sin 30 }{ \frac{ 1 }{ \cos30 } - \cos30} = \frac{ \sin 30 }{ \frac{ 1- \cos ^{2} 30 }{ \cos 30 }} = \frac{ \sin30 }{ \frac{ \sin ^{2}30 }{ \cos 30 }} = \cot 30 \]
Thus theta +30 = 90 degree
WOW it's done.

Answer 17

@MichelleShum Does it clear.
i had to overcome some steps as they aren't tricky.

Answer 18

ok...very clear...tq...

Answer 19

Please help me this also... @Catch.me
An aircraft is flying in a horizontal circle of radius 80km at a speed of 300m/s. What is the ratio of the lift on the aircraft to its weight?

Answer 20

oh no this problem took me a lot of time and i have to go out now, so ask another one :D
sorry.

Answer 21

ok...nevermine...anyway thanks for your help...

Answer 22

you are welcome.
What is your course??

Answer 23

I'm actually study Form 6...

Answer 24

what is form 6 ??

Answer 25

Something like A-level

Answer 26

OK thanks

@Mashy can you help here??

Answer 27

yes i can help ..

since the aircraft is taking the circle.. it must be banked right?