Solve Elastic Collision Homework: Neutron Mass m0

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Homework Statement


A neutron of mass m0 collides with an immobile atomic nucleus of carbon of mass m=12m0. Collision is considered central and elastic. how many times does the kinetic energy of neutron decreases during the collision.


Homework Equations


m=12m0

m0v0=mu-m0u

(m0v02)/2=(mu2)/2-(m0u02)/2

The Attempt at a Solution


v0= (mu-m0u0)/m0=
=(12m0u-m0u0)/m0=
=12u-u0

u=(v0+u0)/12
u2=(v02+2u0v0+u02)/144


v02=12u2-u02

u2=(v02+u02)/12

(v02+2u0v0+u02)/144=(v02+u02)/12

0=11u02-2u0v0+11v02)

I'm stuck here. I know tht I hv to find x=(u02)/(1v02) because (m0u02)/2=x(m0v02)/2
u02=xv02

(the answer is 1.4 times)
 
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Take this equation
v0 =12u+u0 and square it to get
v0^2 = 144u^2 + 24u0u + u0^2
You also have this equation
v0^2=12u^2 + u0^2
So equate those and that leads to
132 u = - 24u0

Substitute this into the first equation I wrote, to see the relative change in velocity, then square it for change in kinetic energy.
 
but m>m0. Shouldnt u0 be negative in both equ?
 
But it can be negative, you could solve these equations and get u0<0 which would indicate that it is traveling in the other direction.
 
In this equation tho you stated incorrectly.
(m0v02)/2=(mu2)/2-(m0u02)/2

Since velocity is squared kinetic energy is allows positive. So Kinetic energy before must equal sum of kinetic energies after the collision.
(m0v02)/2=(mu2)/2+(m0u02)/2
 
I agree with Mr.A.Gibson in all the above posts.
 
yea2, ur right! got it, thank u very much!