Calculating Kinetic Energy Loss in Neutron Collision

AI Thread Summary
To calculate the fraction of kinetic energy lost by a neutron in a head-on elastic collision with a target particle, apply the principles of conservation of momentum and the work-energy theorem. Start by deriving a relationship between the velocities of the neutron and the target particle after the collision. The equation 0.5MaV^2 + 0.5MbV^2 = 0.5Ma'V^2 + 0.5Mb'V^2 + energy loss can be utilized to find the energy loss. Using specific examples like hydrogen (1.01u) and heavy hydrogen (2.01u) can help illustrate the calculations. This approach effectively determines the kinetic energy loss in neutron collisions.
botab
Messages
7
Reaction score
0
I need help with this problem where we are suppose to determine the fraction of kinetic energy lost by a neutron m=1.01u when it collides head-on and elastically with a target particle at rest which is.

examples used..
Hydrogen = 1.01u
heavy hydrogen = 2.01u

How would you do this?
Do we use .5MaV^2 +.5MbV^2 = .5Ma'V^2 + .5Mb'V^2 + energy loss?


examples used..
Hydrogen = 1.01u
heavy hydrogen = 2.01u

Thanks again,
 
Physics news on Phys.org
hello,
such questions can be solved using Conservation Of Momentum,and Work Energy theorem(that you used)

1st derive a relation between velocities of masses(call v_1\ and\ v_2)

then use the equation you are using


thanks
 
Last edited:
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top