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Sagot :
Answer:
Part A
a) F = -16x + 4, b) x = 0.25 m, c) STABLE
Explanation:
Part A
a) Potential energy and force are related
F = [tex]- \frac{dU}{dx}[/tex]- dU / dx
F = - (8 2x -4)
F = -16x + 4
b) The object is in equilibrium when the forces are zero
0 = -16x + 4
x = 4/16
x = 0.25 m
c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.
In this case there is only one equilibrium point
by changing the position a bit
x ’= x + Δx
we substitute
F ’= - 16 x’ + 4
F ’= - 16 (x + Δx) + 4
F ’= (-16x +4) - 16 Δx
at equilibrium position F = 0
F ’= 0 - 16 Δx
we can see that the body returns to the equilibrium position, therefore it is STABLE
PART B
This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved
initial instant. Before the shock
p₀ = m v
final instant. After the crash
p_f = (m + M) v_f
We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic
p₀ = pf
mv = mv ’+ M v_f
in the case of the elastic collision, the kinetic energy is conserved
K₀ = K_f
½ m v² = ½ m v’² + ½ M v_f²
we write the system of equations
mv = mv ’+ M v_f (1)
m (v² -v'²) = M v_f ²
m (v - v ’) = M v_f
m (v-v ’) (v + v’) = M v_f
v + v ’= v_f
we substitute in equation 1 and solve
v ’=[tex]\frac{m -M }{m+M } \ vo[/tex]
v_f = [tex]\frac{2m}{m+M} \ v_o[/tex]
the mechanical energy of the neutron is
initial
Em₀ = K = ½ m v²
final moment
Em_f = K + U = ½ m v_f ² + U
U is the energy lost in the collision
total energy is conserved
Em₀ = Em_f
½ m v² = ½ m v_f ² + U
U = ½ m (v² -v_f ²)
U = ½ m [v² - ( [tex]\frac{m-M}{m+M}[/tex] v)² ]
U = ½ m v² [1- ( [tex]\frac{m-M}{m+M}[/tex] )² ]
U = ½ m v2 [ [tex]\frac{2M}{m+M}[/tex]]
U = [tex]\frac{2 mM}{m +M } \ v^2[/tex]
Let's do the same calculations for the nucleus
initial Em₀ = 0
final Em_f = K + U = ½ M v_f ² + U
Em₀ = Em_f
0 = K + U
U = -K
U = - ½ M v_f ²
U = - ½ M [ [tex]\frac{2m}{m+M} \ v[/tex] ]²
U = [tex]\frac{2 m M }{m+M} \ v^2[/tex]
We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.
b) the fraction of energy lost
f = U / Em₀
f = 4 m M / m + M
c) let's calculate the fraction of energy lost in a collision
m = 1.67 10⁻²⁷ kg
M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg
f = 4 1.6 20 / (1.6+ 20) 10⁻²⁷
f = 5.92 10⁻²⁷ J
the energy of a fast neutron is greater than 1 eV
Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J
Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo
#_collisions = 0.95 Eo / f
#_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷
#_collisions = 2.7 10⁷ collisions
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