EJERCICIOS SOBRE RELATIVIDAD

EJERCICIOS SOBRE RELATIVIDAD
PRINCIPIO DE RELATIVIDAD GALILEANA
1.-
En el marco de referencia en reposo
Pi= 2000 kg (20 m/s) +1500 kg (0) = 41500 kg m/s
PF= (2000 +1500) Kg Vf
Pi = Pf
Vf= 11,429 m/s
En el marco de referencia en movimiento (V´= +10 m/s)
V´a i = Vai – V´= 20 – (+10) = 10 m/s
V´bi = Vbi – V´ = 0 – (+10) = -10 m/s
V´f = Vf – V´= 11,429 – (+10) = 1,429 m/s
P´i = 2000 kg (10m/s) + 1500 kg (-10 m/s) = ´5000 kg m/s
P´f = ( 2000 +1500) kg (1.429) = 5001 kg m/s
2.-
A) 20
40
B)
V = 40 m/s + 20 m/s = 60 m/s
20
40
V = 40m/s – 20 m/s = 20 m/s
C)
i.
40
20
V = √ [ 40² + 20²) = 44,72 m/s
CONTRACCIÓN DE LA LONGITUD-DILATACIÓN DEL TIEMPO
3.-
L = Lp/ ɣ = Lp √ [ 1 - v²/c]
(Lp/2)²/ Lp² = 1 – v²/ c²
v² = 0,75 c²
v = 0,866 c
[L/Lp] = √ [ 1 - v²/c²]
¼ = 1 – v²/ c²
L²/Lp² = 1 – v²/ c²
v²/ c² = 0,75
v= 0,866 ( 3 x108 m/s) = 2,598 x 108 m/s
4.-
Δt = ɣ Δtp
Δt = Δtp / √ [ 1 - v²/c²]
[Δtp / Δt]² = [ 1 - v²/c²]
Δt = 2 Δtp
[Δtp / 2Δtp]² = [ 1 - v²/c²]
(1/4 ) = [ 1 - v²/c²]
V= 0,866 c
5.-
Lp2 = 3 Lp1
L2²=L1²
y V = 0,350 c
Lp2² (1 - v²/c²) = Lp² (1 - v²/c²)
9 Lp² (1 – v2²/c²) = Lp² (1 – v1²/c²)
9 (1 – v2²/c²) = (1 – (0,35c)²/c²)
Hallar la V2 = 0,95 c
(3Lp1)² (1 - v²/c²) = Lp² (1 - v²/c²)
9 (1 – v2²/c²) = (1 – v1²/c²)
TRANSFORMACIONES DE LORETZ
6.-
U´x = (Ux – v)/[(1 -Uxv)/c²] = (0,95c -0,75c) / (1 – (0,75c 0,95c)/c²)
= 0,20c /(1 – 0,7125)=0,696c
7.-
U´x = (Ux – v)/[(1 -Uxv)/c²] = (-75c – 0,75c) / (1 – (-075c) (0,75c)/c²)
= -1,50c / (1 – (-0,5625)) = - 0,960c
MOMENTUM LINEAL RELATIVISTA
8.-
A)
.p= ɣ mu
u= 0,01 c
m = 9,11 x10-31 kg
c= 3 x 108 m/s
ɣ = 1 / √(1 - u²/c²) = 1 / √(1 – (0,01c)²/c²) = 1 / √(1 – (0,01)²/) = 1,0
.p = (1,0) (9,11 x10-31 kg) (0,01 x 3 x 108 m/s) = 2.73 x 10-24 kg m/s
B)
C)
ENERGÍA RELATIVISTA
9.-
E = ɣ m c²
m = 9,11 x10-31 kg
c= 3 x 108 m/s
La energía en reposo de un electrón = me C² = 9,11 x10-31 kg (3 x 108 m/s)²=
= 0,511 MeV
1 eV= 1,6 x 10-19 Julios
ΔE = E2 -E1 = ɣ2 m c² - ɣ1 m c² = (ɣ2 - ɣ1 ) m c² =
= [1 / √ [ 1 – v2²/c²] - 1/ √ [ 1 – v1²/c²]] ] m c² =
= [1 / √ [ 1 – (0,9c)²/c²] – [1/ √ [ 1 – (0,5c)²/c²]] ] 0,511 MeV=
=
10.-
A)
Eo = mp c²
mp= 1,67 x 10-27 kg
c= 3 x 108 m/s
E0 = (1,67 x 10-27 kg) (3 x 108 m/s)² = 938 MeV
B.) Energía total = ɣ mp c² = (1 / √ [ 1 – v²/c²) mp c²
= (1 / √ [ 1 – (0,95c)²/c²) 938 MeV = 3 x 109 eV
C.) E = K + Ereposo = K + m c²
K = E - m c² = 3 x 109 eV – 0,938 x 109 eV = 2,062 MeV