DS 1 Physique-Chimie : Corrigé Problème 1 : Étude du mouvement

ex
y
e
v
0
Oθ
0
m
−→
P=m−→ g=−mg−→ ez
md−→
v
dt =m−→
g⇔d−→
v
dt =−→
g⇔
dvx
dt = 0
dvy
dt =−g
⇔vx=cste
vy=−gt +cste0
vx(t= 0)=v0cos(θ0)
vy(t= 0)=v0sin(θ0)⇒cste =v0cos(θ0) cste0=v0sin(θ0)
vx=v0cos(θ0)
vy=−gt +v0sin(θ0)
vxvyx y (x=v0cos(θ0)t+cste
y=−1
2gt2+v0sin(θ0)t+cste0)
x(t= 0)=0
y(t= 0)=0 ⇒cste =0 cste0=0 
x=v0cos(θ0)t
y=−1
2gt2+v0sin(θ0)t
tfy(t=tf)=0
y(t=tf)=0⇔ − 1
2gt2
f+v0sin(α)tf= 0 ⇒tf= 0 −1
2gtf+v0sin(α) = 0
tf=2v0sin(θ0)
g
L0t=tfL0=x(tf) = v2
0
g2 cos(θ0) sin(θ0)⇔L0=v2
0
gsin(2θ0)
[v0] = L.T −1[g] = L.T −2⇒[L0] = L2.T −2
L.T −2=L L0
•v0= 0 L0= 0
•g→0L0→ ∞
•θ0= 0 ⇒L0= 0
•θ0=π/2⇒L0= 0 O
sin(2θ0)=1 θ0=π
4
v0= 55 −2θ0= 52o⇒tf= 8,8L0= 299
•v0x= 1,4y= 1,75
d=p1,42+ 1,752= 2,24

v0v0=2,24
0,05 = 44,8−1
•v∞o
d=p(10 −10)2+ (0,15 −0,45)2= 0,3
v∞v∞=0,3
0,05 = 6 −1
[Cx] = [F]
[ρ][S][v2]=M.L.T −2 M.L−3L2L2.T −2=∅ ⇒ Cx
L=R= 4cm L = 2R
t= 0 v=v0= 55 −1⇒Re,max =Rv0
ν= 1,47.105
v∞
⇒Re,min =Rv∞
ν= 1,7.104
Re>103
Re∈[103,2.105]Cx
Cx= 0,4
ex
y
ev
0
Oθ
0
phase ascendante
(trajectoire rectiligne)
phase intermédiaire
(trajectoire parabolique)
phase descendante
(trajectoire rectiligne)
−→
F=−1
2ρSCxv2−→
u
md−→
v
dt =−1 2ρSCxv2−→
u⇔mdv
dt
−→
u=−1
2ρSCxv2−→
u
−→
u mdv
dt =−1 2ρSCxv2⇔dv dt +αv2= 0 α=ρSCx 2m
−→
v(t) = v(t)−→ u−→ u−→ v0−→ u=cos θ0
sin θ0
−→
v=vx
vy=v0
1 + αv0tcos θ0
sin θ0

•x(t)vx=dx
dt =v0 1 + αv0tcos θ0⇒x(t) = 1 αln(1 + αv0t) cos θ0+cste
x(t= 0) = 0 ⇒cste = 0 ⇒x(t) = 1
αln(1 + αv0t) cos θ0
•y(t)vy=dy
dt =v0 1 + αv0tsin θ0y(t) = 1 αln(1 + αv0t) sin θ0
−→
F=−mαv2−→ u=−mα v2
0
(1 + αv0t)2cos θ0
sin θ0=Fx Fy
FyFy=−mg
⇔αv2
0
(1 + αv0t)2sin θ0=g⇔t=1 αrαsin θ0 g−1 v0=t1t1= 0,52
ot=
v(t1) = v0
1 + v0αt1
= 7,52 −1
x(t1) = 5,57 y(t1)=7,12
ox= 5,65 y= 6,5
t1
(x1, y1)
v1
t0t0=2v1sin θ0
gt0= 1,21
L0L0=v2
1
gsin(2θ0)L0= 5,60
t2=t1+t0= 1,63
x2=x1+L0= 11,17 y2=y1= 7,12
x2= 11,17
11,17 −10
10 = 0,117 = 11,7%
−→
P=m−→ g=−mg−→ ey
−→
F=−mαv2−→ u
md−→
v
dt =−αmv2−→
u+m−→ g⇔d−→
v
dt =−αv2−→
u+−→ g
−→
v=−−→ cste
−→
d−→ v
dt =−→
0−→ 0 = −αv2−→ u+−→ g⇔αv2−→ u=−→ g
⇔αv2−→ u=−g−→ ey⇔−→
u=−−→ ey
αv2=g⇔
−→
u=−−→ ey
v=rg
α
⇔−→ v=−rg
α
−→
ey=−→ v∞
k−→ v∞k= 6,68 −1
y2= 7,12
t00 =y2
k−→ v∞k=y2
pg/α t00 = 1,07
t3=t2+t00 = 2,7
47 ×0,05 = 2,35

•−→ P=−mg−→ uz
•−→ F=−k(z−z0)−→ uz
−→
T+−→ P=−→ 0
ze=z0−mg
k
−→
T+−→ P=m−→ a
(Oz)−→ a= ¨z−→ uz
⇔ −k(z−z0)−→ uz−mg−→ uz=m¨z−→ uz
−→
uz−k(z−z0)−mg =m¨z⇔¨z+k
mz=k m(z0−mg k)⇔¨z+k mz=k mze
¨z+k
mz=k mze
•
¨z+k
mz= 0 zh(t) = Acos(ω0t) + Bsin(ω0t)A B ω0=rk m
•zp=cste
⇒0 + k
mzp=ze⇒zp=ze
•z(t) = zh(t) + zp(t) = Acos(ω0t) + Bsin(ω0t) + ze
T0=2π
ω0
= 2πrm
k
ω0=r105
103= 10 −1T0=2π 10 = 6,28
z(t= 0) = z0˙z(t)=0
•z(t= 0) = z0⇔A+ze=z0⇔A=z0−ze
•˙z(t) = −ω0Asin(ω0t) + ω0Bcos(ω0t) ˙z(t= 0) = 0 ⇔ω0B= 0 ⇔B= 0
z(t)=(z0−ze) cos(ω0t) + ze
z(t)t
•z(t= 0) = z0
•z(t=T0/4) = ze
•z(t=T0/2) = 2ze−z0
•z(t= 3T0/4) = ze
•z(t=T0) = z0
zmin =z0zmax = 2ze−z0
zmoy =ze
[h] = [F]
[v]=kg.m.s−2 m.s−1= kg.s−1

•−→ P=−mg−→ uz
•−→ FR=−k(z−z0)−→ uz
•−→ F=−h−→ v=−h˙z−→ uz
−→
P+−→ FR+−→ F=−→ 0−→
F=−→ 0−→ P+−→ FR=−→ 0
ze
−→
P+−→ FR+−→ F=m−→ a⇔ −mg−→ uz−k(z−z0)−→ uz−h˙z−→ uz=m¨z−→ uz
−→
uz−mg +−k(z−z0)−h˙z=m¨z⇔¨z+h
m˙z+k mz=k mze
r2+h
mr+k m= 0
∆ = h
m2
−4k
m=h2−4km m2
•h2<4km ∆<0
•h2= 4km ∆=0
•h2>4km ∆>0
m0m
h2= 4km0
m>m04km > h2
4km < h2
m0< m 4km0< h2