Nom: Pr´enom: No¯ carte d’´etudiant: Sectionhomepages.vub.ac.be/~pnardon/Janvier2014.pdf ·...
Transcript of Nom: Pr´enom: No¯ carte d’´etudiant: Sectionhomepages.vub.ac.be/~pnardon/Janvier2014.pdf ·...
Nom:
Prenom:
No¯ carte d’etudiant:
Section:
Physique, PhysG101 1er Bachelier
Janvier 2014
Il y a 10 questions.
Le teste dure de 13h a 17h
Essayez d’aller le plus loin possible en :
1 Justifiant brievement toutes vos reponses;
2 En donnant, en plus des valeurs numeriques, les unites physiques.
3 Prenez partout g = 10 m s−2
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Q 1: [5pt] 120 km
40 km/h 20 km/h
50 km/h
30 km/h
50 km/h x ! t
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xJ = 40 t xK = 120!20 t
40 t = 120 ! 20 t t = 2 h
TA total TR total
TA + TR = 2 h N
N
80 km 50kmh TA ! 30kmh TB = 80 km
!TA + TR = 2 h
50 TA ! 30 TB = 80 h!
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TA =7
4h
TB =1
4h
50kmh
7
4h + 30
kmh
1
4h = 95 km
21/01/2014 : 1
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Q 2: [5pt]
!
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H R
n r
h(t) t H R n
r R H πR2H/3
r 4πr3/3
t n t
t n t4
3π r
3
π
3R(t)2 h(t) = n t
4
3π r
3
R(t)
h(t)=
R
HR(t) =
R
Hh(t)
π
3
R2
H2h(t)3 = n t
4
3π r
3 → h =
�4 n r
3H
2
R2t
�1/3
21/01/2014 : 2
!
Q 3: [10pt] 200g 10 g F
µs µs = 0.9
µd µd = 0.6 maximale F
�−mVg+NC/V = 0
mV aV = fC/V
fC/V ≤ µs NC/V
NC/V fC/V
�−mCg−NV/C +NT/C = 0
mC aC = F− fV/C − fT/C
fT/C = µd NT/C
NT/C fT/C
aC = aV = a
NC/V = mV g NT/C = (mV +mC) g
�mV a = fC/V
mC a = F− fV/C − µd(mV +mC) g
a fC/V
a =F
mV +mC− µd g
fC/V
fC/V = mVF
mV +mC− µd mV g ≤ µsmV g F ≤ (µd + µs)(mV +mC) g = 3.15 N
21/01/2014 : 3
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Q 4: [8pt] F
µ 0.7 m 300g
Q4-1m g N
f x
N f x ≤ 10cm
Q4-2
�N−m g = 0
F− f = 0
N = m g = 3 N f = µdN = F = 2.1 N
Q4-3 h
d x ≤ d
h F+d
2m g− xN = 0 x =
h µdm g+ d2m g
mg= hµd +
d
2≤ d → h ≤ d
2µd=
5 cm
0.7= 7.14 cm
21/01/2014 : 4
Q 5: [7pt]60.7 T 0.5 m s−2
50 km/h km
kWh/km
T
VF L
V = a t X =1
2a t
2T =
VF
aL =
V2F
2a
E =1
2mV
2F
E
L=
12mV
2F
V2F
2a
= m a = 60.7 103 kg 0.5 m s−2 = 30.35 10
3 N
kWh/km
1kWh
km=
1033600 J
103 m= 3600 N
E
L= 30.35 10
3 1
3600
kWh
km= 8.43
kWh
km
21/01/2014 : 5
Q 6: [3pt] 27 M = 7.35 1022 kg
G = 6.67 10−11 m3 kg−1 s−2
G ML
R2=
�2π
T
�2
R → R3 =
T2
4π2GML
T
T = 27× 24× 3600 s = 2.33 106 s → T
2 = 5.44 1012 s2
GML = 6.67 10−11 × 7.35 10
22 = 4.90 1012 m3 s−2 T
2
4π2GML = 6.76 10
23 m3
R = 8.77 107 m
21/01/2014 : 6
Q 7: [6pt] 60 Tonnes 50 km h−180 km h−1
Q7-1
mTVT −mVVV = (mT +mV )Vfinal Vfinale =60× 50− 1× 80
61= 47.87 km h−1
Q7-2E1 = 1
2mTV2T + 1
2mVV2V E2 = 1
2 (mT + mV )V2finale
E1 − E2 =1
2mTV
2T +
1
2mVV
2V −
1
2(mT +mV )V
2finale = 6.03 10
6 J− 5.39 106 J = 6.41 10
5 J
Q7-3 10 m
F
E = F L F =5.39 10
6 J
10 m= 5.39 10
5 N
Q7-4(mT +mV )a = F
a =5.39 10
5 N
61 103 kg= 8.84 m s−2
21/01/2014 : 7
Q 8: [3pt] R
33
0.2 Rmax
N m g
m a = f → mω2R = f N = m g
f ≤ µsN
mω2R ≤ µsm g R ≤ µs g
ω2=
0.2× 10
(33×2π60 )2
= 16.75 cm
21/01/2014 : 8
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Q 9: [19pt] m1 = 100 g m2 = 50 g
m = 10 g R
I I = 5mR2
m2
µd µd = 0.6
Q9-1 Situation statique
µs m2
N = m2g f = T2 T1 = m1g T2 = T1 → f = m1g ≤ µs m2g µs ≥ m1
m2= 2
Q9-2 Situation dynamique t = 0
V(0) = 0
Q9-3
m1a1 = m1g− T1
m2a2 = T2 − f
Iα = R(T1 − T2)
N = m2g
f = µdm2g
a1 = a2 = Rα
→
T2 = 0.475 N
T1 = 0.65 N
N = 0.5 N
f = 0.3 N
Q9-4 a1 = a2 = 3.5 m s−2
Q9-5 t = 1 s totale
Ec =1
2m1V
21 +
1
2m2V
22 +
1
2Iω2 =
1
2(m1 +m2 + 5m)V2 =
1
2(m1 +m2 + 5m) (at)2 = 1.225 J
Q9-6 t = 1 s m1
h = a1t2/2
Ep = m1gh h =1
2at
2 = 1.75 m → Ep = 1.75 J
Q9-7 toutes
f T = f h =
0.3× 1.75 = 0.525 J Ec = Ep − T 1.225 = 1.75− 0.525
21/01/2014 : 9
-50 2 4 6 8
0
5
10
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Q 10: [9pt] Vx(t) t
X(0) = 5 m
Q10-1 X(8) t = 8 s
X(8) = 5+2× 10
2−
2× 5
2− 2× 5−
1× 5
2+
1× 5
2= 0
Q10-2 X(5) t = 5 s
X(5) = 5+2× 10
2−
2× 5
2− 1× 5 = 5
Q10-3 t t = 2 s V = 0
> 0 < 0
Q10-4 0 s < t < 2 s a = −5 m s−2
Q10-5 6 s < t < 8 s a = 5 m s−2 Q10-6 t1 t2
4 s < t < 6 s
a = 0
21/01/2014 : 10