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

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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

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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