Općinsko natjecanje iz fizike Rješenja i smjernice za bodovanje – 1. grupa srednje škole

22. 01. 2019.

1. zadatak (10 bodova)

(a) Neka je x udaljenost pretrčana s ubrzanjem a dok nije postignuta maksimalna brzina v. Ukoliko je to postignuto u vremenu t1 možemo napisati sljedeće tri jednadžbe:

, , i . [3 boda]

Prve dvije daju:

[1 bod]

Za Maricu:

Za Šteficu: [1 bod]

(b) [1 bod] Za Maricu: . Za Šteficu: . [1 bod]

(c) Nakon 6 s vrijedi:

. [1 bod]

Za Maricu: .

Za Šteficu: . [1 bod]

Marica je u prednosti za 2.62 m. [1 bod]

2. zadatak (10 bodova) Kada Stipe pliva nizvodno, njegova brzina je , a kada pliva uzvodno tada je .

[2,5 boda] Stoga je ukupno Stipino vrijeme jednako:

. [2 boda]

Brzina kojom Cvita pliva (i u jednom i u drugom smjeru) je . [2,5 boda] Njeno ukupno vrijeme je:

. [2 boda]

x = 12v + v1( )t1 100 − x = v 10.2 − t1( ) v = v1 + at1

100 = 10.2 − 12t1

⎛⎝⎜

⎞⎠⎟ v = 10.2 − 1

2t1

⎛⎝⎜

⎞⎠⎟ at1

a = 20020.4 − t1( )t1

.

a = 20018.4( ) 2.00( ) = 5.43m/s

2.

a = 20017.4( ) 3.00( ) = 3.83m/s

2.

v = a1tv = 5.43( ) 2.00( ) = 10.9m/sv = 3.83( ) 3.00( ) = 11.5m/s

x = 12at1

2 + v 6.00 − t1( )

x = 125.43( ) 2.00( )2 + 10.90( ) 4.00( ) = 54.3m

x = 123.83( ) 3.00( )2 + 11.50( ) 3.00( ) = 51.7m

c + v c − v

t1 =L

c + v+ Lc − v

=2Lc

1− v2

c2

c2 − v2

t2 =2Lc2 − v2

=

2Lc

1− v2

c2

Budući da je , , odnosno Cvita će se prije vratiti u početni položaj. [1 bod]

3. zadatak (10 bodova) (a) Skica:

[3 boda] (b)

(za blok #2) [1 bod] (za blok #1) [1 bod]

Iz toga dobivamo:

[1 bod]

[2 boda]

. [2 boda]

4. zadatak (10 bodova) Skica:

[2 boda] (a) Kolotura P1 ima ubrzanje a2.

Budući da m1 prijeđe dvostruko veću udaljenost u odnosu na koloturu P1 u istom vremenu, ubrzanje od m1 je dvostruko veće od ubrzanja P1, odnosno . [1 bod]

(b) Koristeći oznake sa slike, možemo napisati: [1 bod]

[1 bod] [1 bod] To možemo srediti i napisati:

1− v2

c2<1 t1 > t2

138 The Laws of Motion

P5.44 Let a represent the positive magnitude of the acceleration −a�j of

m1 , of the acceleration −a�i of m2 , and of the acceleration +a�j of m3 .Call T12 the tension in the left rope and T23 the tension in the cordon the right.

For m1 , F may y∑ = + − =−T m g m a12 1 1

For m2 , F max x∑ = − + + =−T n T m ak12 23 2µ

and F may y∑ = n m g− =2 0

for m3 , F may y∑ = T m g m a23 3 3− =+

we have three simultaneous equations

− + =

+ − − =

+ − =

T a

T T a

T a

12

12 23

23

39 2 4 00

0 350 9 80 1 00

19 6 2 00

. .

. . .

. . .

N kg

N kg

N kg

b ga f b g

b g

(a) Add them up:

n

T12 T23

m g 2

f = n k µ

m g 1

T12

m g 3

T23

FIG. P5.44

+ − − =39 2 3 43 19 6 7 00. . . . N N N kga fa

a m m m= 2 31 1 2 3. , m s , down for , left for and up for 2 .

(b) Now − + =T12 39 2 4 00 2 31. . . N kg m s2a fc h

T12 30 0= . N

and T23 19 6 2 00 2 31− =. . . N kg m s2a fc h

T23 24 2= . N .

P5.45 (a)

(b)

See Figure to the right

68 0 2 2

1 1

. − − =− =

T m g m aT m g m a

µµ

(Block #2)(Block #1)

Adding,

68 0

68 01 29

27 2

1 2 1 2

1 2

1 1

.

..

.

− + = +

=+

− =

= + =

µ

µ

µ

m m g m m a

am m

g

T m a m g

b g b g

b g m s

N

2

T

m 1m 2

T F

m 1

n 1

T

m g 1 = 118 N

f = n k µ 1 1

m 2

n 2

F

m g 2 = 176 N

f = n k µ 2 2

FIG. P5.45

138 The Laws of Motion

P5.44 Let a represent the positive magnitude of the acceleration −a�j of

m1 , of the acceleration −a�i of m2 , and of the acceleration +a�j of m3 .Call T12 the tension in the left rope and T23 the tension in the cordon the right.

For m1 , F may y∑ = + − =−T m g m a12 1 1

For m2 , F max x∑ = − + + =−T n T m ak12 23 2µ

and F may y∑ = n m g− =2 0

for m3 , F may y∑ = T m g m a23 3 3− =+

we have three simultaneous equations

− + =

+ − − =

+ − =

T a

T T a

T a

12

12 23

23

39 2 4 00

0 350 9 80 1 00

19 6 2 00

. .

. . .

. . .

N kg

N kg

N kg

b ga f b g

b g

(a) Add them up:

n

T12 T23

m g 2

f = n k µ

m g 1

T12

m g 3

T23

FIG. P5.44

+ − − =39 2 3 43 19 6 7 00. . . . N N N kga fa

a m m m= 2 31 1 2 3. , m s , down for , left for and up for 2 .

(b) Now − + =T12 39 2 4 00 2 31. . . N kg m s2a fc h

T12 30 0= . N

and T23 19 6 2 00 2 31− =. . . N kg m s2a fc h

T23 24 2= . N .

P5.45 (a)

(b)

See Figure to the right

68 0 2 2

1 1

. − − =− =

T m g m aT m g m a

µµ

(Block #2)(Block #1)

Adding,

68 0

68 01 29

27 2

1 2 1 2

1 2

1 1

.

..

.

− + = +

=+

− =

= + =

µ

µ

µ

m m g m m a

am m

g

T m a m g

b g b g

b g m s

N

2

T

m 1m 2

T F

m 1

n 1

T

m g 1 = 118 N

f = n k µ 1 1

m 2

n 2

F

m g 2 = 176 N

f = n k µ 2 2

FIG. P5.45

68,0 −T − µm2g = m2aT − µm1g = m1a

68,0 − µ(m1 +m2 )g = m1 +m2( )aa = 68,0

m1 +m2( ) − µg = 1,29m s2

T = m1a + µm1g = 27,2N

132 The Laws of Motion

P5.33 First, we will compute the needed accelerations:

1 0

21 20 0

0 8001 50

3 0

40 1 20

1 500 800

a fa f

a fa f

Before it starts to move:

During the first 0.800 s: m s

s m s

While moving at constant velocity:

During the last 1.50 s: m s s

m s

2

2

a

av v

t

a

av v

t

y

yyf yi

y

yyf yi

=

=−

=−

==

=−

=−

= −

..

.

.

..

FIG. P5.33Newton’s second law is: F may y∑ =

+ − =

= +

S a

S a

y

y

72 0 9 80 72 0

706 72 0

. . .

. .

kg m s kg

N kg

2b ge j b gb g

(a) When ay = 0 , S= 706 N .

(b) When ay =1 50. m s2 , S= 814 N .

(c) When ay = 0 , S= 706 N .

(d) When ay =−0 800. m s2 , S= 648 N .

P5.34 (a) Pulley P1 has acceleration a2 .Since m1 moves twice the distance P1 moves in the sametime, m1 has twice the acceleration of P1 , i.e., a a1 22= .

(b) From the figure, and using

F ma m g T m aT m a m a

T T

∑ = − == =

− =

: 2 2 2 2

1 1 1 1 2

2 1

12 2

2 0 3

a fa fa f FIG. P5.34

Equation (1) becomes m g T m a2 1 2 22− = . This equation combined with Equation (2) yields

Tm

mm

m g1

11

222

2+F

HGIKJ=

Tm mm m

g11 2

112 22

=+

and Tm m

m mg2

1 2

114 2

=+

.

(c) From the values of T1 and T2 we find that

aTm

m gm m1

1

1

2

112 22

= =+

and a am g

m m2 12

1 2

12 4

= =+

.

a1 = 2a2

F = ma :∑ m2g −T2 = m2a2T1 = m1a1 = 2m1a2T2 − 2T1 = 0

. [1 bod]

Time dobivamo:

i . [1 bod]

(c) Iz T1 i T2 dobivamo:

i . [2 boda]

5. zadatak (10 bodova)

(a) Ukupan pomak je jednak površini ispod krivulje (v, t) od t = 0 do 50 s. [1 bod]

.

. [1 bod] (b) Između t = 10 s i t = 40 s, pomak je

. [1 bod]

(c) . [1 bod]

. [1 bod]

. [1 bod]

Skica:

[1 bod]

(d) (i) . [1 bod]

(ii) . [1 bod]

(iii) Za :

,

odnosno

,

T1m1

2m1 +m2

2⎛⎝⎜

⎞⎠⎟ = m2g

T1 =m1m2

2m1 + 12 m2

g T2 =m1m2

m1 + 14 m2

g

a1 =T1m 1

= m2g2m1 + 1

2 m2

a2 =12a1 =

m2g4m1 +m2

Δx = 1250 m s( ) 15s( )+ 50 m s( ) 40 −15( )s+ 1

250 m s( ) 10s( )

Δx = 1875m

Δx = 1250 m s+33.3m s( ) 5s( )+ 50 m s( ) 25s( ) = 1458m

0 ≤ t ≤15s; a1 =ΔvΔt

=50 − 0( )m s15s− 0

= 3,3m s2

15 ≤ t ≤ 40s; a2 = 0m s2

40 ≤ t ≤ 50s; a3 =ΔvΔt

=0 − 50( )m s50s− 40s

= −5m s2

30 Motion in One Dimension

*P2.23 (a) Choose the initial point where the pilot reduces the throttle and the final point where theboat passes the buoy:

xi = 0 , x f =100 m , vxi = 30 m s, vxf = ?, ax =−3 5. m s2 , t= ?

x x v t a tf i xi x= + + 12

2:

100 0 3012

3 5 2 m m s m s2= + + −a f c ht t.

1 75 30 100 02. m s m s m2c h a ft t− + = .

We use the quadratic formula:

tb b ac

a=− ± −2 4

2

t=± − ( )

=±

=30 900 4 1 75 100

2 1 75

30 14 13 5

12 6 m s m s m s m

m s

m s m s m s

s2 2 2

2 2

.

.

..

.c h

c h or 4 53. s .

The smaller value is the physical answer. If the boat kept moving with the same acceleration,it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s.

(b) v v a txf xi x= + = − =30 3 5 4 53 14 1 m s m s s m s2. . .e j

P2.24 (a) Total displacement = area under the v t,a f curve from t= 0to 50 s.

∆

∆

x

x

= + −

+

=

12

50 15 50 40 15

12

50 10

1 875

m s s m s s

m s s

m

b ga f b ga f

b ga f

(b) From t= 10 s to t= 40 s , displacement is

∆x = + + =12

50 33 5 50 25 1 457 m s m s s m s s mb ga f b ga f .

FIG. P2.24

(c) 0 15≤ ≤t s : avt1

50 015 0

3 3= =−( )

−=∆

∆ m s

s m s2.

15 40 s s< <t : a2 0=

40 50 s s≤ ≤t : avt3

0 5050 40

5 0= =−( )

−= −∆

∆ m s

s s m s2.

continued on next page

x1 =12a1t

2 = 123,3m s2( )t 2 = 1,67m s2( )t 2

x2 =1215s( ) 50m s− 0( )+ 50m s( ) t −15s( ) = 50m s( ) t − 375m

40s ≤ t ≤ 50s

x3 =povrsina ispod v − t grafaod t = 0 do 40s

⎛⎝⎜

⎞⎠⎟+ 12a3 t − 40s( )2 + 50m s( ) t − 40s( )

x3 = 375m +1250m + 12

−5,0m s2( ) t − 40s( )2 + 50m s( ) t − 40s( )

što se u konačnici svodi na: . [1 bod] x3 = 250m s( )t − 2,5m s2( )t 2 − 4375m