Monbukagakusho Physics 2010 Solution

Iskandar Setiadi 2012. All Rights Reserved. | STEI ITB 2011

1

Physics

STEP BY STEP SOLUTION

BY : Iskandar Setiadi


2

Solution:

Note that A BP P= . Hence '

' ..

A B

A B A A

F F F F F n FA A A n A

= → = → =

Total volume is always constant. Thus, A BV V↓= ↑ , so ' ' '. . . . .A B A AhA h A h A h n A h hn

= → = → =

The answer is option 5.


3

Solution:

The pole and the box do not change form, so total force exerted at point A must equal total force exerted at point B. Normal force (N) at point A is in the same direction as AF , so F at point A =

.sin .cos .sinAN F W Fθ θ θ+ = + while F at point B = .cos .sinW Fθ θ− .

Hence .cos . .sin cos sin 1 tan.cos . .sin cos sin 1 tan

A

B

F W m gF W m g

θ θ θ θ θθ θ θ θ θ+ + +

= = =− − −

.


F.sin


4

Solution:

Assume that, velocity when the ball is launched = v, we can conclude that xv = constant, while

yv is defined by .( ) . .( )2 2y

T g Tv f t t g g t= = − = −

How could we arrive on this conclusion? On t = 2T , 0yv = (mid-point)

Hence t = 0, 2oyTv g= , while at lim ( ) then

2yt T

Tf t v g→

= − .

The height h is defined by:

2

2

1. . .2

1( . ). . .2 2

1 . . .( )2

oyh v t g t

Th g t g t

h g t T t

= −

= −

= −

.



5


6

Solution:

Let determine the constant ( k∑ ) from two springs

. .k x m g∆ =∑

2.(0, 2 ) (0,5 ).(9,8 / )k m kg m s=∑

24,5 /k N m=∑

When A is pulled downward and released gently:

2 mTk

π=

0,5 2224,5 7

T π π= =

Note that: time elapses from release until A reaches its maximum height, t = 2T

1 0,449

2 7Tt sπ= = =

The nearest/best answer is 0,45



7

Solution: As mechanic energy is constant, at point P, , 0p m kE E E= = . While at point Q, 0, p k mE E E= = .

Each point at x-axis from graph p kE E+ must be constant. The answer is option 4.

constant ( )


8

m.g


9

Solution:

This is similar to Q4. First, we need to determine k (constant) from the lightweight rubber.

. .(0,1) (0,5).(9,8) 49 /exertF m g k k N m= → = → =

When the weight is pulled down by 20 cm (from Fig.2), we’ll have:

2 21 1. .( ) .49.(0,2) 2,2052 2rubberE k x J= ∆ = =

Hence,

rubber ballE E=

2,205 . . (0,5).(9,8).( )J m g h h= =

0,45 45h m cm= =



10


11

Solution:

It is known that 2 31 1

2 2

( ) ( )T RT R

= and 2 lTg

π=

Hence, 2 3

2

2( ) ( )

Rg R

T r

π=

2 3

2

2( ) ( )

Rg R

T r

π=

32

22 ( ) .R R Tg r

π =

32

2 2 .( )R rTg R

π=

2 2 r rTR g

π=



12

Solution:

Using Black’s principle:

. . . . . .c c w w p pm c T m c T m c T∆ + ∆ = ∆

(100).(0.4).( 20) (200).(4, 2).( 20) (150).(0, 4).(80 )T T T− + − = − 40 800 840 16800 4800 60T T T− + − = − 940 22400T = 23,83T Cο=

The nearest / best answer is 24.



13

Solution:

1 1

2 2

. . .

. . .PV n R TPV n R T

=

In such a temperature, monoatomic ideal gas at constant pressure and same type of gas can be

written as: 1 1

2 2

(10 ) 57 273 10 12 1,2(10 ) 273

V T x x xV T x

+ += → = → + = −

−

2, 2 2 0,91x x cm= → = . The nearest/best answer is 1.



14

Solution:

It is obvious that chemical energy from gasoline is converted to kinetic energy of the automobile. While energy which is lost to the environment and cannot be reused heat.



15

Solution:

Speed of waves in Earth increases with increasing depth. As the wave-front pattern takes on the same shape as the shoreline, we can infer that wave speed plays a major role in this observation.



16

Solution:

In night, ground releases heat faster than sea. Heat from ground is moving through upper sky towards sea, hence, air temperature in the upper sky > air temperature near the ground. Speed of sound (moving particle) is temperature-dependent; as temperature increases the average molecular velocity increases.



17

Solution:



18


19

Solution:

V (electrical potential) is defined by qV kr

= .

At point A: V = 2 2 2 2 2 2 2 2

( )(2 ) ( ) (2 ) ( ) (3 ) (2 ) (3 ) (2 )

Q Q Q Qka a a a a a a a+ − + −

+ + ++ + + +

= 0.

At point B: V = 2 2 2 2 2 2 2 2

( )(2 ) (2 ) (2 ) (2 ) (2 ) (2 ) (2 ) (2 )


+ + ++ + + +

= 0.

At point C: V = 2 2 2 2 2 2 2 2

( )( ) ( ) (3 ) ( ) (3 ) (3 ) (3 ) ( )


+ + ++ + + +

V = ( ) 02 10 18 10

Q Q Q Qka a a a+ − + −

+ + + >

Therefore, A B CV V V= < .



20

Solution:

Fig.1 0 ACdεκ=

Fig.2 0 01 2 0

1 1 1 1 1 2

(2 ) (2 )

dA AC C C A

d d d d

κε ε εκ κ= + = + =

− −

0

2ACd

εκ=

0

2

01

122

AFig d

AFigd

εκ

εκ= = .



21

Solution:

We can have several approaches here, but first, note that the bulb is connected with 5Ω resistor, so the total current should be less than 2,4 A. From Fig.1, when I = 2A, V = 2V, and R = 1Ω .

.2 6 2figR I A= Ω→ =∑ . Thus the amount of power consumed: 2 2. 2 .1 4P I R W= = = .



22


23

Solution:

First, we know that F = 1 22

.q qkr

while E = 2

qkr

F = E.q

It is known that F = B.q.v

. . .E q B q v=

E vB=

Law of conservation states that

argkinetics ch eE E=

21 .2

mv qV=

2 2. .v qVm

=

2qVvm

=

Since E vB= and 2qVv

m= , thus 2E qV

B m= .



24

Solution:

Using Maxwell Law, 02

. .4

i d s rdBr

µπ

×=

Blvε =

. .. B l vi R Blv iR

= → =


25

02

0

.( . .(sin . )). .4

r b

r b

I dr d l vr

iR

π µ θ θπ

+

−=∫ ∫

[infinite straight conducting wire]

02

0

.( (sin . ). . ). .4

r b

r b

Id dr l vr

iR

π µθπ

+

−=∫ ∫

[ sin .dθ θ is independent, we can separate it from equation]

02

.(2. . ). .4

r b

r b

I dr l vr

iR

µπ

+

−=∫

0. 1[ ] . .( )2

r br b

I l vriR

µπ

+−−

=

0. 1 1[ ]. .( )2 ( ) ( )

I l vr b r bi

R

µπ

− ++ −=

0. ( ) ( )[ ]. .( )2 ( )( )

I b r b r l vr b r bi

R

µπ

+ + −+ −=

0. ( ) ( )[ ].(2 ).( )2 ( )( )

I b r b r a vr b r bi

R

µπ

+ + −+ −=

02 2

. 2[ ].(2 ).( )2 ( )

I b a vr bi

R

µπ −=

0

2 2

. (2 ).(2 ).( )2

.( )

I b a vi

R r b

µ

π=

−

2 2

2. . . . ..( )

o a b v IiR r bµ

π=

−



26

Solution:

The electromotive force (emf) opposes the change in flux. ddt

ε Φ= −


Monbukagakusho Physics 2010 Solution

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