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SKKU-CS-009
기초자연과학영역(BSM) 공동시험문제
Basic Science and Mathematics Common Examination
학기
Semester
학과/계열
Dept./Division
과목명
Subject
수업정보
Class Time
유의사항
Note
2018 년
1 학기
일반물리학1
요일
Day
시간
Time
시험구분
Exam Type
중간 Mid Term
기말 Final
학번
Student PIN
학수번호
Course Code
수업담당교수
Instructor
GEDB008
☑
□
시험시간
Exam Time
60 분/Minutes
성명
Name
분반
Section Number
감독자확인
Proctor’s Signature
1. Electronic calculator is not allowed!!
2. Answer the questions in English or Korean.
[1] [25 points] A self-driving car is equipped with a device that
detects an object by measuring the distance from the car to the
object. After an object is detected, it takes 0.1 sec for the car to
decide whether it stops or not. With the decision to stop made,
the car immediately applies the brake and keeps the brake applied
until it stops. The acceleration applied by the brake is   m/sec2.
Assume the air resistance is negligible.
(a) [10 points] While the car cruises at the speed of 72 km/h, it
detects a standing obstacle 50 m ahead and decides to stop. What
is the distance between the car and obstacle when the car stops?
(b) [15 points] Again the car cruises at the speed of 72 km/h. It
catches up to a van (whose height is 2 m) driving in front at the
speed of 54 km/h. When the van is 50 m ahead of the car, an
object (originally fixed on the van's top) suddenly gets released and
starts the projectile motion with the initial velocity the same as the
van's velocity. The object crashes to pieces after it hits the road,
and the last detection of the full object is made when it hits the
road. The car decides to stop after this last detection. What is the
distance between the car and object's road-hitting position when
 =3.
the car stops? Assume g=9.8 m/sec2 and 
(a) a = -5 m/sec2, vi = 72 km/h = 72×(1000/3600) m/s = 20 m/s
0 = vf = vi + a×tstop → tstop = 4 sec
dstop = xf - xi = vi×tstop + (1/2)at2stop → dstop = 40 m
tpre-brake = 0.1 s
dcar = vi×tpre-brake + dstop = 2 m + 40 m = 42 m
Δd = dobstacle - dcar = 50 m - 42 m = 8 m
(b) The initial speed of the object is voi = vvan = 54 km/h =
54×(1000/3600) m/s = 15 m/s.
tflight: the flight time of the object


 


Δy = -2 m = -(1/2)gt2flight → tflight =      s



dflight: the horizontal flight distance of the object
dflight = voi×tflight = 15×(2/3) = 10 m


[2] [25 points] A block of mass  on a rough, horizontal
surface is connected to a ball of mass  by a lightweight cord
over a lightweight, frictionless pulley as shown in Figure below.
(a) [15 points] A force of magnitude  at an angle  with the
horizontal is applied to the block as shown, and the block slides
to the right. The coefficient of kinetic friction between the block
and surface is  . Determine the magnitude of the acceleration
of the two objects.
(b) [5 points]
If the block of mass  slides to the left,
determine the magnitude of the acceleration of the two objects.
(c) [5 points] Determine the magnitude of the acceleration of

the two objects if the force  is removed and the surface
becomes frictionless.
예제 5.13
(a) Newton's second law of motion for
m1 gives
Newton's second law of motion for m2
gives
Therefore the normal force and friction are:
Then we obtain
Finally
Th position where the object hits the road is
dobj = 50 + dflight = 50 + 10 = 60 m.
The position of the car is
dcar = vi×(tflight+tpre-brake) + dstop
(b) If the direction of velocity is reversed, the sign of  in the

formula for the acceleration obtained in (a) must be reversed
= 20×(  +0.1) + 40 = 55.3 m (또는 because the kinetic friction opposes the motion. Then the

55 m).
The distance between the car and object's road-hitting position is
Δd = dobj - dcar = 60 – 55.3 = 4.7m (또는 60-55=5m).
acceleration for the reversed motion is
or
(c)          → a= -[m1/(m1+m2)]g or a=[m1/(m1+m2)]g
In this situation, the net force acting on the system of m1 + m2
is the gravitational force on m1. Therefore m1 falls and m2 moves
to the left with the acceleration .
Copyright ⓒ 2018 College of Science, Sungkyunkwan University (SKKU)
SKKU-CS-009
Alternative solutions for 2(b) & 2(c)
(b)
Newton's second law of motion for m1 gives
Newton's second law of motion for m2 gives
Combining equations (2) and (3)' gives
Substituting equation (1)' to equation (4), we obtain
Finally
(c)
         → a=[m1/(m1+m2)]g
Copyright ⓒ 2018 College of Science, Sungkyunkwan University (SKKU)
SKKU-CS-009
[3] [25 pts]
A bullet (with mass  = 0.10 kg) is fired
horizontally into two blocks (which are at rest initially) on a
frictionless table. The bullet passes through block 1 (mass  =
1.00 kg) and embeds itself in block 2 (mass  = 1.25 kg). The
blocks end up with speeds    m/s for block 1 and   
m/s for block 2 as shown in Figure below. The material removed
from block 1 by the bullet is negligible.
[4] [25 pts] A star rotates with a period of 10 days about an axis
through its center. The period is the time interval required for
one complete revolution. After the star undergoes a supernova
explosion, the stellar core, which had a radius of 10,000 km,
collapses into a neutron star of radius 10 km.
(a) [15 pts] Determine the period of rotation of the neutron star.
(Hint: 10 days = 864,000 seconds. You may assume that the
moment of inertia of a stellar core of mass  and radius  is

     , and that the mass of the core is not changed by the

explosion.)
(b) [10 pts] Calculate  , where  and  are the rotational
kinetic energies of the core before and after the explosion.
(a) [15 pts] Find the speed of the bullet before entering block 2.
예제 11.7
(b) [10 pts] Find the initial speed of the bullet before entering
block 1.
[Solution of 4-a, 15 pts] 0.864 seconds.
The law of conservation of angular momentum says that
        

  ,

Since


          (10 pts).



or
T
is

 
      

 

the
period,
 ×  sec   sec
pts)
[Solution of 3-b, 10 pts]
Let   be the initial speed of the bullet. Then the momentum
conservation law for block 1 reads
                ,


 
           , thus we have

have
. ( 5

[Solution of 4-b, 10 pts]        (10 pts, 3 pts for the
answer "the kinetic energy is increased", and 5 pts for the correct
answer but without any proper explanation).

         

 
          ×       m/s.

 
thus

[Solution of 3-a, 15 pts]
Let  be the speed of the bullet as it enters block 2. Then momentum
conservation says
        where m is the mass of the bullet.
We thus have
  
   
       ×    m/s

 
we

 


                    


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Copyright ⓒ 2018 College of Science, Sungkyunkwan University (SKKU)
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