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 Page 2/2 Copyright ⓒ 2018 College of Science, Sungkyunkwan University (SKKU)
