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4. Momentum and Impulse The linear momentum of an object is defined as the product of its mass and its velocity. The conservation law of momentum is another most important physical laws in physics. 1 Momentum (动量) The (linear) momentum of an object is defined to be the product of the mass and velocity: p mv Momentum is a vector quantity and its direction is the same as that for velocity; And it has dimension ML/T. In SI system, the momentum has the units kg·m/s. p mv , x, y , z . 2 Momentum and Force As pointed out before, the Newton’s second law can be rewritten as: dp F dt From above equation, we see that if the net force on an object is zero, the time derivative of the momentum is zero, and thus the momentum of the object must be constant. Of course, if the particle is isolated, then no forces act on it and the momentum remains unchanged—this is Newton’s first law of motion. 3 Momentum and Isolated Systems The momentum of an isolated system is a constant. pt pi The total momentum i for an isolated system Ft 0 Thus, we have dpt 0 p p =Const t i dt i pt Const, x, y, z. The law of conservation of linear momentum! (动量守恒定律) 4 Example 4.1 A 10000-kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed afterward? Solution: The initial momentum is simply m1v1. After the collision, the total momentum will be the same. Since the twe two cars attached, they will have the same speed v '. Then: m1 m2 v ' m1v1 , which gives out 1 v ' v1 12 m/s . 2 5 Impulse and Momentum Assuming a net varying force acts on a particle, then dp Ft dp Ft dt dt the change in the momentum of the particle during the time interval Δt = tf − ti reads: tf p p f pi Ft dt ti The Impulse of a force is defined as: tf I Ft dt p ti impulse-momentum theorem (冲量-动量定理) Also valid for a system of particles. 6 Example 4.2 Calculate the final speed of a 110 kg football player initially running at 8.00 m/s who collides head on with a padded goalpost and experiences a backward force of 1.76 104 N for 5.5 102 s. 7 The Impulse is an interaction between the system and its environment. As a result of this interaction, the momentum of the system changes. The impulse approximation: We assume that one of the forces exerted on a particle acts for a short time but is much greater than other force present. this simplification model allows us to ignore the effects of other forces, because these effects are be small during the short time during which the large force acts. 8 Collisions When two objects collide, it is a good approximate in many cases to assume that the forces due to the collision are much larger than any external forces present, so we can use the simplified model: the impulse approximation. Collisions Elastic collision Inelastic collision Completely inelastic collision 9 Momentum is conserved in all cases, but kinetic energy is conserved only in elastic collisions! (动量在所有的碰撞过程中守恒,而动能仅在弹性碰撞中守恒) 10 Example 4.3 A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3.50 10 m/s and makes a glancing collis-ion with the second proton, as in the figure. After the collision, one proton moves off at angle of 37.0 to the original direction of motion, and the second deflects at an angle of to the same axis. Find the final speeds of the two protons and the angle . 5 11 x component: m1v1i 0 m1v1 f cos m2v2 f cos y component: 0 0 m1v1 f sin m2v2 f sin For the collision is elastic, we get the third equation for conservation of kinetic energy: 1 1 1 2 2 2 m1v1i m1v1 f m2v2 f 2 2 2 Solve the set of equations composed of above three equations, we obtain: v1 f sin tan v1 f cos v1i 2 m 1 1 m1v12i m1v12f 1 v12i v12f 2v1i v1 f cos 2 2 2m2 12 The Centre of Mass The center of mass of a system of particles is defined as: rCM m r i i i xCM i yCM j zCM k M i mi xi j mi yi k mi zi i For an extended object reads: i i M rCM 1 rdm M 13 The centre of mass of a homogeneous, symmetric body must lie on an axis of symmetry. The centre of mass is different from its centre of gravity. Each portion of a system is acted on by the gravitational force. The net effect of all of these forces is equivalent to the effect of a single force Mg acting at a special point called the center of gravity. The centre of gravity is the average position of the gravitational force on all parts of the object. If g is uniform over the system, the centre of gravity coincides with the centre of mass. In most cases, for objects or systems of reasonable size, the two points can be considered to be coincident. 14 E.g. 0.1 A system consists of three particles located at the corners of a right triangle as in the figure. Find the centre of mass of the system. 15 E.g. 0.2 A rod of length 30.0 cm has a linear density: 50.0 g/m 20.0 x g/m 2 where x is the distance from one end, measured in meters. (a)What is the mass of the rod? (b)How far from the x = 0 end is its center of mass? 16 Motion of a System of Particles rCM MaCM mi ri i M dpt mi ai Ft dt i dri mi drCM dt i vCM dt M MvCM dri mi pt dt i Newton’s second law for a system of particles 17 The sum of all the forces acting on the system is equal to the total mass of the system times the acceleration of its center of mass. The center of mass of a system of particles (or bodies) with total mass M moves like a single particle of mass M acted upon by the same net external force. The general motion of a system of particles can be considered as the sum of the translational motion of the CM, plus rotational, vibrational, or other types of motion about the CM. 18 Motion of a System of Particles rCM MaCM mi ri dri mi drCM dt i vCM dt M i M dpt mi ai Ft dt i MvCM dri mi pt dt i 质心运动定律 19 Outline of Rotational Motion 20 Rigid body model: A rigid body is any system of particles in which the particles remain fixed in position with respect to one another. Rotation about a fixed axis: Every particles on a rigid body has the same angular speed and angular acceleration. d lim t t dt d lim t t dt rigid body 刚体 rotation about a fixed axis 定轴转动 21 Rotational kinematics d d 0 , , ; dt dt t t t0 t0 0 dt , 0 dt. Rotations with constant angular acceleration Const =0 t 1 2 0 0 t t 2 22 Rotational and Translational Quantities ds d tangential velocity: v r v r . dt dt dv d tangential acceleration: a r a r . dt dt 2 v 2 2 radial acceleration: ac r ac r . r 23 24 Rotational Kinetic Energy (转动动能) I mi ri the moment of inertial 2 i 1 rotational kinetic energy K R I 2 2 For an extended system: I lim mi 0 r m r dm r dV 2 i 2 2 i i 25 The Rigid Body under a Net Torque The net torque acting on the rigid body is proportional to its angular acceleration. t I (转动定律) 26 The Rigid Body in Equilibrium The torque vector (力矩) r F Two conditions for complete equilibrium of an object: Ft 0; t 0 translational equilibrium rotational equilibrium 27 Work and Energy in Rotational Motion Work done by a torque dW d Id dW The power of a torque dt f f i i W d 1 2 1 2 I d I f I i . 2 2 28 Angular Momentum The angular momentum of the particle relative to the origin is defined as: Lrp dL dp r dt dt 29 Conservation of Angular Momentum (角动量守恒) The total angular momentum of a system remains constant if the net external torque acting on the system is zero. 30 Eg. 01 In the classical model of the hydrogen atom, the electron (mass m) is held in a circular orbit about the nucleus (a proton) by the electric force on it. Show the angular momentum of the electron with respect to the necleus is conserved. 31 32 一个刚体的运动,可以视为二种运动所组成,即质量 中心,受到所有外力作用所引起的运动,加上物体在外力作用 下,绕质心的转动。一个刚体的动能,系质心的平移运动的动 能,加上绕质心运动的转动动能。 引自《古典动力学》,吴大猷 刚体的运动总可以分解为质心的平动刚体的转动的合成。 33 A Brief Summary of Part I ① Kinematics of a Particle ② Newton’s Laws of Motion ③ Work and Energy ④ Momentum and Impulse ⑤ Motion of a System of Particles ⑥ Rotations of a Rigid Body about a Fixed Axis 34 The End 35 Problems P255 5, 21, 40, 47, 54, 70, 74 36