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Chapter 9 Center of Mass and Linear Momentum In this chapter we will introduce the following new concepts: -Center of mass (com) for a system of particles -The velocity and acceleration of the center of mass -Linear momentum for a single particle and a system of particles We will derive the equation of motion for the center of mass, and discuss the principle of conservation of linear momentum. Finally, we will use the conservation of linear momentum to study collisions in one and two dimensions and derive the equation of motion for rockets. (9-1) The Center of Mass : Consider a system of two particles of masses m1 and m2 at positions x1 and x2 , respectively. We define the position of the center of mass (com) as follows: m x m2 x2 xcom 1 1 m1 m2 We can generalize the above definition for a system of n particles as follows: xcom m1 x1 m2 x2 m3 x3 ... mn xn m1 x1 m2 x2 m3 x3 ... mn xn 1 m1 m2 m3 ... mn M M n m x i 1 Here M is the total mass of all the particles M m1 m2 m3 ... mn . We can further generalize the definition for the center of mass of a system of particles in three-dimensional space. We assume that the ith particle ( mass mi ) has position vector ri . rcom 1 M n m r i 1 i i (9-2) i i The position vector for the center of mass is given by the equation rcom 1 M ˆ The position vector can be written as rcom xcom ˆi ycom ˆj zcom k. The components of rcom are given by the equations xcom 1 M n m x i 1 i i ycom 1 M n m y i 1 i i zcom 1 M n m z i 1 i i The center of mass has been defined using the quations given above so that it has the following property: The center of mass of a system of particles moves as though all the system's mass were concentrated there, and that the vector sum of all the external forces were applied there The above statement will be proved later. An example is given in the figure. A baseball bat is flipped into the air and moves under the influence of the gravitation force. The center of mass is indicated by the black dot. It follows a parabolic path as discussed in Chapter 4 (projectile motion). All the other points of the bat follow more complicated paths. (9-3) n m r. i 1 i i Example: A 3.00 kg particle is located on the x axis at x = −5.00 m and a 4.00 kg particle is on the x axis at x = 3.00 m. Find the center of mass of this two–particle system. Example: A 3.00 kg particle is located on the x axis at x = −5.00 m and a 4.00 kg particle is on the x axis at x = 3.00 m. Find the center of mass of this two–particle system. x CM m1x 1 m 2 x 2 (3.00 kg)( 5.00 m) (4.00 kg)(3.00 m) m1 m 2 (3.00 kg 4.00 kg) 0.429 m. The Center of Mass for Solid Bodies : Solid bodies can be considered as systems with continuous distribution of matter. The sums that are used for the calculation of the center of mass of systems with discrete distribution of mass become integrals: 1 1 1 xcom xdm y ydm z zdm com com M M M The integrals above are rather complicated. A simpler special case is that of dm M uniform objects in which the mass density is constant and equal to : dV V 1 1 1 xcom xdV ycom ydV zcom zdV V V V In objects with symmetry elements (symmetry point, symmetry line, symmetry plane) it is not necessary to evaluate the integrals. The center of mass lies on the symmetry element. For example, the com of a uniform sphere coincides with the sphere center. In a uniform rectangular object the com lies at the intersection of the diagonals. C . C (9-4) z F1 m2 m1 m3 F2 x Newton's Second Law for a System of Particles : Consider a system of n particles of masses m1 , m2 , m3, ..., mn F2 O and position vectors r1 , r2 , r3 ,..., rn , respectively. y The position vector of the center of mass is given by Mrcom m1r1 m2 r2 m3 r3 ... mn rn . We take the time derivative of both sides d d d d d M rcom m1 r1 m2 r2 m3 r3 ... mn rn dt dt dt dt dt Mvcom m1v1 m2 v2 m3v3 ... mn vn . Here vcom is the velocity of the com and vi is the velocity of the ith particle. We take the time derivative once more d d d d d vcom m1 v1 m2 v2 m3 v3 ... mn vn dt dt dt dt dt Macom m1a1 m2 a2 m3a3 ... mn an . Here acom is the acceleration of the com M and ai is the acceleration of the ith particle. (9-5) Macom m1a1 m2 a2 m3 a3 ... mn an . z m1 F1 m2 F2 mi ai Fi . Here Fi is the net force on the ith particle, m3 F2 x We apply Newton's second law for the ith particle: O Macom F1 F2 F3 ... Fn . The force Fi can be decomposed into two components: applied and internal: y Fi Fi app Fi int . The above equation takes the form: F ... F F Macom F1app F1int F2app F2int F3app F3int ... Fnapp Fnint Macom app 1 F2app F3app app n int 1 F2int F3int ... Fnint The sum in the first set of parentheses on the RHS of the equation above is just Fnet . The sum in the second set of parentheses on the RHS vanishes by virtue of Newton's third law. The equation of motion for the center of mass becomes Macom Fnet . In terms of components we have: Fnet, x Macom, x Fnet, y Macom, y Fnet, z Macom, z (9-6) Macom Fnet Fnet, x Macom, x Fnet, y Macom, y Fnet, z Macom, z The equations above show that the center of mass of a system of particles moves as though all the system's mass were concentrated there, and that the vector sum of all the external forces were applied there. A dramatic example is given in the figure. In a fireworks display a rocket is launched and moves under the influence of gravity on a parabolic path (projectile motion). At a certain point the rocket explodes into fragments. If the explosion had not occurred, the rocket would have continued to move on the parabolic trajectory (dashed line). The forces of the explosion, even though large, are all internal and as such cancel out. The only external force is that of gravity and this remains the same before and after the explosion. This means that the center of mass of the fragments follows the same parabolic trajectory that the rocket would have followed had it not exploded. (9-7) v m p p mv Linear Momentum : Linear momentum p of a particle of mass m and velocity v is defined as p mv . The SI unit for linear momentum is the kg.m/s. Below we will prove the following statement: The time rate of change of the linear momentum of a particle is equal to the magnitude of net force acting on the particle and has the direction of the force. dp In equation form: Fnet . We will prove this equation using dt Newton's second law: dp d dv p mv mv m ma Fnet dt dt dt This equation is stating that the linear momentum of a particle can be changed only by an external force. If the net external force is zero, the linear momentum cannot change: dp (9-8) Fnet dt Example: A 3.00 kg particle has a velocity of (3.0 i − 4.0 j) m/s. Find its x and y components of momentum and the magnitude of its total momentum. Using the definition of momentum and the given values of m and v we have: p = mv = (3.00 kg)(3.0 i − 4.0 j) = (9.0 i − 12. j) So the particle has momentum components px = +9.0 and py = −12. The Linear Momentum of a System of Particles z m1 p1 m2 p3 linear momentum to a system of particles. The m3 p2 x In this section we will extend the definition of O y ith particle has mass mi , velocity vi , and linear momentum pi . We define the linear momentum of a system of n particles as follows: P p1 p2 p3 ... pn m1v1 m2v2 m3v3 ... mn vn Mvcom . The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity vcom of the center of mass. dP d Mvcom Macom Fnet . dt dt The linear momentum P of a system of particles can be changed only by a The time rate of change of P is net external force Fnet . If the net external force Fnet is zero, P cannot change. P p1 p2 p3 ... pn Mvcom dP Fnet dt (9-9) Collision and Impulse : We have seen in the previous discussion that the momentum of an object can change if there is a nonzero external force acting on the object. Such forces exist during the collision of two objects. These forces act for a brief time interval, they are large, and they are responsible for the changes in the linear momentum of the colliding objects. Consider the collision of a baseball with a baseball bat. The collision starts at time ti when the ball touches the bat and ends at t f when the two objects separate. The ball is acted upon by a force F (t ) during the collision. The magnitude F (t ) of the force is plotted versus t in fig. a. The force is nonzero only for the time interval ti t t f . F (t ) dp . Here p is the linear momentum of the ball, dt tf tf dp F (t )dt dp F (t )dt ti ti (9-10) tf tf tf dp F (t )dt dp p ti ti f pi p = change in momentum ti tf F (t )dt is known as the impulse J of the collision. ti tf J F (t )dt The magnitude of J is equal to the area ti under the F versus t plot of fig. a p J . In many situations we do not know how the force changes with time but we know the average magnitude Fave of the collision force. The magnitude of the impulse is given by J Fave t , where t t f ti . p J J Fave t Geometrically this means that the area under the F versus t plot (fig. a) is equal to the area under the Fave versus t plot (fig. b). (9-11) Example: A child bounces a ball on the sidewalk. The linear impulse delivered by the sidewalk is 2.00 N· s during the 1/800 s of contact. What is the magnitude of the average force exerted on the ball by the sidewalk? I 2.00 3 F 1.60 10 N t 1/ 800 Example: A 3.0 kg steel ball strikes a wall with a speed of 10 m/s at an angle of 60o with the surface. It bounces off with the same speed and angle, as shown in the figure. If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the wall by the ball? Since has no x component, the average force has magnitude 2.6 ×102 N and points in the y direction (away from the wall). Fave Series of Collisions Consider a target that collides with a steady stream of identical particles of mass m and velocity v along the x-axis. A number n of the particles collides with the target during a time interval t. Each particle undergoes a change p in momentum due to the collision with the target. During each collision a momentum change p is imparted on the target. The impulse on the target during the time interval t is J np. J np n mv. t t t Here v is the change in the velocity of each particle along the x-axis due m to the collision with the target Fave v. t m Here is the rate at which mass collides with the target. t If the particles stop after the collision, then v 0 v v. (9-12) If the particles bounce backwards, then v v v 2v. The average force on the target is Fave Conservation of Linear Momentum : z m1 p1 m2 p3 Consider a system of particles for which Fnet 0 dP Fnet 0 P Constant O dt y x If no net external force acts on a system of particles, the total linear momentum p2 m3 P cannot change. total linear momentum total linear momentum at some initial time t at some later time t f i The conservation of linear momentum is an important principle in physics. It also provides a powerful rule we can use to solve problems in mechanics such as collisions. Note 1: In systems in which Fnet 0 we can always apply conservation of linear momentum even when the internal forces are very large as in the case of colliding objects. Note 2: We will encounter problems (e.g., inelastic collisions) in which the (9-13) energy is not conserved but the linear momentum is. Example: A 4.5 kg gun fires a 0.1 kg bullet with a muzzle velocity of +150 m/s. What is the recoil velocity of the gun? Pi Pf m v j j j,i m k v k,f k mGv G ,i mbv b ,i mGv G ,f m bv b ,f 0 0 mGv G ,f m bv b ,f mGv G ,f mbv b ,f v G ,f (0.1)(150) 3.3 m/s. 4 Momentum and Kinetic Energy in Collisions Consider two colliding objects with masses m1 and m2 , initial velocities v1i and v2i , and final velocities v1 f and v2 f , respectively. If the system is isolated, i.e., the net force Fnet 0, linear momentum is conserved. The conservation of linear momentum is true regardless of the collision type. This is a powerful rule that allows us to determine the results of a collision without knowing the details. Collisions are divided into two broad classes: elastic and inelastic. A collision is elastic if there is no loss of kinetic energy, i.e., K i K f . A collision is inelastic if kinetic energy is lost during the collision due to conversion into other forms of energy. In this case we have K f K i . A special case of inelastic collisions are known as completely inelastic. In these collisions the two colliding objects stick together and they move as a single body. In these collisions the loss of kinetic energy is maximum. (9-14) One - Dimensional Inelastic Collisions : In these collisions the linear momentum of the colliding objects is conserved p1i p2i p1 f p2 f . m1v1i m1v2i m1v1 f m1v2 f One - Dimensional Completely Inelastic Collisions : In these collisions the two colliding objects stick together and move as a single body. In the figure to the left we show a special case in which v2i 0. m1v1i m1V m1V m1 V v1i m1 m2 The velocity of the center of mass in this collision is vcom p1i p2i m1v1i P . m1 m2 m1 m2 m1 m2 In the picture to the left we show some freeze-frames of a totally inelastic collision. (9-15) Example: An object of 12.0 kg at rest explodes into two pieces of masses 4.00 kg and 8.00 kg. The velocity of the 8.00 kg mass is 6.00 m/s in the +ve x-direction. The change in the kinetic is: Example: An object of 12.0 kg at rest explodes into two pieces of masses 4.00 kg and 8.00 kg. The velocity of the 8.00 kg mass is 6.00 m/s in the +ve x-direction. The change in the kinetic is: M v m 4v 4 m 8v 8 0 4 v 4 8 6iˆ v 12iˆ, 4 K 432 J 1 1 4 8 2 2 m 4 v 4 m 8 v 8 (12) 2 (6) 2 2 2 2 2 Inelastic collisions in 1D Example: A 10.0 g bullet is stopped in a block of wood (m = 5.00 kg). The speed of the bullet–plus–wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet? Elastic collisions in 1D One - Dimensional Elastic Collisions Consider two colliding objects with masses m1 and m2 , initial velocities v1i and v2i , and final velocities v1 f and v2 f , respectively. Both linear momentum and kinetic energy are conserved. Linear momentum conservation: m1v1i m1v2i m1v1 f m1v2 f 2 2 m1v12i m1v22i m1v1 f m2v2 f Kinetic energy conservation: 2 2 2 2 We have two equations and two unknowns, v1 f and v1 f . (eq. 1) (eq. 2) If we solve equations 1 and 2 for v1 f and v1 f we get the following solutions: m1 m2 2m2 v1 f v1i v2i m1 m2 m1 m2 v2 f 2m1 m m1 v1i 2 v2i m1 m2 m1 m2 (9-16) Example: A 10 kg ball (m1) with a velocity of +10 m/s collides head on in an elastic manner with a 5 kg ball (m2) at rest. What are the velocities after the collision? Example: A 10 kg ball (m1) with a velocity of +10 m/s collides head on in an elastic manner with a 5 kg ball (m2) at rest. What are the velocities after the collision? Special Case of Elastic Collisions - Stationary Target v 2 i = 0 : We substitute v2i 0 in the two solutions for v1 f and v1 f : v1 f v2 f m1 m2 2m2 m m2 v1i v2i v1 f 1 v1i m1 m2 m1 m2 m1 m2 2m1 m2 m1 2m1 v1i v2i v2 f v1i m1 m2 m1 m2 m1 m2 Below we examine several special cases for which we know the outcome of the collision from experience. v1i v2i = 0 1. Equal masses m1 m2 m v1 f v2 f m1 m2 mm v1i v1i 0 m1 m2 mm x m m 2m1 2m v1i v1i v1i m1 m2 mm v1f = 0 v2f x m m The two colliding objects have exchanged velocities. (9-17) v1i 2. A massive target v2i = 0 x m1 m2 v2f v1f m 1 x m2 m2 m1 m1 m2 1 m1 1 m m2 m v1 f 1 v1i 2 v1i v1i m1 m1 m2 1 m2 v2 f m 2 1 m2 m1 2m1 v1i v1i 2 v1i m m1 m2 1 m2 1 m2 Body 1 (small mass) bounces back along the incoming path with its speed practically unchanged. m1 Body 2 (large mass) moves forward with a very small speed because 1. m2 (9-18) 2. A massive projectile m1 v1i m2 v2i = 0 m1 m2 x v1f v2f m1 m2 x m2 m1 m2 m m2 m1 v1 f 1 v1i v1i v1i m m1 m2 1 2 m1 1 v2 f 2m1 2 v1i v1i 2v1i m m1 m2 1 2 m1 Body 1 (large mass) keeps on going, scarcely slowed by the collision. Body 2 (small mass) charges ahead at twice the speed of body 1. (9-19) 1 Collisions in 2-D Collisions in Two Dimensions : In this section we will remove the restriction that the colliding objects move along one axis. Instead we assume that the two bodies that participate in the collision move in the xy -plane. Their masses are m1 and m2 . The linear momentum of the system is conserved: p1i p2i p1 f p2 f . If the system is elastic the kinetic energy is also conserved: K1i K 2i K1 f K 2 f . We assume that m2 is stationary and that after the collision particle 1 and particle 2 move at angles 1 and 2 with the initial direction of motion of m1. In this case the conservation of momentum and kinetic energy take the form: x axis: m1v1i m1v1 f cos 1 m2 v2 f cos 2 (eq. 1) y axis: 0 m1v1 f sin 1 m2v2 f sin 2 (eq. 2) 1 1 1 m1v12i m1v22 f m2v22 f (eq. 3) We have three equations and seven variables: 2 2 2 Two masses: m1 , m2 ; three speeds: v1i , v1 f , v2 f ; and two angles: 1 , 2 . If we know the values of four of these parameters we can calculate the remaining three. (9-20) Example: An unstable nucleus of mass 17 × 10**27 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.0×10**27 kg, moves along the y axis with a speed of 6.0 × 10**6 m/s . Another particle of mass 8.4 × 10**27 kg, moves along the x axis with a speed of 4.0 × 10**6 m/s . Find (a) the velocity of the third particle and (b) the total energy given off in the process. The parent nucleus is at rest , so that the total momentum was (and remains) zero: Pi = 0. Example: A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33 m/s at an angle of 30.0degree with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball’s velocity. Then the condition that the total x momentum be conserved gives us: We can similarly find vy by using the condition that the total y momentum be conserved in the collision. This gives us: A 10.0 kg toy car is moving along the x axis. The only force Fx acting on the car is shown in Fig. 5 as a function of time (t). At time t = 0 s the car has a speed of 4.0 m/s. What is its speed at time t = 6.0 s? (Ans: 8.0 m/s ) A 10.0 kg toy car is moving along the x axis. The only force Fx acting on the car is shown in Fig. 5 as a function of time (t). At time t = 0 s the car has a speed of 4.0 m/s. What is its speed at time t = 6.0 s? (Ans: 8.0 m/s ) dp p Fdt Area under the curve dt 10 (6 2) 40 2 p m v f v i F v f p m v i 40 4 8 m/s 10 An impulsive force Fx as a function of time (in ms) is shown in the figure as applied to an object (m = 5.0 kg) at rest. What will be its final speed? An impulsive force Fx as a function of time (in ms) is shown in the figure as applied to an object (m = 5.0 kg) at rest. What will be its final speed? dp F p Fdt Area under the curve dt 5000 4 103 10 2 p 10 m v 2 m 5 s If the masses of m1 and m3 in Fig. 5 are 1.0 kg each and m2 is 2.0 kg, what are the coordinates of the center of mass? (Ans: (1.00, 0.50) m ) If the masses of m1 and m3 in Fig. 5 are 1.0 kg each and m2 is 2.0 kg, what are the coordinates of the center of mass? (Ans: (1.00, 0.50) m ) x cm m1x 1 m 2 x 2 m 3x 3 1.0 0.0 1.0 2.0 2.0 1.0 1.0 m m1 m 2 m 3 4.0 y cm m1 y 1 m 2 y 2 m 3 y 3 1.0 0.0 2.0 1.0 1.0 0 0.5 m m1 m 2 m 3 4.0 A 1500 kg car traveling at 90.0 km/h east collides with a 3000 kg car traveling at 60.0 km/h south. The two cars stick together after the collision (see Fig 2). What is the speed of the cars after collision? A 1500 kg car traveling at 90.0 km/h east collides with a 3000 kg car traveling at 60.0 km/h south. The two cars stick together after the collision (see Fig 2). What is the speed of the cars after collision? M v m1v 1 m 2v 2 (1500 3000) v 1500 90 iˆ 3000 60 jˆ km v 30 40 50 13.9 m/s hour 2 2 A 2.00 kg pistol is loaded with a bullet of mass 3.00 g. The pistol fires the bullet at a speed of 400 m/s. The recoil speed of the pistol when the bullet was fired is: (Ans: 0.600 m/s) A 2.00 kg pistol is loaded with a bullet of mass 3.00 g. The pistol fires the bullet at a speed of 400 m/s. The recoil speed of the pistol when the bullet was fired is: (Ans: 0.600 m/s) 0 m1v 1 m 2v 2 m 0 2 v 0.003 400 v 0.6 . s A 1.0 kg ball falling vertically hits a floor with a velocity of 3.0 m/s and bounces vertically up with a velocity of 2.0 m/s . If the ball is in contact with the floor for 0.10 s, the average force on the floor by the ball is: A 1.0 kg ball falling vertically hits a floor with a velocity of 3.0 m/s and bounces vertically up with a velocity of 2.0 m/s . If the ball is in contact with the floor for 0.10 s, the average force on the floor by the ball is: The change of momentum of the particle, p M v m v f v i (1) 2 (3) 5 The average force on the floor by the ball F p 5 50 N (down) t .1 A 20-g bullet is fired into a 100-g wooden block initially at rest on a horizontal frictionless surface. If the initial speed of the bullet is 10 m/s and it comes out of the block with a speed of 5.0 m/s, find the speed of the block immediately after the collision. A 20-g bullet is fired into a 100-g wooden block initially at rest on a horizontal frictionless surface. If the initial speed of the bullet is 10 m/s and it comes out of the block with a speed of 5.0 m/s, find the speed of the block immediately after the collision. Consrevation of momentum m bv 1 m bv f m B V m b v 1 v f V mB 0.02 (10 5) 1.0 m/s 0.1 A 1.0-kg block at rest on a horizontal frictionless surface is connected to a spring (k = 200 N/m) whose other end is fixed (see figure). A 2.0kg block moving at 4.0 m/s collides with the 1.0-kg block. If the two blocks stick together after the one-dimensional collision, what maximum compression of the spring does occur when the blocks momentarily stop? A 1.0-kg block at rest on a horizontal frictionless surface is connected to a spring (k = 200 N/m) whose other end is fixed (see figure). A 2.0kg block moving at 4.0 m/s collides with the 1.0-kg block. If the two blocks stick together after the one-dimensional collision, what maximum compression of the spring does occur when the blocks momentarily stop? Consrevation of momentum mv ( m M )V 2 4 m 2.67 (2 1) s 1 1 Conservation of K.E. after collision ( m M )V 2 k x 2 2 2 (m M ) 3 x V 2.67 0.33 m k 200 V