AP Physics Chapter 9 Center of Mass and Linear Momentum 1 AP Physics Lecture: Chapter 9 Q&A 2 Standards: (AP Physics C: Mechanics) Newtonian Mechanics D. Systems of particles, linear momentum (12%) 1. Center of mass 2. Impulse and momentum 3. Conservation of linear momentum, collisions 3 System • A system is a collection of objects. • A system includes the objects, but not the space in between them. • In the system All objects that belong to the system Not necessary the space enclosed by the system 4 Center of Mass Center of mass (of a body or a system of bodies): the point that moves as though all of the mass were concentrated there and all external forces were applied there. Position of Center of Mass: 1D: x cm ˆ ˆ ˆ r x , y , z x i y j z k 3D: cm cm cm cm cm cm cm 5 Position of Center of Mass: xcm (2 objects, 1D) x cm m1 x1 m 2 x 2 m1 m 2 xcm: position of center of mass of two objects m1: mass of Object 1 m2: mass of Object 2 x1: position of (center of mass of) Object 1 x2: position of (center of mass of) Object 2 • x1 m1 m2 • x2 6 Position of Center of Mass: rcm (n objects, 3D) x cm y cm z cm 1 M 1 M 1 M n i 1 n 1 m i y i rcm M i 1 n m i z i i 1 mx i M m1 m 2 i n mr i i i 1 m n m i T o tal M ass 7 Position of Center of Mass: (Continuous Mass Distribution, 3D) x cm y cm z cm 1 M 1 M 1 M xdm ydm zdm uniform density 1 x cm V 1 y cm V 1 z cm V xdV ydV zdV V: volume of object 8 Example: Pg238-102 The distance between the centers of the carbon (C) and oxygen (O) atoms in a carbon monoxide (CO) gas molecule is 1.131 10-10 m. Locate the center of mass of a CO molecule relative to the carbon atom. (Find the masses of C and O in Appendix D.) C O x 0 Let C = 1, O = 2, then m1 = 12.01115u, m2 = 15.9994u. Also, let x = 0 at C, then x1 = 0, and x2 = 1.131 10-10m. xcm = ? x cm m 1 x1 m 2 x 2 m1 m 2 15.9994 u 1.131 10 10 12.01115 u 15.9994 u m 6.46 10 11 m The center of mass is 6.46 10-11 m from the carbon atom, in between the two atoms. Center of mass is closer to the object with larger mass. 9 Example: Pg229-3 In Fig. 9-37, three uniform thin rods, each of length L = 22 cm, form an inverted U. The vertical rods each have a mass of 14 g; the horizontal rod has a mass of 42 g. a) What are the x coordinate and b) the y coordinate of the system’s center of mass? Set up coordinates, then m3 y m1 = 14g at (0, 11cm) m2 = 14g at (22cm, 11cm) m1 • cm m 2 m3 = 42g at (11cm, 22cm) x cm y cm m 1 x1 m 2 x 2 m 3 x 3 m1 m 2 m 3 m 1 y1 m 2 y 2 m 3 y 3 m1 m 2 m 3 x 14 g 0 14 g 22 cm 42 g 11cm 14 g 14 g 42 g 11cm 14 g 11cm 14 g 11cm 42 g 22 cm 14 g 14 g 42 g 17.6 cm rcm 11cm ,17.6 cm Symmetric? 10 Newton’s Second Law for a System of Particles Fext M a cm • Fext: external force, force from object not included in the system • M: total mass of system • acm: acceleration of center of mass 3D Fext , x M a cm , x Fext , y M a cm , y Fext , z M a cm , z 11 Special Cases and Application What if the total external force is zero? acm = 0 vcm = constant If vcm, i = 0 vcm remains 0 CM does not move. 12 Velocity of Center of Mass, vcm x cm m 1 x1 m 2 x 2 m1 m 2 v cm m1 v1 m 2 v 2 m1 m 2 13 Example: Pg238-104 An old Chrysler with mass 2400 kg is moving along a straight stretch of road at 80 km/h. It is followed by a Ford with mass 1600 kg moving at 60 km/h. How fast is the center of mass of the two cars moving? Let m1 = 2400kg, v1 = 80km/h, m2 = 1600kg, v2 = 60 km/h vcm = ? v cm km 2400 kg 80 1600 kg m 1 v1 m 2 v 2 h m1 m 2 2400 kg 1600 kg km 60 h 72 km h 14 Practice: Pg240-138 Two particles P and Q are initially at rest 1.0 m apart. P has a mass of 0.10 kg and Q a mass of 0.30 kg. P and Q attract each other with a constant force of 1.0 10-2 N. No external forces act on the system. a) Describe the motion of the center of mass. b) At what distance from P’s original position do the particles collide? Let x1 = 0, then x2 = 1m. Also, m1 = 0.10 kg, m2 = 0.30kg. a) Fext = 0 CM stays at rest. b) What are we actually looking for? What happens to the center of mass? x cm ? x cm m1 x1 m 2 x 2 m1 m 2 0.10 kg 0 0.30 kg 1m 0.10 kg 0.30 kg 0.75 m 15 Newton’s Second Law F ma m dv dt d mv dP dt dt Define: (Linear) Momentum: F dP dt P mv P is a vector. Unit: Same direction as v. P m v kg m s 16 Momentum of System of Particles P M v cm m1 v1 m 2 v 2 ... F ext M a cm dP dt 17 Practice: Pg230-18 A 0.70 kg ball is moving horizontally with a speed of 5.0 m/s when it strikes a vertical wall. The ball rebounds with a speed of 2.0 m/s. What is the magnitude of the change in linear momentum of the mass? Let toward the wall be the positive direction, then m = 0.70kg, vi = 5.0m/s, vf = -2.0 m/s |P| = ? P P f Pi m v f m v i m v f v i m m m 0.70 kg 2.0 5.0 4.9 kg s s s P 4.9 kg m s 18 Closed, isolated System Closed System: n = constant n = constant No particles leave or enter the system. Isolated System: Fext = 0 The sum of the external forces acting on the system of particles is zero. There could be external forces acting on the system, but the net external force is zero. 19 Law of Conservation of Linear Momentum In closed and isolated system: P = constant Pi = Pf P: total momentum of system P1i + P2i = P1f + P2f If a component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change. 20 Example: Pg232-35 A 91 kg man lying on a surface of negligible friction shoves a 68 g stone away from himself, giving it a speed of 4.0 m/s. What speed does the man acquire as a result? m 1 91kg , m 2 68 g 0.068 kg , v1 i v 2 i 0, v 2 f 4.0 m s v1 f ? Pi P f P1i P2 i P1 f P2 f 0 m1 v1 f m 2 v 2 f v1 f m 2 v2 f m1 0.068 kg 4.0 9 1kg m s 2.9 10 3 m s Notice the negative sign. When we define v2 = 4.0 m/s, we already define the direction of v2f to be the positive direction. 21 Practice: Pg232-37 A space vehicle is traveling at 4300 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 82 km/h relative to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to earth after the separation? Motor = 1, Module = 2, v1i =v2i= 4300 km/h, m1 = 4m2, v1f – v2f = -82 km/h v1f = v2f – 82 km/h = v2f - vR, v2f = ? Pi P f m 1 m 2 v i m 1 v1 f m 2 v 2 f m1 m 2 v i m1 v 2 f v R m 2 v 2 f m1 m 2 v 2 f m1v R m1 m 2 v 2 f m1 m 2 v i m1v R v2 f m1 m 2 v i m1 v R m1 m 2 5 vi 4 v R 5 4m2 m 2 vi 4 m 2 v R 4m2 m2 5 4300 km / h 4 82 km / h 5 5 m 2 vi 4 m 2 v R 5m2 4.4 10 km / h 3 22 Practice A railroad flatcar of mass M can roll without friction along a straight horizontal track. Initially, a man of mass m is standing on the car, which is moving to the right with speed v0. What is the change in velocity of the car if the man runs to the left so that his speed relative to the car is vrel? Man’s motion Flatcar’s motion 23 Solution m1 = M, m2 = m, v1i = v2i = vo, v2f = v1f - vrel, v1=v1f - v1i = ? Pi P f m1 m 2 vi m 1 v1 f m 2 v 2 f m 1 m 2 v o m 1 v1 f m 2 v1 f v rel v1 f m1 m 2 v o m 2 v rel m1 m 2 v1 v1 f M m 1 m 2 v1 f m v o m v rel M m v rel v0 v1 i v o M m m m 2 v rel vo m v rel M m m v rel M m 24 Practice: Pg236-85 The last stage of a rocket is traveling at a speed of 7600 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 290.0 kg and a payload capsule with a mass of 150.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 910.0 m/s. a) What are the speeds of the two parts after they have separated? Assume that all velocities are along the same direction. b) Find the total kinetic energy of the two parts before and after they separate; account for any difference. 25 Solution: Pg236-85 case 1, capsule 2, m1 = 290.0kg, m2 = 150.0kg, v1i=v2i=vi = 7600 m/s, v2f – v1f = vR = 910.0 m/s a ) v1 f ?, v 2 f ? Pi P f m1 m 2 v i v1 f m 1 v1 f m 2 v 2 f m 1 v1 f m 2 v1 f v R m 1 m 2 v1 f m2vR m1 m 2 v i m 2 v R m1 m 2 m m 290.0 kg 150.0 kg 7600 150.0 kg 910.0 m s s 7290 s 290.0 kg 150.0 kg v 2 f v1 f v R 7290 m s 910.0 m s 8200 m s 26 Solution: Pg236-85 (2) b) Ki = ?, Kf = ? Ki K f 1 2 m1 m 2 v i 2 K1 f K 2 f 1 2 2 m 290.0 kg 150.0 kg 7600 1.271 10 10 J 2 s 1 m1 v1 f 2 2 1 2 m 2 v2 f 2 2 m 1 m 10 290.0 kg 7290 150.0 kg 8200 1.275 10 J 2 s 2 s 1 K f Ki The increase in the KE comes from the spring, when the spring does positive work to separate the two parts. 27 System with Varying Mass: Rocket Rocket accelerates forward by burning and ejecting fuel backward. Mass of rocket keeps changing. Newton’s Second Law and Conservation of Momentum still apply, though more complicated. 28 First Rocket Equation R v rel M a R = |dM/dt|, rate at which rocket losing mass (fuel) vrel: exhaust speed, backward speed of fuel relative to the rocket M: mass of rocket at that moment a: acceleration of rocket at that moment 29 Second Rocket Equation v f v i v rel ln Mi M f vf: final speed of rocket after some fuel consumption vi: initial speed of rocket vrel: exhaust speed of fuel Mi: initial mass of rocket Mf: final mass of rocket after ejecting some fuel 30 Example: Pg238-112 A rocket is moving away from the solar system at a speed of 6.0 103 m/s. It fires its engine, which ejects exhaust with a velocity of 3.0 103 m/s relative to the rocket. The mass of the rocket at this time is 4.0 104 kg, and its acceleration is 2.0 m/s2. a) What is the thrust of the engine? b) At what rate, in kilograms per second, was exhaust ejected during the firing? vi = 6.0 103 m/s, vrel = 3.0 103 m/s, M = 4.0 104 kg, a = 2.0 m/s2. a)F ? m 4 m a 4.0 10 kg 2.0 8.0 10 N s2 F 4 b)R ? Rv rel M a m 4 4 .0 1 0 kg 2 .0 2 Ma kg s R 27 m v rel s 3 3 .0 1 0 2 s 31 Practice: P238-114 During a lunar mission, it is necessary to increase the speed of a spacecraft by 2.2 m/s when it is moving at 400 m/s. The exhaust speed of rocket engine is 1000 m/s. What fraction of the initial mass of the spacecraft must be burned and ejected for the increase? vi = 400 m/s, v = 2.2 m/s,vrel = 1000 m/s |M/Mi| = ? v v f v i v rel ln Mi M e v rel M f v ln v rel 2.2 m / s M i M f M i 1.0022 M e 1000 m / s 1.0022 f f M Mi v Mi M Mi M f Mi Mi M f 1.0022 M 1.0022 M f f 0.0022 M 1.0022 M f 0.0022 f 0 .0 0 2 2 32 Impulse Newton’s Second Law F ma m v t mv t mv t P t F t P Area under curve of F-t Define Impulse: J F t F t J P P f Pi F dt t Impulse-Momentum Theorem 33 Example: A cue stick strikes a stationary pool ball, exerting an average force of 50 N over a time of 10 ms. If the ball has mass 0.20 kg, what speed does it have after impact? F 50 N , t 10 10 3 s , m 0 . 20 kg , v i 0 vf ? J P P f Pi F t m v f m vi m v f vf F t m 50 N 10 10 s 3 0.20 kg 2.5 m s 34 Practice: A ball of mass m and speed v strikes a wall perpendicularly and rebounds in the opposite direction with the same speed. a) If the time of collision is t, what is the average force exerted by the ball on the wall? b) Evaluate this average force numerically for a rubber ball with mass 140 g moving at 7.8 m/s; the duration of the collision is 3.8 ms. m , vi v , v f v a )t, F ? F t P f Pi m v f m v i m v m v 2 m v F 2m v t 2mv t 3 b ) m 0.140 kg , v 7.8 m / s , t 3.8 10 s F ? m 2 0.140 kg 7.8 2mv s F 570 N 3 t 3.8 10 s 35 Practice: Pg231-26 A 1.2 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an initial speed of 10 m/s. a) What impulse acts on the ball during the contact? b) If the ball is in contact with the floor for 0.020 s, what is the average force exerted on the floor? U pw ard , m 1.2 kg , v i 25 m s , v f 10 m s a)J ? J P P f Pi m v f m v i m v f v i m m m 1.2 kg 10 25 42 kg s s s b ) t 0.020 s , F ? J F t F J t 42 kg m s 2100 N 0.020 s 36 Series of Collisions F t t v Impulse-Momentum Theorem Average force on body being hit F : m m : Rate at which mass collides with the body v : Change in velocity of hitting bodies 37 Example: Pg237-99 A pellet gun fires ten 2.0 g pellets per second with a speed of 500 m/s. The pellets are stopped by a rigid wall. a) What is the momentum of each pellet? b) What is the kinetic energy of each pellet? c) What is the average force exerted by the stream of pellets on the wall? d) If each pellet is in contact with the wall for 0.6 ms, what is the average force exerted on the wall by each pellet during contact? Why is this average force so different from the average force calculated in (c)? 38 Solution: Pg237-99 n 10 , m 2 . 0 10 3 kg , t 1 s , v i 500 m / s , v f 0 m m 3 a ) P m v 2.0 10 kg 500 1.0 kg s s 2 m 2 3 b ) K m v 2.0 10 kg 500 250 J 2 2 s 1 c)F 1 m t v d ) t 0.6 10 3 nm t v f vi nm v i t 10 2.0 10 m kg 500 s 10 N 1s 3 s, F ? J P F t P f Pi mv f mv i mv i F m vi t m kg 5 0 0 s 3 1 .7 1 0 N 3 0 .6 1 0 s 2 .0 1 0 3 1.7 10 N 3 Most of the time no bullet is in contact with the wall, average is much smaller. 39 Practice: A movie set machine gun fires 50 g bullets at a speed of 1000 m/s. An actor, holding the machine gun in his hands, can exert an average force of 180 N against the gun. Determine the maximum number of bullets he can fire per minute while still holding the gun steady? m 0.050 kg , v i 0, v f 1000 m , F 180 N , t 60 s s n? F n m t F t mv f v nm t v f vi 180 N 60 s m 0 .0 5 0 kg 1 0 0 0 s nm t vf 216 40 Collisions Total momentum is always conserved. Pi = Pf Total kinetic energy: Conserved Elastic Collision: Ki = Kf Not conserved Inelastic Collision Extreme case: Stick and move together Completely Inelastic Collision (No other collision loses more portion of the kinetic energy than this.) 41 Elastic Collisions in 1D Conservation of P: Conservation of K: m1 v1i m 2 v 2 i m1 v1 f m 2 v 2 f 1 2 m 1 v1i 2 m1 m 2 2m2 v v v2i 1i 1f m1 m 2 m1 m 2 v 2 m1 v m 2 m1 v 2 f 1i 2i m m m m 1 2 1 2 1 2 m 2 v2i 2 if v2i = 0 1 2 m 1 v1 f 2 1 2 m 2v2 f 2 m1 m 2 v1 f m m v1 i 1 2 v 2 m1 v 2 f 1i m1 m 2 42 Special Cases (of v2i = 0) Equal masses: m1 = m2 v1 f 0 v 2 f v1i Massive target: m2 >> m1 v1 f v1i v 2 f 0 Light target: m2 << m1 v1 f v1i v 2 f 2 v1 i Motion of CM: v cm m 1 v1 i m 2 v 2 i m1 m 2 m1 m1 m 2 v1 i m1 m 2 v1 f m m v1 i 1 2 v 2 m1 v 2 f 1i m1 m 2 v cm P m 43 Example: A hovering fly is approached by an enraged elephant charging at 2.1 m/s. Assuming that the collision is elastic, at what speed does the fly rebound? Note that the projectile (the elephant) is much more massive than the stationary target (the fly). E lephant 1, F ly 2, v1 i 2.1 m s , m1 m 2 v2 f ? v 2 f 2 v1i 2 2.1 m s 4.2 m s Notice that the elephant is chosen as mass 1 because of the formula we have. 44 Practice: Pg234-55 A cart with mass 343 g moving on a frictionless linear air track at an initial speed of 1.2 m/s strikes a second cart of unknown mass at rest. The collision between the carts is elastic. After the collision, the first cart continues in its original direction at 0.66 m/s. a) What is the mass of the second cart? b) What is its speed after impact? c) What is the speed of the two-cart center of mass? 45 Solution: Pg234-55 m1 0.340 kg , v1i 1.2 m s , v 2 i 0, v1 f 0.66 m s a )m2 ? v1 f m1 m 2 m1 m 2 v1i m1 v1 f m 2 v1 f m 1 v1i m 2 v1i m 2 v1 f m 2 v1i m 1 v1i m1 v1 f m 2 v1i v1 f m2 b )v2 f c ) v cm m 1 v1 i v1 f v1 i v1 f m v 1 1i v1 f m m 0.340 kg 1.2 0.66 s s 0.099 kg 99 g m m 1.2 0.66 s s 2 0.340 kg m m v1i 1.2 1.9 m1 m 2 0.340 kg 0.099 kg s s 2 m1 m 0 .3 4 0 kg 1 .2 m 1 v1 i m s 0 .9 3 m1 m 2 0 .3 4 0 kg 0 .0 9 9 kg s 46 Practice: Pg234-59 A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with onefourth of its original speed. a) What is the mass of the struck body? b) What is the speed of the two-body center of mass if the initial speed of the 2.0 kg body was 4.0 m/s? 47 Solution: Pg234-59 m 1 2 . 0 kg , v1i v , v1 f v / 4 , v 2 i 0 a )m2 ? v1 f m1 m 2 v1i m1 m 2 v1 f v1i m1 m 2 m1 m 2 4 m1 m 2 m1 m 2 v/4 v 1 4 4 m1 4 m 2 m1 m 2 4 m1 m1 m 2 4 m 2 3 m1 5 m 2 m 2 b ) v1i 4.0 v cm m s 3 5 m1 3 5 2.0 kg 1.2 kg , v cm ? m1 m1 m 2 v1i m m 4.0 2.5 2.0 kg 1.2 kg s s 2.0 kg 48 Practice A platform scale is calibrated to indicated the mass in kilograms of an object placed on it. Particles fall from a height of 3.5 m and collide with the platform of the scale. The collisions are elastic; the particles rebound upward with the same speed they had before hitting the pan. If each particle has a mass of 110 g and collisions occur at the rate of 42 s-1, what is the average scale reading? 49 Solution: up , y ground 0, y 3.5 m , v f v i , v 2 f 0, m 0.110 kg , n 1 t 42 s , F ? 1 m gy 2 m m v i 2 gy 2 9.8 2 3.5 m 8.28 s s m vi v f 8.28 2 m s m m v v f v i 8.28 8.28 16.56 s s s y=0 m F m t v nm t v n t m v 42 1 s 0.110 kg 16.56 m 76.5 N s But that is force on scale. By Newton’s Third Law, the force of scale on particle is 76.5N, and this is a normal force. Remember that scale reading (apparent weight) is the normal force. m read N g 76.5 N 9.8 m s 7.8 kg 2 50 Inelastic Collision P is conserved. K is not conserved (lost). Where does the energy go? – – – Sound Heat Deformation, change in shape 51 Completely Inelastic Collision in 1D If m1, initially moving with v1i, collides with m2, initially moving at v2i, and the two stick and move together with a combined velocity of Vf, then Pi P f P1i P2 i P1 f P2 f m 1 v1 i m 2 v 2 i m 1 m 2 V f Vf m1 v1i m 2 v 2 i m1 m 2 Reverse CIC: Decay and explosion Similar analysis 52 Example: A 6.0 kg box sled is coasting across frictionless ice at a speed of 9.0 m/s when a 12 kg package is dropped into it from above. What is the new speed of the sled? Motion on horizontal direction only. m 1 6.0 kg , v1i 9.0 m s , m 2 12 kg , v 2 i 0 vf ? m 6.0 kg 9.0 m s m 1 v1i 3.0 vf 6.0 kg 12 kg s m1 m 2 53 Practice: Pg240-136 A 3000 kg weight falls vertically through 6.0 m and then collides with a 500 kg pile, driving it 3.0 cm into bedrock. Assuming that the weight-pile collision is completely inelastic, find the magnitude of the average force on the pile by the bedrock during the 3.0 cm descent. dow n , m 1 3000 kg , y1 6.0 m , m 2 500 kg , v 2 i 0, d 0.030 m , 1 F ? E1 E 2 m gy1 1 2 m v2 2 v2 2 gy1 m m 2 9.8 2 6.0 m 10.8 s s m 3000 kg 10.8 m 1 v1 i m s V3 9.26 m1 m 2 3000 kg 500 kg s W K K 4 K 3 K 3 m1 m 2 g d F d F m1 m 2 V 3 2 2d 1 2 m1 m 2 V 3 2 y=0 2 3 4 m1 m 2 g 2 m 3000 kg 500 kg 9.26 m s 6 3000 kg 500 kg 9.8 2 5.0 10 N 2 0.030 m s 54 Practice: Pg234-54 In Fig. 9-62, block 2 (mass 1.0 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 200 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 2.0 kg), traveling at speed v1 = 4.0 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed? 1 v1 2 55 Solution: Pg234-54 m 1 2.0 kg , m 2 1.0 kg , k 200 N / m , v1 i 4.0 x m ax ? m s , v2i 0 m 2 .0 kg 4 .0 m 1 v1 m s vf 2 .6 7 m1 m 2 1 .0 kg 2 .0 kg s CIC: Conservation of E during compression of spring: Ei E f 1 2 Mv f x m ax 2 1 2 M k kx m ax vf 2 1.0 kg 2.0 kg m 2.67 0.33 m 200 N / m s 56 Decay Reverse process of completely inelastic collision. – Same analysis using conservation of momentum. Momentum is conserved. Kinetic energy is not conserved, but total massenergy is conserved. Gain in KE is given by loss in mass. – – – Energy released: Q = ΔK = -Δm c2 atomic mass unit: u = 1.66 10-27 kg 1 eV = 1.60 10-19 J 57 Example: Pg237-75E A particle called - (sigma minus) is initially at rest and decays spontaneously into two other particles according to - - + n. The masses are m = 2340.5 me, m = 273.2me, mn = 1838.65me, Where me (9.1110-31 kg or 0.511 MeV/c2) is the electron mass. a) How much energy is transferred to kinetic energy in this process? b) How do the linear momenta of the decay products (and n) compare? c) Which product gets the larger share of the kinetic 58 energy? Solution: Pg237-75E m 2340 . 5 m e , m 273 . 2 m e , m n 1838 . 65 m e a )K ? K m c m m m n c 2 2 2340.5 m e 228.65 m e c 228.65 0.511 M eV / c 2 2 c 2 273.2 m e 1838.65 m e c 117 M eV b) The two particles should have momentum equal in magnitude and opposite in direction. c) K 1 2 mv 2 mv 2m 2 P 2 2m K Kn mn m 183865 m e 6.730 273.2 m e Smaller particle has larger kinetic energy. 59 2 Collision in 2D Glancing collision (as compared to head-on collision.) Scattering (Not really hitting each other) 60 Collision in 2D (only for v2i = 0) Conservation of linear momentum Px: Py: m 1 v1 i m 1 v1 f cos 1 m 2 v 2 f cos 2 0 m1 v1 f sin 1 m 2 v 2 f sin 2 If kinetic energy is also conserved. 1 2 2 m 1 v1 i 1 2 2 m 1 v1 f 1 2 m 2v2 f 2 v2f y m2 If m1 = m2: 1 2 90 2 v1i o m1 m2 x 1 m1 v1f 61 Example: Two vehicles A and B are traveling west and south, respectively, toward the same intersection, where they collide and lock together. Before the collision, A (total weight 2700 lb) is moving with a speed of 40 MPH and B (total weight 3600 lb) has a speed of 60 MPH. Find the magnitude and direction of the velocity of the (interlocked) vehicles immediately after the collision. East x , North y , m A 2700 lb , m B 3600 lb , v Ai 40 mph iˆ , v Bi 60 mph ˆj vf ? Pix P fx m A v A ix m A m B v f x v f x Piy P fy m B v B iy m A m B v f vf v fx v fy tan 2 1 vf y vf x 2 1 7 .1m p h tan 1 34.3 m ph 17.1m ph 2 y v f y m A v Aix mA mB m B v Biy mA mB 2700 lb 40 m ph 2700 lb 3600 lb 3600 lb 60 m ph 2700 lb 3600 lb 3 4 .3 m p h 3 8 m p h 64 , S outh of W est 34.3 m ph y 2 o 17.1m ph B A vf x 62 External Force and Internal Energy Change E int Fext d cm cos Eint: change in internal energy of system that causes the external force Fext: external force that is caused by system dcm: displacement of center of mass of system : angle between Fext and dcm In this case, instead of thinking Fextdcmcos as the work done by the external force, it is better to understand it as the work done by the system. If this work is positive, the Eint is negative, and the system loses internal energy. 63 then kinetic energy of CM is changed: K cm Fext d cm cos If there is also potential energy of any object within the system involved, then the mechanical energy of the system (which is expanded to include the earth and/or spring) will be changed: E K cm U cm Fext d cm cos 64 Collision An isolated event in which two or more bodies exert relatively strong forces on each other for a relatively short time. If not isolated, not applicable, except when Fext << Fint Fext = 0 at the direction of consideration 65