Chapter 11 Collision Theory

Transcription

1 Chapter Collision Theory Introduction. Center o Mass Reerence Frame Consider two particles o masses m and m interacting ia some orce. Figure. Center o Mass o a system o two interacting particles Choose a coordinate system (Figure.) in which the position ector o body is gien by r and the position ector o body is gien by r. The relatie position o body with respect to body is gien by r, = r r. Figure. Center o mass coordinate system. During the course o the interaction, body is displaced by dr and body is displaced by dr, so the relatie displacement o the two bodies during the interaction is gien by dr = dr dr. The relatie elocity between the particles is, dr, dr dr, = = =. (..) dt dt dt

2 We shall now show that the relatie elocity between the two particles is independent o the choice o reerence rame. Let R be the ector rom the origin o rame S to the origin o reerence rame S. Denote the position ector o particle i with respect to origin o reerence rame S by r i and similarly, denote the position ector o particle i with respect to origin o reerence rame S by r i (Figure.4). Figure.4 Position ector o th i particle in two reerence rames. The position ectors are related by r = r + R i i. (..) The relatie elocity (call this the boost elocity) between the two reerence rames is gien by V = d R. (..3) dt Assume the boost elocity between the two reerence rames is constant. Then, the relatie acceleration between the two reerence rames is zero, A= dv = 0. (..4) dt th Suppose the i particle in Figure.4 is moing; then obserers in the dierent th reerence rames will measure dierent elocities. Denote the elocity o i particle in rame S by i = dr i dt, and the elocity o the same particle in rame S by i = dr i dt. Since the deriatie o the position is elocity, the elocities o the particles in two dierent reerence rames are related according to = + V i i. (..5) In rame S, the relatie elocity is

3 , = (..6) The relatie elocity in reerence rame S can be determined rom using Equation (..5) to express Equation (..6) in terms o the primed quantities, = = + V + V = = ( ) ( ),, (..7) and is equal to the relatie elocity in rame S. For a two-particle interaction, the relatie elocity between the two ectors is independent o the choice o reerence rame. In Appendix 8.B, we showed that when two particles o masses m and m interact, the change o kinetic energy between the inal state B and the initial state A due to the interaction orce only is equal to Δ K = μ ( B A ) (..8) where μ = mm / m + m ( ) is the reduced mass o the two-particle system. (I Equation (..4) did not hold, Equation (..8) would not be alid in all rames.) In Equation (..8), the square o the inal relatie elocity ( ) ( ) B B is gien by ( ) = (( ) ( ) ), (( ) ( ) ) B B B B B (..9) A and the square o the initial relatie elocity ( ) ( ) A is gien by ( ) = (( ) ( ) ), (( ) ( ) ) A A A A A. (..0) By expressing the change o kinetic energy in terms o the relatie elocity, a quantity that is independent o the reerence rame, the change in kinetic energy is thereore independent o the choice o reerence rame. Characterizing Collisions In a collision, the ratio o the magnitudes o the initial and inal relatie elocities is called the coeicient o restitution and denoted by the symbol e, e B =. (..) A 3

4 I the magnitude o the relatie elocity does not change during a collision, e =, then the change in kinetic energy is zero, (Equation (..8)). Collisions in which there is no change in kinetic energy are called elastic collisions, ΔK = 0, elastic collision. (..) I the magnitude o the inal relatie elocity is less than the magnitude o the initial relatie elocity, e <, then the change in kinetic energy is negatie. Collisions in which the kinetic energy decreases are called inelastic collisions, ΔK < 0, inelastic collision. (..3) I the two objects stick together ater the collision, then the relatie inal elocity is zero, e = 0. Such collisions are called totally inelastic. The change in kinetic energy can be ound rom Equation (..8), Δ = μ = mm K A A, totally inelastic collision m+ m. (..4) I the magnitude o the inal relatie elocity is greater than the magnitude o the initial relatie elocity, e >, then the change in kinetic energy is positie. Collisions in which the kinetic energy increases are called superelastic collisions,. Worked Examples.. Example: Elastic One-Dimensional Collision ΔK > 0, superelastic collision. (..5) Consider the elastic collision o two carts along a track; the incident cart has mass m and moes with initial speed,0. The target cart has mass m = m and is initially at rest,,0 = 0. Immediately ater the collision, the incident cart has inal speed and the target cart has inal speed o the initial speed. Solution,0,,. Calculate the inal elocities o the carts as a unction Draw a momentum low diagram or the objects beore (initial state) and ater (inal state) the collision (Figure.5, with a greatly simpliied rendering o a cart ). 4

5 Figure.5 Momentum low diagram or elastic one-dimensional collision There are no external orces acting on the system, so the component o the momentum along the direction o the collision is the same beore and ater the collision, m,0 = m, + m,. (..) Note that in the aboe igure and Equation (..), the incident cart is taken to be moing backwards i, > 0. Equation (..) simpliies to, 0 =, +,. (..) The collision is elastic; the kinetic energy is the same beore and ater the collision, m =, 0 m, + m,, (..3) which simpliies to, 0 =, +,. (..4) There are many ways to manipulate Equations (..) and (..4) to sole or the inal elocities in terms o. One straightorward way is to sole Equation (..) or the inal speed o cart,,0, =,, 0 (..5) and substitute Equation (..5) into Equation (..4) yielding ( ) Equation (..6) simpliies to = + = (..6),0,,0,,,0,,0, 5

6 0 = 6, 4,, 0. (..7) We can now sole or solution is,. There are two solutions to this quadratic equation. One, = 3, 0. (..8) The inal elocity o the incident cart is then, =,, 0 = 4 3, 0,0 = 3,0. (..9) The other solution is, = 0, in which the inal state o the target is the same as the initial state; in addition, Equation (..5) would gie, =,0, corresponding to the incident cart moing at its initial elocity. That is, the carts did not collide. This situation is not physical or carts on a track o inite length, but could certainly happen in other circumstances, as beginning pool players know well. Or, i in the top igure in Figure.5 the initially moing cart were placed to the right o the stationary cart, Equations (..) and (..3) would still be alid or no collision. A technique that physicists know well is to rewrite Equations (..) and (..4) as,0 +, =,,0, =,. (..0) I,0 +, 0 ( 0 + = 0,, is the second case o a miss, discussed aboe), the second equation in (..0) can be diided by the irst to yield,0, =,. (..) Adding Equation (..) to the irst expression in (..0) yields Equation (..8). The more general case o a one-dimensional collision is discussed in Appendix.A... Example: Elastic Two-Dimensional Collision Object with mass is initially moing with a speed,0 = 3.0m s and collides m elastically with object that has the same mass, m = m, and is initially at rest. Ater the collision, object moes with an unknown speed, at an angle θ, = 30 o with respect 6

7 to its initial direction o motion and object moes with an unknown speed unknown angle θ, objects and the angle., at an (as shown in the Figure.6). Find the inal speeds o each o the θ,, Solution: Figure.6 Momentum low diagram or two-dimensional elastic collision Choose a set o positie unit ectors or the initial and inal states as shown in Figure.7. We designate the respectie speeds o each o the particles on the momentum low diagrams. Figure.7 Choice o unit ectors or momentum low diagram Initial State: The components o the total momentum state are gien by total p x,0 = m, 0 total p y,0 = 0. p = m + m,0 total 0,0 in the initial (..) Final State: The components o the momentum p total = m, + m, are gien by in the inal state 7

8 p = m cosθ + m cosθ total x,,,,, p = m sinθ m sin θ. total y,,,,, (..3) There are no any external orces acting on the system, so each component o the total momentum remains constant during the collision, total total p x,0 = p x, (..4) p total total y,0 = p y,. (..5) These two equations become m = m cosθ + m cosθ,0,,,, 0= m sinθ m sin θ.,,,, (..6) The collision is elastic; the kinetic energy is the same beore and ater the collision, or K total total 0 = K, (..7) m =, 0 m, + m,. (..8) We hae three equations, two momentum equations and one energy equation, with three unknown quantities,,,, and θ, since we are already gien that, 0 = 3.0 m s and θ, = 30 o. We irst rewrite the expressions in Equation (..6), canceling the actors o, as m cosθ =,,,0,,,,,, cosθ sinθ = sin θ. (..9) Add the squares o the expressions in Equation (..9), yielding ( ), cos θ,, sin θ,,0, cosθ,, sin, + = + θ. (..0) We can use the identities cos θ, + sin θ, = and cos θ, + sin θ, = to simpliy Equation (..0), yielding 8

9 = cosθ +. (..),,0,0,,, Substituting Equation (..) into Equation (..8) yields m,0= m, + m(,0,0, cosθ, +, ). (..) Equation (..) simpliies to 0= cosθ, (..3),,0,, which may be soled or the inal speed o object,,,0, ( ) = cosθ = 3.0 m s cos30 =.6 m s. (..4) Diide the expressions in Equation (..9), yielding sinθ = sinθ,,,,, cosθ,,0, cosθ,. (..5) Equation (..5) simpliies to tanθ, = sinθ,, cosθ,0,,. (..6) Thus object moes at an angle θ θ, tan sinθ,, = = tan cosθ,0,, (.6 m s ) sin30 ( ) 3.0 m s.6 m s cos30, = 60. (..7) The aboe results or, and θ, may be substituted into either o the expressions in Equation (..9), or Equation (..8), to ind., =.5m s Beore going on, it must be noted that the act that θ, + θ, = 90, that is, the objects moe away rom the collision point at right angles, is not a coincidence. A ector 9

10 deriation is presented below. We can see this result algebraically rom the aboe result. Using the result o Equation (..4),, =,0 cosθ,, in Equation (..6) yields tanθ cosθ sinθ = = cotθ,,,, cosθ, ; (..8) the angles θ, and θ, are complements. It should be noted that Equation (..3) also has the solution, = 0, which would correspond to the incident particle missing the target completely. We can proe that the particles emerge rom the collision at right angles by making explicit use o the act that momentum is a ector quantity. Since there are no external orces acting on the two objects during the collision (the collision orces are all internal), momentum is constant. Thereore p = p total total 0 (..9) which becomes m = m + m,0,, (..30) Equation (..30) simpliies to = +, 0,,. (..3) Recall the ector identity that the square o the speed is gien by the dot product =. (..3) With this identity in mind, we take the dot product o each side o Equation (..3) with itsel, This becomes, 0, 0 = (, +, ) (, +, ) (..33) = + +.,,,,,, = + +, 0,,,,. (..34) 0

11 Recall that kinetic energy is the same beore and ater an elastic collision, and the masses o the two objects are equal, so Equation (..8) simpliies to = +. (..35), 0,, Comparing Equation (..34) with Equation (..35), we see that,, = 0. (..36) The dot product o two nonzero ectors is zero when the two ectors are at right angles to each other. Since = 30 o, the inal angle o object must be θ, θ =, 60. (..37) A geometric demonstration or obtaining the result o Equation (..37) is gien in Appendix.B to this chapter. Note that Equation (..36) allows the algebraic solutions, = 0 and, = 0. Had we not been gien a nonzero alue or, the, = 0 solution would represent a collision, in which all o the momentum and energy o object is transerred to object. In this case, the gien angle o θ = is not well deined, but is immaterial or a stationary, 30 particle. Had we not been gien a nonzero alue or θ, the, = 0 solution would indicate a complete miss...3 Example: Bouncing Superballs Two superballs are dropped rom a height aboe the ground, one on top o the other. The ball on top has a mass m, and the ball on the bottom has a mass m. Assume that the when the lower ball collides with the ground there is no loss o kinetic energy. Then, as the lower ball starts to moe upward, it collides with the upper ball that is still moing downwards. Assume again that the total energy o the two balls remains the same ater the collision. How high will the upper ball rebound in the air? Assume m >> m. Hint: Consider this set o collisions rom an inertial reerence rame that moes upward with the same speed as the lower ball has ater it collides with ground. What speed does the upper ball hae in this reerence rame ater it collides with the lower ball?,

12 Figure.8 Fie stages o motion (not necessarily to scale) as seen by obserer at rest on ground. Solution: The system consists o the two superballs. There are ie special states or this motion (Figure.8). Initial State (time t 0 ): the superballs are released rom rest at a height h 0 aboe the ground. State (time ): the superballs just reach the ground with the same speed =. t,0,0 State (time t ): immediately beore the collision o the large and small superballs but ater the larger superball has collided with the ground and reersed direction with the same speed,,0 =. The smaller superball is still moing down with speed,0., 0 State 3 (time t 3 ): immediately ater the collision o the superballs. The smaller superball moes upward with speed,. The larger superball moes upward with speed., Final State (time ): the smaller superball reaches maximum height aboe the t ground. Choice o Reerence Frame: This collision is best analyzed rom the reerence rame o the obserer moing upward with speed,0 =,0, the elocity o the larger superball just ater it rebounded with the ground. In this rame immediately beore the collision, the smaller superball is moing h

13 downward with a speed, 0 that is twice the speed as seen by an obserer at rest on the ground (lab reerence rame),,0 =, 0. (..38) The momentum low diagrams between States and 3 or the two dierent reerence rames are shown in Figure.9. Figure.9 Momentum low diagrams or States and 3 as determined by an obserer at rest on ground (let igure) and an obserer moing upwards with speed o large superball,0 (right igure). Model: The mass o the larger superball is much larger than the mass o the smaller superball, m >> m. This enables us to consider the collision (between States and 3) to be equialent to the small superball bouncing o a hard wall, while the larger ball experiences irtually no recoil. Hence the large superball remains at rest in the reerence rame moing upwards with speed with respect to obserers at rest on ground.,0 Beore the collision, the smaller superball has speed, 0 =, 0. Since there is no loss o kinetic energy during the collision, the result o the collision is that the smaller superball changes direction but maintains the same speed, =, 0 =, 0 (..39) Howeer, according to an obserer at rest on the ground, ater the collision the smaller superball is moing upwards with speed. (..40), =,0 +, 0 = 3, 0 Ater the collision, the mechanical energy o the smaller superball is constant and hence between State 3 and the Final State, 3

14 ΔK + ΔU = 0. (..4) The change in kinetic energy is Δ K = m( 3,0). (..4) The change in potential energy is ΔU = m gh. (..43) Equation (..4), the condition that mechanical energy is constant, becomes m( 3,0) + mgh = 0 mgh = 9 m(,0). (..44) Recall that we can also use the act that the mechanical energy doesn t change between the Initial State and State, yielding an equation similar to Equation, (..44), m m ( ),0 0 ( ) m gh = 0 = m gh.,0 0 (..45) Comparing the second expressions in (..44) and (..45), the smaller superball is seen to reach a maximum height h = 9h. (..46) 0..4 Example Pendulums and Collisions A simple pendulum consists o a bob o mass m that is suspended rom a piot by a string o negligible mass. The bob is pulled out and released rom a height h 0 as measured rom the bob s lowest point directly under the piot point and then swings downward in a circular orbit (Figure.0a). At the bottom o the swing, the bob collides with a block o mass m that is initially at rest on a rictionless table. Assume that there is no riction at the piot point. a) What is the speed o the bob immediately beore the collision at the bottom o the swing? 4

15 b) Assume that the kinetic energy o the bob beore the collision is equal to the kinetic energy o the bob and the block ater the collision (the collision is elastic). Also assume that the bob and the block moe in opposite directions but with the same speed ater the collision (Figure.0b). What is the mass o the block? m c) Suppose the bob and block stick together ater the collision due to some putty that is placed on the block. What is the speed o the combined system immediately ater the collision? (Assume now that is the combined mass o the block and putty.) m d) What is the change in kinetic energy o the block and bob due to the collision in part c)? What is the ratio o the change in kinetic energy to the kinetic energy beore the collision? e) Ater the collision in part d), the bob and block moe together in circular motion. What is the height h aboe the low point o the bob s swing when they both irst come to rest ater the collision (Figure.0c)? Ignore any air resistance. (Hint: The relatie sizes o the heights h and h in Figure.0 (a), (c) are not correct.) 0 (a) (b) (c) Figure.0 (a) Simple pendulum released on a collision course with a stationary block (b) Collision at the bottom o the swing (c) Bob and block rise to the top o the swing Solution: a) The mechanical energy o the bob is constant between when it is released and the bottom o the swing. We can use m = mgh (..47),0 0 to calculate the speed o the bob at the low point o the swing just beore the collision,, 0 = gh 0. (..48) 5

16 b) Consider the bob and the block as the system. Although tension in the string and the graitation orce are now acting as external orces, both are particular to the motion o the bob and block during the collision. I we additionally assume that the collision is nearly instantaneous, then the momentum is constant in the direction o the bob s motion, m,0 = m, m,, (..49) where, is the speed o the block immediately ater the collision. Since the bob and block are gien to hae the same speeds ater the collision, deine, =, and rewrite Equation (..49) as ( ) m = m m. (..50),0 Sole Equation (..50) or the speed o the bob and block ater the collision, =,0 m m m (..5) (at this point we see explicitly what we might hae guessed, that m > m ). The collision is gien to be elastic, m,0 = ( m+ m). (..5) Substituting Equation (..5) into Equation (..5) yields m m = ( m+ m),0,0 m m. (..53) Canceling the common actor o m, 0 rom both sides o Equation (..53) and rearranging gies ( ) ( ) m m = m + m m. (..54) Expanding the square and canceling m yields m (m 3 m ) = 0, (..55) 6

17 and so the block has mass m = 3m (..56) and the inal speed is gh,0 0 = =. (..57) c) The bob and block stick together and moe with a speed ater the collision. The external orces are still perpendicular to the motion, and i we assume that the collision time is negligible, then the momentum in the direction o the motion is constant, ( ) m = m+ m (..58),0 The speed immediately ater the collision is (recalling that m ) = 3m m,0 = = m+ m,0. (..59) 4 Using Equation (..48) in Equation (..59) yields gh gh =,0 = 0 =. (..60) d) The change in kinetic energy o the bob and block due to the collision in part c) is gien by Δ K = K K = ( m + m ) m ater beore,0 Using Equation (..59), (..48) and (..59) in Equation (..6),. (..6) gh0 Δ K = ( 4m) mg 8 3 = mgh 0. 4 h 0 (..6) The kinetic energy beore the collision was mgh 0, and so the ratio o the change in kinetic energy to the kinetic energy beore the collision is 7

18 ΔK K beore = 3/4. (..63) See Appendix.C or a more general result, ΔK m = K m + m beore (..64) or completely inelastic collisions when the target object (the block o mass example) is initially stationary. m in this e) Ater the collision, the tension is acting on the bob-block system but the tension orce is perpendicular to the motion so does no work on the bob-block system and the mechanical energy ater the collision is the same as when the bob-block combination reaches its highest point, K = ( m + m ) gh mgh 4 ater 0 = 4mgh h 6 (..65) 0 h =. (..66) 8