1 Collisions: All around us (cars, buses) Grand astronomical scale (stars, galexies) Nuclear, atomic and molecular physics The impulse-momentum theorem relates the change in the momentum of a system to the total force on it and the time interval during which this total forces acts. § 9.1 Momentum and Newton’s second law of motion 2 1. momentum (1)a single particle of mass m Define: t r mvmp d d r rr == (2)a group of particles of mass m 1 , m 2 , …, m i , … ∑∑ == i ii i i vmpp rrr 2. the center of mass ? ? ? ? ? ? ? ? === ∑∑∑ i ii i i i i i M rm t M t r mpp rr rr d d d d The total momentum of a system of particle： t r Mp d d CM r r = ? Position vector of center of mass Total mass § 9.1 Momentum and Newton’s second law of motion x y z 1 r r N r r 1 m 2 m N m O 2 r r (1)The position vector of mass center (weighted average) N NN mmm rmrmrm r +++ +++ = L r L rr r 21 2211 CM That is N N i ii r M m r M m r M m M rm r r L rr r r +++= = ∑ 2 2 1 1 CM Define: N mmmmM +++== L 21total Total mass of the particle system: C c r r § 9.1 Momentum and Newton’s second law of motion 3 The components in the Cartesian coordinate system: M zm z M ym y M xm x N i ii N i ii N i ii ∑ ∑ ∑ = = = = = = 1 CM 1 CM 1 CM x y z 1 r r N r r 1 m 2 m N m O C c r r 2 r r For the system composed of single particles M rm r N i ii∑ = = 1 CM r r § 9.1 Momentum and Newton’s second law of motion M mz z M my y M mx x ∫ ∫ ∫ = = = d d d CM CM CM o x z y M ()zyxm ,,d r r For the common objects as essentially continuous distribution of matter: M mr r ∫ = d CM r r § 9.1 Momentum and Newton’s second law of motion 4 Example 1: P420 56(b) l l l 2m m 3m x y 12 33 6 60sin3020 12 7 6 2/320 l m mlmm y l m mlmlm x CM CM = +×+× = = ++× = o Solution: § 9.1 Momentum and Newton’s second law of motion Example 2: A thin strip of material is bent into the shape of a semicircle of radius R. Find its center of mass. Solution: since the symmetry, we obtained R RR M R M my M y x CM CM 637.0 2 dsin d)sin( 1 d 1 0 0 0 === == = ∫ ∫∫ π π π φφ π φ π φ x y R φ dm dφ § 9.1 Momentum and Newton’s second law of motion 5 (2) Momentum of system of particles CM CM total d d vM t r Mp r r r == t r M M rm t M t r mpp i ii i i i i i d d d d d d CM rrr rr = ? ? ? ? ? ? ? ? === ∑∑∑ (3) Kinetic energy of system of particles ii ii vvv rrr ′ += ′ += rrr rrr CM CM § 9.1 Momentum and Newton’s second law of motion x y z i m O C CM r r i r r i r ′ r The total Kinetic energy of the whole system i i ii i i i i iii i iii i iii i ii i vmvvmvm vvvvm vvvvm vvmvmKE ′?+′+= ′?+′+= ′+?′+= ?== ∑∑∑ ∑ ∑ ∑∑ rr rr rrrr rr CM 22 CM CM 22 CM CMCM 2 total 2 1 2 1 )2( 2 1 )()( 2 1 2 1 2 1 0=′ ∑ i ii vm r 22 CMtotal 2 1 2 1 ii i vmMvKE ′ += ∑ § 9.1 Momentum and Newton’s second law of motion 6 3. Momentum and Newton’s second law (1)dynamics of a single particle am t vm t p F r rr r === d )(d d d total § 9.1 Momentum and Newton’s second law of motion (2)dynamics of a system of particles M mv M vm t r M m M rm tt r v i ii ii i iiCM CM ∫ ∑ ∑∑ ==== d d d d d d d r r rrr r 或 (a)The velocity of the center of mass: M ma M am t r t v a i ii CMCM CM ∫ ∑ === d or d d d d 2 2 r r rr r (b)The acceleration of the center of mass: (d) Newton’s second law for the system of particles § 9.1 Momentum and Newton’s second law of motion CM i ii aM t v M t vm t p F r r r r r ==== ∑ d d d )(d d d CMtotal total The longer the total force acts on a system, the longer the system is subjected to an acceleration and the greater the change in its velocity. CM CM vM t r Mp r r r == d d total (c)Momentum of a system of particles 7 The forces from other particles within the system of particles is called internal force. The forces from sources outside the system of particles is called external force. The total force of the system: (e) Internal and external forces L r L rr L rrrr r L r L rrr +++++= ++++= +++++= ex iex2ex1 ex2in2ex1in1 21total FFF FFFF FFFFF Ni § 9.1 Momentum and Newton’s second law of motion § 9.1 Momentum and Newton’s second law of motion 1 m 2 m 3 m 12 F r 21 F r 13 F r 31 F r 32 F r 23 F r ex F 1 r ex F 3 r ex F 2 r According to Newton’s third law 0 inin ≡= ∑ i i FF rr t p FFF t p FFF t p FFF N NNN d d d d d d inex 2 in2ex22 1 in1ex11 r rrr LLL r rrr r rrr =+= =+= =+= because 8 Consider The sum of the equations all above is )( d d 21exex2ex1 NN ppp t FFF r L rr r L rr +++=+++ 0 inin ≡= ∑ i i FF rr § 9.1 Momentum and Newton’s second law of motion CM aM t v M t p F r rr r === d d d d CMtotal total § 9.2 Impulse-momentum theorem 1. Impulse-momentum theorem of a single particle ptF r r dd total = total d d F t p r r = because tFI dd total rr =let --differential impulse The impulse of the total force on the particle pppptFII if p p t t f i f i rrrr rrr r r ?=?==== ∫∫∫ d(t)dd total 9 z t t zz y t t yy x t t xx ptFI ptFI ptFI f i f i f i ? ? ? == == == ∫ ∫ ∫ d d d total total total tFItFItFtFI zzyy t t xxx ?=?=?== ∫ aveaveaveave 2 1 d O t totalx F avex F i t f t tFtFI t t ? avetotal 2 1 d rrr == ∫ 2. The Cartesian components of the impulse 3. Impulse and average impulse force § 9.2 Impulse-momentum theorem totaltotaltotal dd total pptFI f i f i p p t t rr rr r r ?=== ∫∫ totaltotaltotal totaltotaltotal totaltotaltotal d d d z t t zz y t t yy x t t xx ptFI ptFI ptFI f i f i f i ? ? ? == == == ∫ ∫ ∫ 4. Impulse-momentum theorem of a system of particles total total d d F t p r r = For a particle system Components in the Cartesian coordinate system: 0 1 intotalin ≡= ∑ = N i i FF rr 0d totalintotalin ≡= ∫ tFI f i t t rr notice： § 9.2 Impulse-momentum theorem 10 Example 1: throw a ball of mass 0.40kg against a brick wall. Find (a)the impulse of the force on the ball;(b)the average horizontal force exerted on the ball, if the ball is in contact with the wall for 0.01s. m/s30 1 ?=v m/s20 2 =v x Solution: (a) xixfx ppI ?= m/skg20)12(0.8 m/skg0.8 2040.0 m/skg12 )30(40.0 2 1 ?=?=?= ?= ×== ??= ?×== xixfx xf xi ppI mvp mvp § 9.2 Impulse-momentum theorem (b) tFtFI x t t xx f i ? ave d == ∫ N2000 01.0 20 ave === t I F x x ? § 9.2 Impulse-momentum theorem Example 2: a base ball of mass 0.14kg is moving horizontally at speed of 42m/s when it is truck by the bat. It leaves the bat in a direction at an angle φ=35o above its incident path and with a speed of 50m/s. (a)Find the impulse of the force exerted on the ball. (b) Assuming the collision last for 1.5 ms, what is the average force? (c)Find the change in the momentum of the bat. o 35 11 Solution : m/skg6.11cos )( ?=+= ??==?= if ixfxxixfxx mvmv mvmvpppI φ ? m/skg0.40sin ?=?= ?==?= φ ? f iyfyyiyfyy mv mvmvpppI m/skg3.12 22 ?=+= yx III Magnitude o 19tan 1 == ? x y I I θ direction pppI if rrr r ?=?= (a) § 9.2 Impulse-momentum theorem φ x y i p r f p r I r θ (b) N8200 0015.0 3.12 === t I F av ? r r o 19tan 1 == ? x y I I θ (c) According to the Newton’s third law, the change in momentum of the bat is equal and opposite to that of the ball. m/skg3.12 22 ?=+== yx IIIp?Magnitude oo 171tan180 1 =?= ? x y I I α direction § 9.2 Impulse-momentum theorem α φ x y i p r f p r I r θ bat p r ? 12 h A v r Example 3: Coal drops from a stationary hopper of height 2.0 m at rate of 40 kg/s on to conveyer belt moving with a speed of 2.0 m/s, find the average force exerted on the belt by the coal in the process of the transportation. § 9.2 Impulse-momentum theorem ghm vmpppp mvpppp yyy xxx 2 112 212 ?= ′?==?=? ?==?=? According to the impulse-momentum theorem ptFtFI av t t f i r rrr ?? === ∫ d Solution: Choose the mass element of the coal as the particle m/s0.2=v 1 p r 2 p r x y △ m ppp rrr ?=? 12 α β § 9.2 Impulse-momentum theorem 13 )(2.1252 )(80 1 av 2 av Nghq t m t p t p F Nqv t mv t p t p F y y x x ===== ===== ? ? ?? ? ? ? ?? ? )(149 avavav NFFF yx =+= o 4.57 80 2.125 tantan -1 av av -1 === x y F F α 1 p r 2 p r x y △ m ppp rrr ?=? 12 α β The angle with respect to the x axis § 9.2 Impulse-momentum theorem ooo 6.1224.57180 )(149 =?= = ′ β NF the average force exerted on the belt by the coal 1 p r 2 p r x y △ m ppp rrr ?=? 12 α β § 9.2 Impulse-momentum theorem 14 § 9.3 the rocket: a system with variable mass A rocket is propelled forward by rearward ejection of burned fuel (gases)that initially was in rocket. The forward force in the rocket is the reaction to the backward force on the ejected material. The total mass is constant , but the mass of the rocket itself decreases as material is ejected. Assuming: initial mass of the system is m 0 ; initial speed of the system is v 0 ; final mass of the system is m′; speed of ejected gases with respect to the rocket is v e . vv rr d+ e v r mm d+ md? The system: rocket and exhausted gases m v r vmp i rr = At instant t: The total mass of the rocket: m The total momentum of the rocket: )0d(d <+ mmm md? vv rr d+ vv e rr + At instant t+dt: mass of the rocket Mass of the exhausted gases Velocity of the rocket Velocity of the exhausted gases § 9.3 the rocket: a system with variable mass 15 mvvmvm vmmvvmvmvmvm vvmvvmmp e e ef dd dddddd ))(d()d)(d( rrr rrrrrr rrrrr ?+= ??+++= +?+++= vv rr d+ e v r mm d+ md? The total momentum of the system mvvmp e ddd rrr ?= The change of the momentum in time interval dt Ignore the resistant force of air, according to the impulse –momentum theorem ptFI r rr ddd total == § 9.3 the rocket: a system with variable mass t m v t v m t p F e d d d d d d total r rr r ?== t v m t m vF e d d d d total r r r =+ t m vF e d d thrust r r = --thrust of the rocket 1For a vertical flying jvv jvv jmgF ee ? dd ? ? total = ?= ?= r r r ptFI r rr ddd total == § 9.3 the rocket: a system with variable mass 16 mvvmtmg e ddd +=?then ∫∫∫ ′′ +=? m m e v v t m m vvtg 0 m 0 d dd 0 Integrate the both sides of the equation (assuming that all fuel is exhausted at instant t ′) tg m m vvv e ′ ? ′ += 0 0m ln The maximum speed of the rocket tg m m vvv e ′ ? ′ =? 0 0m ln § 9.3 the rocket: a system with variable mass 2For a horizontal flying m m vvv e ′ += 0 0m ln ivvivvF eex ? dd, ? ,0 total =?== rr r t v m t m vF e d d d d total r r r Q =+ e v v m m dd =?∴ m m vv e ′ = 0 m ln If v 0 =0, then § 9.3 the rocket: a system with variable mass 17 3Step rocket: nenee NvNvNvvv lnlnln 22110m ++++= L if -13 m 321 -1 321 0 sm13440ln62500 6 sm2500 0 ?=?= === ?=== = v NNN vvv v eee then It is enough for the launch of a satellite. Question: What will be the effect of the air resistant force? § 9.3 the rocket: a system with variable mass § 9.3 the rocket: a system with variable mass 三 18 光荣的长征火箭家族 中国已经自行研制了四大系列12 种型号的运载火箭： 长征1号系列：发射近地轨道小 卫星. 长征2号系列：发射近地轨道中 、大型卫星，和其它航天器. 长征3号系列：发射地球同步高 轨道卫星和航天器. 长征4号系列：发射太阳同步轨 道卫星. 长征 2号 C火箭 § 9.3 the rocket: a system with variable mass 1970年4月 ……2003年5月：发射70次，将54颗国产卫 星，27颗外国卫星，4艘神舟号无人飞船送入太空。 成功率91％（美国德尔塔火箭：94％，欧空局阿丽亚 娜火箭：93％，俄罗斯质子号火箭：90％）。 2003年10月15日：长征2号F运载火箭成功发射神舟5 号载人飞船。 长征 3号 A火箭发射的东 方红三号通信卫星 宇航员杨力伟 § 9.3 the rocket: a system with variable mass 19 神州 1号 神州 2号 神州 3号 神州 4号 神 州 5 号 升 空 § 9.3 the rocket: a system with variable mass 10月15日，我国在 酒泉卫星发射中心进行 首次载人航天飞行。９ 时整，“神舟”五号载 人飞船发射升空。 10月１６日６时２３ 分，“神舟”五号载人飞船 在内蒙古主着陆场成功着 陆，实际着陆点与理论着陆 点相差４.8公里。返回舱完 好无损。航天英雄杨利伟自 主出舱。我国首次载人航天 飞行取得圆满成功。 § 9.3 the rocket: a system with variable mass 20 沙漠化的中国 !!! 从图片上我 们可以清晰看 到我国大部分 土地没有被绿 色植被所覆 盖，而是以赤 裸裸的黄色直 接面向宇宙， 多年的干旱和 毫无节制的滥 砍滥伐使我们 的绿色极度匮 乏！ 神州 5号拍摄的中国版图 2005年 10月 12日：长征 2号 F型运载火箭成功发射神舟 6号载人飞船。 报道： “我们在神舟五号的基础上继续攻克多项载人航天的基本技术， 第一次进行了真正有人参与的空间科学实验。 ” 神舟 6号矗立在发射台上 宇航员费俊龙、聂海胜 § 9.3 the rocket: a system with variable mass 21 § 9.3 the rocket: a system with variable mass 神舟载人飞 船在组装调 试。最上部 为轨道舱、 中部灰黑色 圆柱体为返 回舱、下部 白色段为推 进舱。 § 9.3 the rocket: a system with variable mass 22 Example 1: A spaceship with a total mass of 13600 kg is moving relative to a certain inertial reference frame with a speed of 960 m/s in a region of space of negligible gravity. It fires its rocket engines to give an acceleration parallel to the initial velocity. The rocket eject gas at a constant rate of 146 kg/s with a constant speed (relative to the spaceship) of 1520 m/s, and they are fired until 9100 kg of fuel has been burned and ejected. (a) what is the thrust produced by the rockets? (b) What is the velocity of the spaceship after the rockets have fired? § 9.3 the rocket: a system with variable mass Solution: (a) The thrust is given by (N)1022.21461520 d d 5 thrust ×=×== t m vF e (b)choose the positive x direction to be that of the spaceship’s initial velocity, then we have m/s2640 13600 4500 ln)1520(960 ln d d, d d d d =?+= =? == ∫∫ f i f eif m m e v v e v m m vvv m m vv t m v t v m f i f i § 9.3 the rocket: a system with variable mass 23 § 9.4 conservation of momentum totaltotaltotal dd total pptFI f i f i p p t t rr rr r r ?=== ∫∫ total total d d F t p r r =For a particle system If 0 d d then,0 total total == t p F r r constant total =p r or 0dd totaltotaltotal total ==== ∫∫ pptFI f i f i p p t t rr rr r r ? ftotal,itotal, pp rr = Notice: 1The condition of conservation!!! The law of conservation : 2for an isolated particle system constant total == c vMp rr The center of mass will remain constant velocity. 3the component forms of conservation of momentum constant0 constant0 constant0 totaltotal totaltotal totaltotal === === === ∑ ∑ ∑ nz n nzz ny n nyy nx n nxx vmpF vmpF vmpF 4if the internal force is much larger than the external forces § 9.4 conservation of momentum 24 β M m θ 1 v r 2 v r v r Solution: For the system of alpha and Oxygen atom, there is no external forces, the total momentum is conserved. Before collision: After collision: 0 O1a == pvmp rrr vMpvmp rrrr == O2a § 9.4 conservation of momentum Example 1: The scattering of alpha. The alpha collide with the Oxygen atom. Find the speed ratio of alpha before and after the scattering, . oo 41,72 == βθ In Cartesian coordinate system βθ βθ sinsin0 coscos 2 21 Mvmv Mvmvmv ?= += x y θ β 1 vm r 2 vm r o vM r vMvmvm rrr += 21 Conservation of momentum ()() 71.0 4172sin 41sin sin sin 1 2 = + = + = oo o βθ β v v the speed ratio of alpha before and after the scattering § 9.4 conservation of momentum 25 Example 2: The ballistic pendulum was used to measure the speed of the bullets before electronic timing devices were developed. The version shown in figure consists of a large block of wood of mass M=5.4 kg, A bullet of mass m= 9.5 g is fired into the block, coming quickly to rest. The block and bullet then swing upward, their center of mass rising a vertical maximum distance h=6.3 cm before the pendulum comes to rest, what is the speed of the bullet just prior to the collision? § 9.4 conservation of momentum Solution: Two steps: The bullet-block collision The bullet-block rise Step 1: the horizontal total force is zero, the momentum of system is conserved. v Mm m VVMmmv + =?+= )( § 9.4 conservation of momentum Step 2: the mechanical energy of the bullet- block-earth system is conserved. m/s6302 )()( 2 1 2 = + = +=+ gh m Mm v ghMmVMm then 26 Example 3: on a frictionless horizontal table, both ends of a spring connect block A and B respectively. The blocks have same mass M. a bullet with mass m and the initial speed v 0 impacted into the block A and stopped in it, find the maximum compression distance of the spring. 0 v r A B § 9.4 conservation of momentum Solution: 1The collision of the bullet and A Mm mv VVMmmv + =?+= 0 000 )( 2 The collision of the bullet with A and B Mm mv VVMmmv 2 )2( 0 0 + =?+= 3the instant when A and B get the same speed 21 0max 2 max 22 0 ] )2)(( [ 2 1 )2( 2 1 )( 2 1 MmMmk M mvx kxVMmVMm ++ = ++=+ § 9.4 conservation of momentum 27 一绳跨过一定滑轮，两端分别系有质量 m 及 M的物体， 且 M>m。最初 M 静止在桌上，抬高 m,使绳处于松弛状 态。当 m自由下落距离 h后，绳才被拉紧，求此时两物 体的速率 v 和 M 所能上升的最大高度（不计滑轮和绳的 质量、轴承摩擦及绳的伸长）。 h M m 当 m自由下落 h距离，绳被拉紧 的瞬间, m和 M获得相同的运动 速率 v。此后 m向下减速运动, M 向上减速运动。 M上升的最大高度为： 分析运动过程 a v H 2 2 = 分两个阶段求解 问题讨论 : § 9.4 conservation of momentum 第一阶段： 绳拉紧，求共同速率 v 解1： 0= ∴> v ,MmmM 共同速率 不能提起Q 解2： 绳拉紧时冲力很大，忽略重力， 系统动量守恒 Mm + Mm ghm v;v)Mm(ghm + =+= 2 2 解3： 动量是矢量，以向下为正，系统动量守恒： Mm ghm v;)v(Mmvghm ? =?+= 2 2 以上三种解法均不对！ h M m § 9.4 conservation of momentum 28 正确解法： 绳拉紧时冲力很大，轮轴反作 用力 不能忽略 ， 系 统动量不守恒，应分别对它们 用动量定理； Mm + N r h M m x N y N + 设冲力为 ，取向上为正方向 F r Mg F mg F + () () MvMvtMgFI )ghm(mvtmgFI =?=?= ???=?= ∫ ∫ 0d 2d 2 1 § 9.4 conservation of momentum 忽略重力，则有 mM ghm v Mv)ghm(mv + = =??? 2 2 21 II = 第二阶段： 与 有大小相等，方向相反的加速度 ，设 绳拉力为 ，画出 与 的受力图 m M a m MT M m + a Mg T a mg T § 9.4 conservation of momentum 29 由牛顿运动定律: ? ? ? =? =? mamgT MaTMg 解得: mM g)mM( a + ? = 22 2 2 2 22 2 mM hm ) mM g)mM( () mM gh(m ( a v H ? = + ? + == 上升的最大高度为 M a Mg T a mg T § 9.4 conservation of momentum A C B 类似问题： § 9.4 conservation of momentum 30 § 9.5 Collisions If the impulse force is much larger than any external forces (such as gravitational force, friction, ……), as is the case in most of the collision, we can neglect the external forces entirely and treat the system as an isolated system. Then the momentum is conserved in the collision. 1. The conservation of momentum in collision The individual momenta of the particles do change, but the total momenta of the system of colliding particles does not. aft2aft1bfr2bfr1total ppppp rrrrr +=+= If a system has zero total force on it, then 2. Elastic collisions During the brief collision, the kinetic energy is stored in the system as potential energy, then dumped back into kinetic energy after the instant of the collision. (1)If the total kinetic energy of the particles is not changed, then the collision is called the elastic collision. J0 total =KE? (2)Elastic collision occur when the forces between the colliding bodies are conservative. J0 totaltotal == KEW ? § 9.5 Collisions 31 For one dimensional motion, show that two particles with relative speed will leave each other with same relative speed after an elastic collision, that means Solution: 2211 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 BBAABBAA BBAABBAA vmvmvmvm vmvmvmvm rrrr +=+ +=+ The momentum and mechanical energy is conserved in an elastic collision 11 BA vv rr ? )( 2211 BABA vvvv rrrr ??=? § 9.5 Collisions 112 112 2 2 A BA A B BA AB B B BA B A BA BA A v mm m v mm mm v v mm m v mm mm v rrr rrr + + + ? = + + + ? = )( ) 2 ( ) 2 ( 11 1 122 BA A BA A BA BA B BA AB BA B BA vv v mm m mm mm v mm mm mm m vv rr r rrr ??= + ? + ? + + ? ? + =? Solving the equations, one can get The relative speed after collision: § 9.5 Collisions 32 3.One-dimensional collisions in the center-of- mass reference frame The velocity of the CM frame relative to the lab frame SS v ′ x A m B m Ai p r Bi p r The total initial momentum of two bodies in the cm frame )()( SSBiBSSAiAi vvmvvmp ′′ ?+?= Define the CM frame to be the frame in which the initial momentum of the two bodies system is zero. § 9.5 Collisions BA BiBAiA SSi mm vmvm vp + + =?= ′ 0 In the CM frame, before the collision, x A m B m Ai p ′ r Bi p ′ r The total momentum is conserved, then we must have BfAfBfAff ppppp rrrrr ′ ?= ′ = ′ + ′ = ′ ,0 x A m B m Af p ′ r Bf p ′ r § 9.5 Collisions 33 x A m B m Af p ′ r Bf p ′ r x A m B m Af p ′ r Bf p ′ r x A m B m Af p ′ r Bf p ′ r x A m B m elastic inelastic Completely inelastic explosive § 9.5 Collisions For elastic collision(in CM frame): SSAfAfSSAiAiAiAf vvvvvvvv ′′ ?= ′ ?= ′′ ?= ′ ,, Bi BA B Ai BA BA BA BiBAiA Ai SSAi SSSSAiAf v mm m v mm mm mm vmvm v vv vvvv + + + ? = + + +?= +?= +??= ′ ′′ 2 2 2 )( After the collision: § 9.5 Collisions 34 Bi BA AB Ai BA A Bf v mm mm v mm m v + ? + + = 2 Following the same way, § 9.5 Collisions Discussion: 1equal masses AiBfBiAf BA vvvv mm == = 2target particle at rest Ai BA A BfAi BA BA Af Bi v mm m vv mm mm v v + = + ? = = 2 0 3massive target BiBfBiAiAf AB vvvvv mm ≈+?≈ >> 2 4massive projectile BiAiBfAiAf BA vvvvv mm ?≈≈ >> 2 § 9.5 Collisions 35 before after before after v r A A A A B B B B vv A rr ?= B v r x x x x v r B v r A v r Example 1: a ping-pang ball and a bowling ball Example 2: if m A =m B v mm m vv mm mm v BA A B BA BA A + = + ? = 2 , vvv BA == ,0 § 9.5 Collisions Example 3:The mass of Saturn is 5.67×10 26 kg, its speed relative to the sun is 9.6 km/s; one spacecraft of mass 150 kg, its speed relative to the sun is 10.4 km/s. the spacecraft is moving toward the Saturn, due to the gravitation of Saturn, the spacecraft rounds the Saturn and departs in the opposite direction—the slingshot effect. Find the speed of the spacecraft relative to the sun. 1A v r 1B v r2A v r Chapter 9 impulse, momentum, and collisions 36 Solution: Due to mmMm AB =>>= 12 112 2 BB BAA vv vvv rr rrr ≈ +?≈ km/s)(6.29 6.924.102 112 ?= ×??=??≈ BAA vvv 12 AA vv > Chapter 9 impulse, momentum, and collisions + 112 112 2 2 A BA A B BA AB B B BA B A BA BA A v mm m v mm mm v v mm m v mm mm v rrr rrr + + + ? = + + + ? = Example 4: Show that two balls with same mass will separate perpendicularly to each other after an elastic collision which is not head-on, if one ball is at rest before collision. Solution: Conservation of momentum and energy 2 2 2 1 2 0 2 2 2 1 2 0 210210 2 1 2 1 2 1 vvvmvmvmv vvvvmvmvm +=?+= +=?+= rrrrrr 1 2 21 2 2 2 1 2 0 212100 2 )()( vvvvv vvvvvv rr rrrrrr ?++= +?+=? 4 3 § 9.5 Collisions 37 Compare 2 and 4 , 21 2 2 2 1 2 0 2 2 2 1 2 0 2 vvvvv vvv rr ?++= += 4 2 we have 02 21 =?vv rr That means 0cos 2121 ==? θvvvv rr Then 2 π θ = § 9.5 Collisions 4. Inelastic collisions If the total energy of the particles involved in a collision is not conserved, the collision is called an inelastic collision. If the particles stick together after collision, the collision is called a completely inelastic collision. totaltotal KEW ?=From For a collision which there are no external forces, we have 00 totaltotal ≠= KEW ? Why? § 9.5 Collisions 38 Notice: (a)The total momentum of the system is conserved in both elastic and inelastic collisions. (b)The total kinetic energy is conserved only for elastic collisions. (c)If the particles stick together after the collision, it is called a completely inelastic collision. Example 1: P392 9.9 Example 2: P394 9.10 Example 3: P395 9.11 Example 4: disintegrations and explosions(§ 9.6) § 9.5 Collisions