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試題詳情

2009高考物理壓軸題匯編

1、如圖所示,一質(zhì)量為M、長為l的長方形木板B放在光滑的水平地面上,在其右端放一質(zhì)量為m的小木塊A,m〈M,F(xiàn)以地面為參照系,給A和B以大小相等、方向相反的初速度(如圖5),使A開始向左運動、B開始向右運動,但最后A剛好沒有滑離L板。以地面為參照系。
(1)若已知A和B的初速度大小為v0,求它們最后的速度的大小和方向。
(2)若初速度的大小未知,求小木塊A向左運動到達的最遠處(從地面上看)離出發(fā)點的距離。

解法1: (1)A剛好沒有滑離B板,表示當A滑到B板的最左端時,A、B具有相同的速度。設(shè)此速度為,A和B的初速度的大小為,則由動量守恒可得:
     
  解得: , 方向向右   、

(2)A在B板的右端時初速度向左,而到達B板左端時的末速度向右,可見A在運動過程中必經(jīng)歷向左作減速運動直到速度為零,再向右作加速運動直到速度為V的兩個階段。設(shè)為A開始運動到速度變?yōu)榱氵^程中向左運動的路程,為A從速度為零增加到速度為的過程中向右運動的路程,L為A從開始運動到剛到達B的最左端的過程中B運動的路程,如圖6所示。設(shè)A與B之間的滑動摩擦力為f,則由功能關(guān)系可知:

  對于B      ②
  對于A     ③            、

由幾何關(guān)系       、

 由①、②、③、④、⑤式解得               ⑥

解法2: 對木塊A和木板B組成的系統(tǒng),由能量守恒定律得:

  ⑦

由①③⑦式即可解得結(jié)果    

本題第(2)問的解法有很多種,上述解法2只需運用三條獨立方程即可解得結(jié)果,顯然是比較簡捷的解法。

2、如圖所示,長木板A右邊固定一個擋板,包括擋板在內(nèi)的總質(zhì)量為1.5M,靜止在光滑的水平面上,小木塊B質(zhì)量為M,從A的左端開始以初速度在A上滑動,滑到右端與擋板發(fā)生碰撞,已知碰撞過程時間極短,碰后木塊B恰好滑到A的左端停止,已知B與A間的動摩擦因數(shù)為,B在A板上單程滑行長度為,求:

(1)若,在B與擋板碰撞后的運動過程中,摩擦力對木板A做正功還是負功?做多少功?

(2)討論A和B在整個運動過程中,是否有可能在某一段時間里運動方向是向左的,如果不可能,說明理由;如果可能,求出發(fā)生這種情況的條件。

解:(1)B與A碰撞后,B相對A向左運動,A受摩擦力向左,而A的運動方向向右,故摩擦力對A做負功。

設(shè)B與A碰后的瞬間A的速度為,B的速度為,A、B相對靜止時的共同速度為,由動量守恒得:                ①

                                 ②

碰后到相對靜止,對A、B系統(tǒng)由功能關(guān)系得:

            ③

由①②③式解得:(另一解因小于而舍去)

這段過程A克服摩擦力做功為④(2)A在運動過程中不可能向左運動,因為在B未與A碰撞之前,A受摩擦力方向向右,做加速運動,碰后A受摩擦力方向向左,做減速運動,直到最后共同速度仍向右,因此不可能向左運動。

B在碰撞之后,有可能向左運動,即,結(jié)合①②式得:  ⑤

代入③式得:         ⑥

另一方面,整個過程中損失的機械能一定大于或等于系統(tǒng)克服摩擦力做的功,即      ⑦     即

故在某一段時間里B運動方向是向左的條件是

3、光滑水平面上放有如圖所示的用絕緣材料料成的“┙”型滑板,(平面部分足夠長),質(zhì)量為4m,距滑板的A壁為L1距離的B處放有一質(zhì)量為m,電量為+q的大小不計的小物體,物體與板面的摩擦不計,整個裝置處于場強為E的勻強電場中,初始時刻,滑板與物體都靜止,試求:

(1)釋放小物體,第一次與滑板A壁碰前物體的速度v1多大?

(2)若物體與A壁碰后相對水平面的速度大小為碰前的3/5,則物體在第二次跟A壁碰撞之前瞬時,滑板的速度v和物體的速度v2分別為多大?(均指對地速度)

(3)物體從開始運動到第二次碰撞前,電場力做功為多大?(碰撞時間可忽略)

3、解:(1)由動能定理

    ① 

(2)若物體碰后仍沿原方向運動,碰后滑板速度為V,

由動量守恒    得物體速度,故不可能  ②

∴物塊碰后必反彈,由動量守恒 ③  得

由于碰后滑板勻速運動直至與物體第二次碰撞之前,故物體與A壁第二次碰前,滑板速度⑤ 。

物體與A壁第二次碰前,設(shè)物塊速度為v2,   ⑥

由兩物的位移關(guān)系有: ⑦即   ⑧

由⑥⑧代入數(shù)據(jù)可得:    ⑨  

(3)物體在兩次碰撞之間位移為S, 

∴ 物塊從開始到第二次碰撞前電場力做功

文本框: 圖5―154(16分)如圖5―15所示,PR是一塊長為L=4 m的絕緣平板固定在水平地面上,整個空間有一個平行于PR的勻強電場E,在板的右半部分有一個垂直于紙面向外的勻強磁場B,一個質(zhì)量為m=0.1 kg.帶電量為q=0.5 C的物體,從板的P端由靜止開始在電場力和摩擦力的作用下向右做勻加速運動,進入

磁場后恰能做勻速運動.當物體碰到板R端擋板后被彈回,若在碰撞瞬間撤去電場,物體返回時在磁場中仍做勻速運動,離開磁場后做勻減速運動停在C點,PC=L/4,物體與平板間的動摩擦因數(shù)為μ=0.4.求:?

(1)判斷物體帶電性質(zhì),正電荷還是負電荷??

(2)物體與擋板碰撞前后的速度v1v2;?

(3)磁感應(yīng)強度B的大小;?

(4)電場強度E的大小和方向.?

解:(1)由于物體返回后在磁場中無電場,且仍做勻速運動,故知摩擦力為0,所以物體帶正電荷.且:mg=qBv2?

(2)離開電場后,按動能定理,有:?-μmg=0-mv2?得:v2=2 m/s?

(3)代入前式求得:B= T?

(4)由于電荷由P運動到C點做勻加速運動,可知電場強度方向水平向右,且:?

(Eq-μmg)mv12-0?

進入電磁場后做勻速運動,故有:Eq=μ(qBv1+mg)?

由以上兩式得:

5、 在原子核物理中,研究核子與核子關(guān)聯(lián)的最有效途徑是“雙電荷交換反應(yīng)”.這類反應(yīng)的前半部分過程和下述力學(xué)模型類似.兩個小球A和B用輕質(zhì)彈簧相連,在光滑的水平直軌道上處于靜止狀態(tài).在它們左邊有一垂直于軌道的固定擋板P,右邊有一小球C沿軌道以速度V0射向B球,如圖2所示.C與B發(fā)生碰撞并立即結(jié)成一個整體D.在它們繼續(xù)向左運動的過程中,當彈簧長度變到最短時,長度突然鎖定,不再改變.然后,A球與擋板P發(fā)生碰撞,碰后A、D都靜止不動,A與P接觸而不粘連.過一段時間,突然解除鎖定(鎖定及解除鎖定均無機械能損失).已知A、B、C三球的質(zhì)量均為

(1)求彈簧長度剛被鎖定后A球的速度.

(2)求在A球離開擋板P之后的運動過程中,彈簧的最大彈性勢能.

分析:審題過程,①排除干擾信息:“在原子核物理中,研究核子與核子關(guān)聯(lián)的最有效途徑是“雙電荷交換反應(yīng)”.這類反應(yīng)的前半部分過程和下述力學(xué)模型類似.”②挖掘隱含條件:“兩個小球A和B用輕質(zhì)彈簧相連,在光滑的水平直軌道上處于靜止狀態(tài)”,隱含摩擦不計和輕質(zhì)彈簧開始處于自然狀態(tài)(既不伸長,也不壓縮),“C與B發(fā)生碰撞并立即結(jié)成一個整體D”隱含碰撞所經(jīng)歷的時間極短,B球的位移可以忽略,彈簧的長度不變,“A球與擋板P發(fā)生碰撞,碰后A、D都靜止不動”隱含在碰撞中系統(tǒng)的動能由于非彈性碰撞而全部消耗掉,只剩下彈性勢能。

此題若用分析法求解,應(yīng)寫出待求量與已知量的關(guān)系式,顯然比較困難,由于物體所經(jīng)歷的各個子過程比較清楚,因此宜用綜合法求解。在解題前,需要定性分析題目中由A、B、C三個小球和連結(jié)A、B的輕質(zhì)彈簧組成的系統(tǒng)是如何運動的,這個問題搞清楚了,本題的問題就可較容易地得到解答.下面從本題中幾個物理過程發(fā)生的順序出發(fā)求解:

1、球C與B發(fā)生碰撞,并立即結(jié)成一個整體D,根據(jù)動量守恒,有

為D的速度)          ①

2、當彈簧的長度被鎖定時,彈簧壓縮到最短,D與A速度相等,如此時速度為,由動量守恒得                    ② 

當彈簧的長度被鎖定后,D的一部分動能作為彈簧的彈性勢能被貯存起來了.由能量守恒,有      ③

3、撞擊P后,A與D的動能都為0,當突然解除鎖定后(相當于靜止的A、D兩物體中間為用細繩拉緊的彈簧,突然燒斷細繩的狀況,彈簧要對D做正功),當彈簧恢復(fù)到自然長度時,彈簧的彈性勢能全部轉(zhuǎn)變成D的動能,設(shè)D的速度為,則有

4、彈簧繼續(xù)伸長,A球離開擋板P,并獲得速度。當A、D的速度相等時,彈簧伸至最長.此時的勢能為最大,設(shè)此時A、D的速度為,勢能為?由動量守恒定律得

                                ⑤ 

由機械能守恒定律得:         ⑥

由①、②兩式聯(lián)立解得:     、

聯(lián)立①②③④⑤⑥式解得                           ⑧

6、如圖(1)所示為一根豎直懸掛的不可伸長的輕繩,下端掛一小物塊A,上端固定在C點且與一能測量繩的拉力的測力傳感器相連。已知有一質(zhì)量為m0的子彈B沿水平方向以速度v0射入A內(nèi)(未穿透),接著兩者一起繞C點在豎直面內(nèi)做圓周運動。在各種阻力都可忽略的條件下測力傳感器測得繩的拉力F隨時間t的變化關(guān)系如圖(2)所示。已知子彈射入的時間極短,且圖(2)中t=0為A、B開始以相同速度運動的時刻,根據(jù)力學(xué)規(guī)律和題中(包括圖)提供的信息,對反映懸掛系統(tǒng)本身性質(zhì)的物理量(例如A的質(zhì)量)及A、B一起運動過程中的守恒量。你能求得哪些定量的結(jié)果?

解:由圖2可直接看出,A、B一起做周期性運動,運動的周期T=2t0,令 m表示 A的質(zhì)量,L表示繩長,v1表示 B陷入A內(nèi)時即t=0時 A、B的速度(即圓周運動最低點的速度),v2表示運動到最高點時的速度,F(xiàn)1表示運動到最低點時繩的拉力,f2表示運動到最高點時繩的拉力,則根據(jù)動量守恒定律,得mv0=( m0+m)v1,在最低點和最高點處運用牛頓定律可得  

F1-( m0+m)g=( m0+m)v12/L, F2+( m0+m)g=( m0+m)v22/L

    根據(jù)機械能守恒定律可得 2L( m+m0)g=( m+m0) v12/2- ( m+m0) v22/2。

    由圖2可知F2=0 。F1=Fm。由以上各式可解得,反映系統(tǒng)性質(zhì)的物理量是

    m=Fm/6g-m0 ,L =36m02v02 g/5Fm2,                

A、B一起運動過程中的守恒量是機械能E,若以最低點為勢能的零點,則E=(m+m0)v12/2。由幾式解得E=3m0202g/F。

7.(15分)中子星是恒星演化過程的一種可能結(jié)果,它的密度很大。現(xiàn)有一中子星,觀測到它的自轉(zhuǎn)周期為T=1/30s。向該中子星的最小密度應(yīng)是多少才能維持該星體的穩(wěn)定,不致因自轉(zhuǎn)而瓦解。計等時星體可視為均勻球體。(引力常數(shù)G=6.67×1011m3/kg?s2

8.(20分)曾經(jīng)流行過一種向自行車車頭燈供電的小型交流發(fā)電機,圖1為其結(jié)構(gòu)示意圖。圖中N、S是一對固定的磁極,abcd為固定在轉(zhuǎn)軸上的矩形線框,轉(zhuǎn)軸過bc邊中點、與ab邊平行,它的一端有一半徑r01.0cm的摩擦小輪,小輪與自行車車輪的邊緣相接觸,如圖2所示。當車輪轉(zhuǎn)動時,因摩擦而帶動小輪轉(zhuǎn)動,從而使線框在磁極間轉(zhuǎn)動。設(shè)線框由N=800匝導(dǎo)線圈組成,每匝線圈的面積S=20cm2,磁極間的磁場可視作勻強磁場,磁感強度B=0.010T,自行車車輪的半徑R135cm,小齒輪的半徑R2=4.cm,大齒輪的半徑R310.0cm(見圖 2),F(xiàn)從靜止開始使大齒輪加速轉(zhuǎn)動,問大齒輪的角速度為多大才能使發(fā)電機輸出電壓的有效值U=3.2V?(假定摩擦小輪與自行車輪之間無相對滑動)

7.(15分)參考解答:

考慮中子星赤道處一小塊物質(zhì),只有當它受到的萬有引力大于或等于它隨星體一起旋轉(zhuǎn)所需的向心力時,中子星才不會瓦解。

設(shè)中子星的密度為ρ,質(zhì)量為M,半徑為R,自轉(zhuǎn)角速度為ω,位于赤道處的小塊物質(zhì)質(zhì)量為m,則有GMm/R2=mω2R    且ω=2π/T,M=4/3πρR3

由以上各式得:ρ=3π/GT2

代人數(shù)據(jù)解得:ρ=1.27×1014kg/m3

8.(20分)參考解答:

當自行車車輪轉(zhuǎn)動時,通過摩擦小輪使發(fā)電機的線框在勻強磁場內(nèi)轉(zhuǎn)動,線框中產(chǎn)生一正弦交流電動勢,其最大值ε=ω0BSN

式中ω0為線框轉(zhuǎn)動的角速度,即摩擦小輪轉(zhuǎn)動的角速度。

發(fā)電機兩端電壓的有效值U=/2εm

設(shè)自行車車輪轉(zhuǎn)動的角速度為ω1,由于自行車車輪與摩擦小輪之間無相對滑動,有

R1ω1=R0ω0

小齒輪轉(zhuǎn)動的角速度與自行車輪轉(zhuǎn)動的角速度相同,也為ω1。設(shè)大齒輪轉(zhuǎn)動的角速度為ω,有R3ω=R2ω1

由以上各式解得ω=(U/BSN)(R2r0/R3r1)     代入數(shù)據(jù)得ω=3.2s1

9.(22分)一傳送帶裝置示意如圖,其中傳送帶經(jīng)過AB區(qū)域時是水平的,經(jīng)過BC區(qū)域時變?yōu)閳A弧形(圓弧由光滑模板形成,未畫出),經(jīng)過CD區(qū)域時是傾斜的,AB和CD都與BC相切,F(xiàn)將大量的質(zhì)量均為m的小貨箱一個一個在A處放到傳送帶上,放置時初速為零,經(jīng)傳送帶運送到D處,D和A的高度差為h。穩(wěn)定工作時傳送帶速度不變,CD段上各箱等距排列,相鄰兩箱的距離為L。每個箱子在A處投放后,在到達B之前已經(jīng)相對于傳送帶靜止,且以后也不再滑動(忽略經(jīng)BC段時的微小滑動)。已知在一段相當長的時間T內(nèi),共運送小貨箱的數(shù)目為N。這裝置由電動機帶動,傳送帶與輪子間無相對滑動,不計輪軸處的摩擦。求電動機的平均抽出功率

9.(22分)參考解答:

以地面為參考系(下同),設(shè)傳送帶的運動速度為v0,在水平段運輸?shù)倪^程中,小貨箱先在滑動摩擦力作用下做勻加速運動,設(shè)這段路程為s,所用時間為t,加速度為a,則對小箱有  s=1/2at2       ①       v0=at       ②

在這段時間內(nèi),傳送帶運動的路程為s0=v0t      ③

由以上可得s0=2s      ④

用f表示小箱與傳送帶之間的滑動摩擦力,則傳送帶對小箱做功為

A=fs=1/2mv02     ⑤

傳送帶克服小箱對它的摩擦力做功A0=fs0=2?1/2mv02    

兩者之差就是克服摩擦力做功發(fā)出的熱量Q=1/2mv02    ⑦

可見,在小箱加速運動過程中,小箱獲得的動能與發(fā)熱量相等。

T時間內(nèi),電動機輸出的功為W=T     ⑧

此功用于增加小箱的動能、勢能以及克服摩擦力發(fā)熱,即

W=1/2Nmv02+Nmgh+NQ      ⑨

已知相鄰兩小箱的距離為L,所以v0T=NL            ⑩

聯(lián)立⑦⑧⑨⑩,得:*[+gh]

10.(14分)為研究靜電除塵,有人設(shè)計了一個盒狀容器,容器側(cè)面是絕緣的透明有機玻璃,它的上下底面是面積A=0.04m2的金屬板,間距L=0.05m,當連接到U=2500V的高壓電源正負兩極時,能在兩金屬板間產(chǎn)生一個勻強電場,如圖所示。現(xiàn)把一定量均勻分布的煙塵顆粒密閉在容器內(nèi),每立方米有煙塵顆粒1013個,假設(shè)這些顆粒都處于靜止狀態(tài),每個顆粒帶電量為q=+1.0×10-17C,質(zhì)量為m=2.0×10-15kg,不考慮煙塵顆粒之間的相互作用和空氣阻力,并忽略煙塵顆粒所受重力。求合上電鍵后:⑴經(jīng)過多長時間煙塵顆粒可以被全部吸附?⑵除塵過程中電場對煙塵顆粒共做了多少功?⑶經(jīng)過多長時間容器中煙塵顆粒的總動能達到最大?

[⑴當最靠近上表面的煙塵顆粒被吸附到下板時,煙塵就被全部吸附。煙塵顆粒受到的電場力F=qU/L,L=at2/2=qUt2/2mL,故t=0.02s

⑵W=NALqU/2=2.5×10-4J                                                 

⑶設(shè)煙塵顆粒下落距離為x,則當時所有煙塵顆粒的總動能

EK=NA(L-x) mv2/2= NA(L-x)  qUx/L,當x=L/2時EK達最大,而x=at12/2,故t1=0.014s ]

 

11.(12分)風(fēng)洞實驗室中可以產(chǎn)生水平方向的、大小可調(diào)節(jié)的風(fēng)力,現(xiàn)將一套有小球的細直桿放入風(fēng)洞實驗室,小球孔徑略大于細桿直徑。

(1)當桿在水平方向上固定時,調(diào)節(jié)風(fēng)力的大小,使小球在桿上作勻速運動,這時小班干部所受的風(fēng)力為小球所受重力的0.5倍,求小球與桿間的滑動摩擦因數(shù)。

(2)保持小球所受風(fēng)力不變,使桿與水平方向間夾角為37°并固定,則小球從靜止出發(fā)在細桿上滑下距離S所需時間為多少?(sin37°=0.6,cos37°=0.8)

13.(1)設(shè)小球所受的風(fēng)力為F,小球質(zhì)量為

    1       2

(2)設(shè)桿對小球的支持力為N,摩擦力為

沿桿方向3

垂直于桿方向4

       5

可解得6

  7         8

評分標準:(1)3分。正確得出2式,得3分。僅寫出1式,得1分。

(2)9分,正確得出6式,得6分,僅寫出3、4式,各得2分,僅寫出5式,得1分,正確得出8式,得3分,僅寫出7式,得2分,g用數(shù)值代入的不扣分。

12.(13分)閱讀如下資料并回答問題:

自然界中的物體由于具有一定的溫度,會不斷向外輻射電磁波,這種輻射因與溫度有關(guān),稱為勢輻射,勢輻射具有如下特點:1輻射的能量中包含各種波長的電磁波;2物體溫度越高,單位時間從物體表面單位面積上輻射的能量越大;3在輻射的總能量中,各種波長所占的百分比不同。

處于一定溫度的物體在向外輻射電磁能量的同時,也要吸收由其他物體輻射的電磁能量,如果它處在平衡狀態(tài),則能量保持不變,若不考慮物體表面性質(zhì)對輻射與吸收的影響,我們定義一種理想的物體,它能100%地吸收入射到其表面的電磁輻射,這樣的物體稱為黑體,單位時間內(nèi)從黑體表面單位央積輻射的電磁波的總能量與黑體絕對溫度的四次方成正比,即,其中常量瓦/(米2?開4)。

在下面的問題中,把研究對象都簡單地看作黑體。

有關(guān)數(shù)據(jù)及數(shù)學(xué)公式:太陽半徑千米,太陽表面溫度開,火星半徑千米,球面積,,其中R為球半徑。

(1)太陽熱輻射能量的絕大多數(shù)集中在波長為2×109~1×104范圍內(nèi),求相應(yīng)的頻率范圍。

(2)每小量從太陽表面輻射的總能量為多少?

(3)火星受到來自太陽的輻射可認為垂直射可認為垂直身到面積為為火星半徑)的圓盤上,已知太陽到火星的距離約為太陽半徑的400倍,忽略其它天體及宇宙空間的輻射,試估算火星的平均溫度。

解:.(1)     1  (赫)   2

(赫)       3

輻射的頻率范圍為3×1012赫-1.5×1017

(2)每小量從太陽表面輻射的總能量為

           4代入數(shù)所得W=1.38×1010焦        5

(3)設(shè)火星表面溫度為T,太陽到火星距離為,火星單位時間內(nèi)吸收來自太陽的輻射能量為      6 

        7

火星單位時間內(nèi)向外輻射電磁波能量為      8

火星處在平衡狀態(tài)    9即      10

由10式解得火星平均溫度(開)    11

評分標準:全題13分

(1)正確得了1,2,3式,各得1分。(2)正確得出5式,得5分,僅寫出4式,得3分。(3)正確得出10式,得4分,僅寫出6式或7式,得1分;僅寫出8式,得1分,正確得出11式,得1分。

13.(13分)如圖所示,在xoy平面內(nèi)有垂直坐標平面的范圍足夠大的勻強磁場,磁感強度為B,一帶正電荷量Q的粒子,質(zhì)量為,從點以某一初速度垂直射入磁場,其軌跡與x、y軸的交點ABO點的距離分別為、,試求:

(1)初速度方向與軸夾角θ

(2)初速度的大小. 20.參考解答:

(1)磁場方向垂直坐標平面向里時,粒子初速度方向與軸的夾角為θ,射入磁場做勻速圓周運動,由幾何關(guān)系可作出軌跡如圖所示,設(shè)圓半徑為R,由數(shù)學(xué)關(guān)系可得:?

      ①           ②

由①、②解得tgθ   ∴θ=arctg        ③?

當磁場方向垂直坐標平面向外時,粒子初速度方向與軸間的夾角為

π+θ=π+arctg          ④

(2)由①、②解得:

由洛侖茲力提供向心力有:?QvB      ⑥

             ⑦

評分標準:本題13分,第(1)問8分,其中①式2分,②式2分,③式2分,④式2分,θ=arcsinθ=arccos同樣給分.

第(2)問5分,其中⑤式2分,⑥式2分,⑦式1分.

14.(13分)俄羅斯“和平號”空間站在人類航天史上寫下了輝煌的篇章,因不能保障其繼續(xù)運行,3月20號左右將墜入太平洋.設(shè)空間站的總質(zhì)量為,在離地面高度為的軌道上繞地球做勻速圓周運動?墜落時地面指揮系統(tǒng)使空間站在極短時間內(nèi)向前噴出部分高速氣體,使其速度瞬間變小,在萬有引力作用下下墜.設(shè)噴出氣體的質(zhì)量為 m,噴出速度為空間站原來速度的37倍,墜入過程中外力對空間站做功為W.求:

(1)空間站做圓周運動時的線速度.

(2)空間站落到太平洋表面時的速度.

(設(shè)地球表面的重力加速度為,地球半徑為R)

14.參考解答:

(1)  設(shè)空間站做圓周運動的速度為,地球質(zhì)量為M.由牛頓第二定律得:

          ①

地表重力加速度為,則:  ②  由①、②式得:   ③

(2)  噴出氣體后空間站速度變?yōu)?i>v,由動量守恒定律得:

            ④

設(shè)空間站落到太平洋表面時速度為

由動能定理得:            ⑤

由③、④、⑤式得:?              ⑥

15.(14分)如圖甲,A、B兩板間距為,板間電勢差為U,C、D兩板間距離和板長均為L,兩板間加一如圖乙所示的電壓.在S處有一電量為q、質(zhì)量為m的帶電粒子,經(jīng)A、B間電場加速又經(jīng)C、D間電場偏轉(zhuǎn)后進入一個垂直紙面向里的勻強磁場區(qū)域,磁感強度為B.不計重力影響,欲使該帶電粒子經(jīng)過某路徑后能返回S處.求:

(1)勻強磁場的寬度L′至少為多少?

(2)該帶電粒子周期性運動的周期T.

 

 

 

 

 

 

(1)AB加速階段,由動能定理得:     ①

偏轉(zhuǎn)階段,帶電粒子作類平拋運動?偏轉(zhuǎn)時間    ②

側(cè)移量  ③

    設(shè)在偏轉(zhuǎn)電場中,偏轉(zhuǎn)角為θ?

θ  ④?

由幾何關(guān)系:cos45°+′⑤

sin45°=    ⑥   則 ′=    ⑦

注:L′也可由下面方法求得:

粒子從S點射入到出偏轉(zhuǎn)電場,電場力共做功為=2qU  

設(shè)出電場時速度為′,有  解得′=    ⑨

粒子在磁場中做圓周運動的半徑:

      ⑩

(2)設(shè)粒子在加速電場中運動的時間為

              11

帶電粒子在磁場中做圓周運動的周期?   12

實際轉(zhuǎn)過的角度α=2π-2θ=           13

在磁場中運動時間          14

故粒子運動的周期T=2+2=4   15

評分標準:本題14分,第(1)問8分,其中①、②、③式各1分,④式2分,⑤、⑥、⑦式各1分.第(2)問6分,其中 12、13、14、式各1分,15式2分.

 

分析與解:由于微波在大氣層中以光速傳播,所以若能求得從同步衛(wèi)星到嘉峪關(guān)的距離L,則由運動學(xué)就能得到同步衛(wèi)星發(fā)出的微波信號傳到位于嘉峪關(guān)的接收站所需的時間t。如何求得L是解題的關(guān)鍵,首先我們知道同步衛(wèi)星是位于赤道上空的,題中說明,該同步衛(wèi)星的定點位置與東經(jīng)98°的經(jīng)線在同一平面內(nèi),而嘉峪關(guān)處的經(jīng)度和緯度近似取為東經(jīng)98°和北緯α=40°,隱含該同步衛(wèi)星P、嘉峪關(guān)Q和地心O在同一個平面內(nèi),構(gòu)成一個三角形,∠QOP=α=40°,如圖11所示,這樣由余弦定理就可求得L。
  設(shè)m為衛(wèi)星質(zhì)量,M為地球質(zhì)量,r為衛(wèi)星到地球中心的距離,w為衛(wèi)星繞地轉(zhuǎn)動的角速度,由萬有引力定律和牛頓定律有。
 、    ①
  式中G為萬有引力恒量,因同步衛(wèi)星繞地心轉(zhuǎn)動的角速度w與地球自轉(zhuǎn)的角速度相等,有                ②
  因     得    ③
設(shè)嘉峪關(guān)到同步衛(wèi)星的距離為L,由余弦定理
           ④
  所求時間為               ⑤
  由以上各式得

   ⑥

17、“和平號”空間站已于2001年3月23日成功地墜落在南太平洋海域,墜落過程可簡化為從一個近圓軌道(可近似看作圓軌道)開始,經(jīng)過與大氣摩擦,空間站的絕大部分經(jīng)過升溫、熔化,最后汽化而銷毀,剩下的殘片墜入大海。此過程中,空間站原來的機械能中,除一部分用于銷毀和一部分被殘片帶走外,還有一部分能量E′通過其他方式散失(不考慮墜落過程中化學(xué)反應(yīng)的能量)。

(1)              試導(dǎo)出以下列各物理量的符號表示散失能量E′的公式.

(2)              (2)算出E′的數(shù)值(結(jié)果保留兩位有效數(shù)字)

墜落開始時空間站的質(zhì)量M=1.17×105kg;

軌道離地面的高度為h=146km;

地球半徑R=6.4×106m;

墜落空間范圍內(nèi)重力加速度可看作g=10m/s2;

入海殘片的質(zhì)量=1.2×104kg

入海殘片的溫度升高=3000K;

入海殘片的入海速度為聲速=340m/s;

空間站材料每1kg升溫1K平均所需能量c=1.0×103J;每銷毀1kg材料平均所需能量μ=1.0×107J.?

分析與解:本題描述的是2001年世界矚目的一件大事:“和平號”空間站成功地墜落在南太平洋海域。讓繞地球運行的空間站按照預(yù)定的路線成功墜落在預(yù)定的海域,這件事情本身就極富挑戰(zhàn)性,表達了人類征服自然改造自然的雄心和實力。

(1)作為一道信息題,首先我們應(yīng)弄清題目所述的物理過程,建立一個正確的物理模型。 我們將空間站看作一個質(zhì)點,開始時以一定的速度繞地球運行,具有一定的動能和勢能,墜落開始時空間站離開軌道,經(jīng)過摩擦升溫,空間站大部分升溫、熔化,最后汽化而銷毀,剩下的殘片墜落大海,整個過程中,總能量是守恒的。

根據(jù)題述條件,從近圓軌道到地面的空間中重力加速度g=10m/s2,若以地面為重力勢能的零點,墜落過程開始時空間站在近圓軌道的勢能

.   、
  以v表示空間站在軌道上的速度,可得 . 、
  其中r為軌道半徑,若R表示地球半徑,則r=R+h.  、
  由式②、③可得空間站在軌道上的動能
             (R+h)   ④
  由式①、④可得,在近圓軌道上空間站的機械能
  E=Mg(R+h)     ⑤
  在墜落過程中,用于銷毀部分所需要的能量為Q=(M-m)μ.⑥
  用于殘片升溫所需要的能量Q=cmΔT.⑦
  殘片的動能為E=   ⑧
  以E′表示其他方式散失的能量,則由能量守恒定律可得
       E=Q+E+Q+E′.         ⑨

  由此得E′=Mg(R+h)-(M-m)μ--cmΔT   ⑩

(2)將題給數(shù)據(jù)代入得E′=2.9×1012J.?

帶電粒子在電磁場中運動問題

帶電粒子在電磁場中運動問題,實質(zhì)是力學(xué)問題,通常從受力分析,運動情況分析入手,利用力學(xué)規(guī)律,并注意幾何關(guān)系即可求解。下面對兩道高考壓軸題作一簡要分析。  

18、一帶電質(zhì)點質(zhì)量為m電量為q,以平行于ox軸的速度v從y軸上的a點射入圖7中第一象限所示的區(qū)域

試題詳情

1.  平行四邊形ABCD,邊長 AB = a, AD = 1,  角 BAD = A, 已知三角形ABD是一個銳角三角形,求證以A,B,C,D為圓心半徑為1的四個圓能夠覆蓋此平行四邊形的充要條件是

 a ≤ cos A + √3 sin A.

2.  若四面體有且僅有一邊大于1,求證其體積 ≤ 1/8.

3.  k, m, n 是自然數(shù) 且 m + k + 1 是一個大于 n+1 的素數(shù),令cs = s(s+1),求證

(cm+1 - ck)(cm+2 - ck) ... (cm+n - ck)

可被乘積 c1c2 ... cn整除。

4.  任意兩個銳角三角形 A0B0C0 和 A1B1C1 。考慮所有與三角形 A1B1C1相似且外接于三角形 A0B0C0 的所有三角形ABC(即BC邊包含A0,CA邊包含B0,AB邊包含 C0),試構(gòu)造出滿足此條件的面積最大的三角形ABC。

5.  a1, ... , a8 是不全為0的實數(shù),令 cn = a1n + a2n + ... + a8n ( n = 1, 2, 3, ... ),如果數(shù)列{ cn }中有無窮多項等于0,試求出所有使 cn=0 的自然數(shù)n。

6.  在一次運動會中,連續(xù) n 天內(nèi)(n>1)一共頒發(fā)了 m 塊獎牌。在第一天,頒發(fā)了一塊獎牌以及剩下 m-1 個中的 1/7;在第二天頒發(fā)了兩塊獎牌以及剩下的 1/7;依此類推。在最后一天即第 n 天,剩下的n塊獎牌全部頒發(fā)完畢。問該運動會共進行了幾天,一共頒發(fā)了多少塊獎牌?

 

試題詳情

1.  在一次數(shù)學(xué)競賽中共有A、B、C三道題,25名參賽者每人至少答對了一題。在所有沒有答對A的學(xué)生中,答對B的人數(shù)是答對C的人數(shù)的兩倍,只答對問題A的人數(shù)比既答對A又至少答對其他一題的人數(shù)多1。又已知在所有恰好答對一題的參賽者中,有一半沒有答對A。請問有多少學(xué)生只答對B?

2.  三角形ABC,如果,

BC + AC = tan C/2 (BC tan A + AC tan B).

則該三角形為等腰三角形。

3.  求證:從正四面體的內(nèi)切圓圓心到各頂點距離之和小于從空間中任意其他點到各頂點距離之和。

4.  對任何自然數(shù) n以及滿足 sin 2nx 不為 0 的實數(shù)x,求證:

1/sin 2x + 1/sin 4x + ... + 1/sin 2nx = cot x - cot 2nx.

5.   ai (i=1,2,3,4)是互不相同的實數(shù),解方程組(i=1,2,3,4)

|ai - a1| x1 + |ai - a2| x2 + |ai - a3| x3 + |ai - a4| x4 = 1。

6.  在三角形ABC的邊BC、CA、AB上分別任選三內(nèi)點K、L、M,求證三角形AML、BKM、CLK之中至少有一個的面積小于活等于三角形ABC的四分之一。

 

試題詳情

1.  試找出所有位于區(qū)間[0, 2pi] 的x使其滿足

        2 cos x ≤ | √(1 + sin 2x) - √(1 - sin 2x)| ≤ √2 .

2.  如下方程組的系數(shù) aij ,

         a11x1 + a12 x2+ a13 x3 = 0

         a21x1 + a22x2 + a23x3 = 0

         a31x1 + a32x2 + a33x3 = 0

滿足:

求證:該方程組的有唯一的解 x1 = x2 = x3 = 0。

3.  四面體ABCD被平行于AB、CD邊的一個平面分割成兩部分,并且該平面到AB邊的距離是該平面到CD邊距離的 k倍。試求出 這兩部分的體積比。

4.  四個實數(shù),它們中的任何三個的乘積再加上第四個數(shù)都等于2,求出這四個數(shù)的所有可能值。

5.  三角形OAB中的角O是銳角,M是邊AB上任意一點,從M向OA、OB邊引垂線,垂足分別為P、Q。設(shè)三角形OPQ的垂心為,求出當M在AB邊上移動時點H的軌跡;若M在三角形OAB內(nèi)部移動是H的軌跡又是什么?

6.  平面上給定了 n>2個點,任何兩點之間都有線斷相連,這些線斷長度中的最大值被定義為這個點集的直徑,求證:長度為直徑的線斷至多有n條。

 

試題詳情

1.   (a)  求所有正整數(shù) n 使得  2n - 1 能被 7整除;

   (b) 求證不存在正整數(shù) n 使得 2n + 1 能被 7 整除。

2.  假設(shè)a、b、c是某三角形的三邊長,求證:

a2(b + c - a) + b2(c + a - b) + c2(a + b - c) <= 3abc.

3.  三角形ABC的三邊長為別為a、b、c。分別平行于ABC的各邊作三角形ABC內(nèi)切圓的切線,每條切線都在ABC中又切出一個小三角形,再在每個這樣的小三角形中作內(nèi)切圓,求這四個內(nèi)切圓的面積之和(用a,b,c表示)。

4.十七個人互相通信,每一個人都和其他人寫信。在他們的信上一共討論有三個不同的話題,每兩個人只討論一個話題,求證:這些人當中至少有三個人他們所討論的話題是一樣的。

5.平面上有五個點,任意兩點的連線都不平行,也不垂直,現(xiàn)從每一個點向其他四點兩兩連接的直線作垂線,試求出所有這些垂線的交點的最大數(shù)目。

6.四面體ABCD的中心是D0 ,分別過A、B、C作 DD0 的平行線,這些線分別交平面BCD、CAD、ABD于點 A0、 B0、 C0,求證:ABCD的體積是A0B0C0D0的三分之一;再問如果 D0 為三角形ABC內(nèi)的任意一點,結(jié)果是否仍然成立?

 

試題詳情

1.  找出下列方程的所有實數(shù)根(其中 p是實參數(shù)):

 √(x2-p)+2√(x2-1) = x.

2.  給定一點A及線斷BC,設(shè)空間中一點P使得存在線段BC上有一點X滿足 角APX是直角,試求出所有這樣的點P的軌跡。

3.  在一個 n邊形中,所有內(nèi)角都相等,邊長依次是

 a1 >= a2 >= ... >= an,

求證:所有邊長都相等。

4.  設(shè) y是一個參數(shù),試找出方程組 xi + xi+2 = y xi+1 (i = 1, ... , 5)的所有解 x1, ... , x5。

5.  求證

cos pi/7 - cos 2pi/7 + cos 3pi/7 = 1/2.

6.  五個同學(xué)A、B、C、D、E參加競賽,一種猜測說比賽結(jié)果的名次依然是ABCDE。但是實際上沒有一位同學(xué)的名次被猜中,而且預(yù)測中名次相鄰的同學(xué)也沒有真的相鄰(例如,C、D兩位同學(xué)名次不是(1,2)、(2,3)、(3,4)、(4,5)中的任何一種)。還有一種猜測說結(jié)果會是DAECB的順序。實際上是恰好有兩個同學(xué)所得的名次與預(yù)測的一樣;而且有兩對同學(xué)(4個不同的同學(xué))的名次像預(yù)測中的一樣是相連。試討論最后的名次如何?

 

試題詳情

1.  找出具有下列各性質(zhì)的最小正整數(shù) n:它的最后一位數(shù)字是6,如果把最后的6去掉并放在最前面所得到的數(shù)是原來數(shù)的4被。

2.  試找出滿足下列不等式的所有實數(shù) x:

√(3-x)- √(x+1) > 1/2.

3.  正方體 ABCDA'B'C'D'(ABCD、A'B'C'D'分別是上下底)。一點 x沿著正方形ABCD的邊界以方向ABCDA作勻速運動;一點Y以同樣的速度沿著正方形B'C'CB的邊界以方向B'C'CBB'運動。點X、Y在同一時刻分別從點A、B'開始運動。求線斷XY的中點的軌跡。

4.  解方程cos2x + cos22x + cos23x = 1。

5.  在圓K上有三個不同的點A、B、C。試在K上再作出一點D使得這四點所形成的四邊形有一個內(nèi)切圓。

6.  一個等腰三角形,設(shè)R為其外接圓半徑,內(nèi)切圓半徑為 r,求證這兩個圓的圓心的距離是√(R(R-2r))。

7.  求證:正四面體有5個不同的球,每個球都與這六條邊或其延長線相切;

反過來,如果一個四面體有5個這樣的球,則它必然是正四面體。

 

試題詳情


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