带电小球对心碰撞(固定一球) 电气0902李玮仓
带电小球对心碰撞（固定一球） 电气0902 李玮仑
十 十 R k 41q X 919 41q dx= mvdr (d-x) a-x
x d x22 2 1 1 2 ( ) dt d x m d x q q k = − v dx dv m dt dx dx dv m d x q q k 2 1 1 1 2 ( ) = = − = − vv x dx m vdv d x kq q 1 2 2 1 0 1 ( ) 1 1 2 1 11 2 0 2 1 v v v m x m d x kq q = − 1 v R
2 kgg 1 kq, q d-x 2 2kq92 1.2 2kqq2 dx d-x are 2kg, q dx tk dx= dt a-x +k2 令k+k2=m X
2 2 1 2 1 1 2 1 1 2 1 2 v m v m d k q q d x k q q − = − − dt dx m d k q q v m d x k q q v + − = − = 1 2 1 2 1 1 1 2 2 1 2 1 1 2 1 2 m k q q k = m d k q q k v 1 2 1 2 2 1 2 = − 1 2 1 k d x k dt dx + − = = + − x t dx dt k d x k 0 0 1 2 1 1 令 k m d x k + = − 1 2 1
2k, m-dm=2k1 dm积分上下限省写 2 (m2-k2) 令 sec sec 6 2k, √k2see=2kk2 j tan 0 de k. tan 0 2K k2 6d0- csc 0d0 sin 6 =2k,k,2 . 0X cOS 1c1 de- de 2sin0 2J sin e Sin 6 cos e =2kk,2 In(csc 0-cot 0) 2sin-0 2
dm m k dm k m k k m m − = − 2 2 2 1 2 1 ( ) 1 2 ( ) 1 22 2 积分上下限省写 令 m = k2 sec k d k k d k k − = 3 23 2 1 2 2 3 2 1 tan sec sec 2 tan1 2 = − + − = − −− k k d d k k d d sin1 sin1 21 2sin cos 2 csc sin1 2 2 23 1 2 3 23 1 2 = − − − − ln(csc cot ) 21 2sin cos 2 2 23 1 2 k k
代入积分上下限上限1k+k2=√k2sece k(d-x sec 6 ktk, d-x cos e h,(d-x) k1+k2(d-x) k k1+k2(d-x) sin e= CSc 6 k1+k2(d-x) k, cot 6= k 下限x=0有Mk+k2=Vk2sec sec0=k+h,d kd COS 6 k,+k,d
代入积分上下限 上限 sec 1 1 2 2 k k d x k + = − ( ) ( ) sec 2 1 2 k d x k k d x − + − = ( ) ( ) cos 1 2 2 k k d x k d x + − − = ( ) sin 1 2 1 k k d x k + − = 1 1 2 ( ) csc k k + k d − x = 1 2 ( ) cot k k d − x = 下限 x = 0 有 sec 1 1 2 2 k k d k + = k d k k d 2 1 2 sec + = k k d k d 1 2 2 cos + =
sin e cscO+kd k,+k,d cot= k 将积分上下限代入得碰撞前小球的类运动学方程: k2/ d2+k,k,d-vkk,(d-x)+k2(d-x))+kk22n k+k,d-vk,d k+k,(d-x)-vk,(d-x 2kq1g2 I kgn +y d-x 2kq,g2 2 2kg, q
k k d k 1 2 1 sin + = 1 1 2 csc k k + k d = 1 2 cot k k d = 将积分上下限代入得碰撞前小球的类运动学方程： t k k d x k d x k k d k d k k d k k d k k d x k d x k k = + − − − + − + − − + − + − − ( ( ) ( ( ) ( ) ) ln 1 2 2 1 2 2 2 3 1 2 2 2 1 2 1 2 2 2 2 2 2 3 2 ） m d k q q v m d x k q q v 1 2 1 2 1 1 1 2 2 1 2 + − − = 1 1 2 1 2 m kq q k = m d k q q k v 1 2 1 2 2 1 2 = −
值得思考的是积分过程中令m=√k,Secb 那么方程成立时些须满足k2=n 2k1g2>0 X 碰撞后两小球电性中和并平分剩余电荷不妨设 q1>q2那么碰撞后各小球的带电量为q 2
值得思考的是 积分过程中 令 m = k2 sec 那么方程成立时必须满足 0 2 1 2 1 2 2 = 1 − m d k q q k v o x 碰撞后两小球电性中和并平分剩余电荷 不妨设 q1 q2 那么碰撞后各小球的带电量为 2 q1 q2 q − =
k O= k 12 dt dy dx' k 12 d x' dt d x 2R rI 2 ka 代入碰撞前小球速度公式求得碰撞后小球初速 ke 41q 2kqig2 1 mr
2 2 2 1 2 ( ) ( ) dt d d x m d x q k − = − 令 d − x = x' 2 2 2 1 2 ' ' dt d x m x q k = v dx dv m dt dx dx dv m x q k ' ' ' ' 2 1 1 2 = = vdv k q m dx x v v x R = ' 2 1 ' 2 2 1 ' ' 1 m d k q q v m R k q q v 1 2 1 2 1 1 1 2 1 2 ' = + − 代入碰撞前小球速度公式求得碰撞后小球初速
x' 2R kg2 2 2 x' kq kg R R 2,q12 C k R m v k k k x dt
) 2 ' 2 ( 2 1 ' 1 2 1 2 1 2 v v k q m x R − + = − q R d R q q v k q m v k q m x 1 ) 1 2 ( ' 2 2 1 2 1 2 2 1 1 2 2 = − + − + a q R d R q q v k q m + − + = 1 ) 1 2 ( 2 2 1 2 1 1 2 令 1 2 2 ' 2 v k q m a x = − ' 2 1 x q a m k v = − ' ' 2 1 x q a m k dt dx = −
k dx 2R 2 h d 4 dx= 4p 2 2省写积分上下限 p(a-p) a1p 令P=√acos94=- vasin g -B)中=4=√aSm9M3r1 a sin gp sin
令 p x a − = ' 2 dp a p p dx 2 2 ( ) 4 ' − = = − t t x R q dt m k dx x a ' 1 ' 2 ' ' 2 1 − = − • dp a p dp a p p p 2 2 2 2 ( ) 1 4 ( ) 1 4 省写积分上下限 令 p = a cos dp = − a sin d − = − − = − d a d a a dp a p 3 2 3 2 2 2 4 sin 1 4 sin sin 4 ( ) 1 4