19、解:(1)由于PQ部分光滑,滑块A只在电场力作用下加速运动,设经时间tB相碰,AB相遇前的速度大小为v1,结合后的共同速度大小为v2,则

  ····························(2分)

·····························(2分)

解得s······························(1分)

m / s

滑块AB碰撞的过程中动量守恒,即·········(2分)

m / s·····························(1分)

(2)两滑块共同运动,与墙壁发生碰撞后返回,第一次速度为零时,两滑块离开墙壁的距离最大,设为,在这段过程中,由动能定理得

·············(3分)

解得m···························(1分)

(3)由于N,N,,即电场力大于滑动摩擦力,AB向右速度为零后在电场力的作用下向左运动,最终停在墙角O点处,设由于摩擦而产生的热为Q,由能量守恒得

J·····················(2分)

AB第二次与墙壁发生碰撞后返回,滑块离开墙壁的最大距离为,假设L2s,在这段过程中,由动能定理得

解得L2≈0.064m

L2<s=0.15m ,符合假设,即AB第二次与墙壁发生碰撞后返回停在Q点的左侧,以后只在粗糙水平面OQ上运动。························(2分)

设在粗糙水平面OQ部分运动的总路程s1,则······(2分)

s1=0.6m·······························(1分)

AB相碰结合后的运动过程中通过的总路程是s2,则

···························(2分)

m······························(1分)

另解:(2)从Q点到与墙壁发生碰撞后返回Q点,设返回Q点时速度为v3,则

·················(2分)

m/s。    过Q点向右还能运动s1,则··(1分)

s1=0.075m。    m··················(1分)

(3) 设AB靠近墙壁过程的加速度为a1, 远离墙壁过程的加速度为a2,则

5/3m/s235/3m/s2·······················(1分)

设第三次、第四次……第n次与墙壁发生碰撞后返回,滑块离开墙壁的最大距离分别为L3L4Ln,则第二次返回与第三次远离过程中··········(1分)

同理得……,·······(1分)

,……,······(1分)

AB相碰结合后的运动过程中通过的总路程是

……

=m·····························(2分)

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