摘要：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t与B相碰.A与B相遇前的速度大小为v1.结合后的共同速度大小为v2.则 ···························· ····························· 解得s······························ m / s 滑块A.B碰撞的过程中动量守恒.即········· m / s····························· (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 ············· 解得m··························· (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····················· 设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L2＜s.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.························ 设在粗糙水平面OQ部分运动的总路程s1.则······ s1＝0.6m······························· 设AB相碰结合后的运动过程中通过的总路程是s2.则 ··························· m······························ 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 ················· m/s. 过Q点向右还能运动s1.则·· s1＝0.075m. m·················· (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/s2.35/3m/s2······················· 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L3.L4-Ln.则第二次返回与第三次远离过程中·········· 同理得--.······· .--.······ AB相碰结合后的运动过程中通过的总路程是 -- =m·····························

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