(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 ····························(2分) ·····························(2分) 解得s······························(1分) m / s 滑块A、B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) 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。························(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) 设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) |