题目

如右图,一质量为M的物块静止在桌面边缘,桌面离水平地面的高度为h．一质量为m的子弹以水平速度v0射入物块后,以水平速度v0/2射出．重力加速度为g．求: （1）此过程中系统损失的机械能； （2）此后物块落地点离桌面边缘的水平距离． 答案：（10分）  答案   （1）（3-）mv02   （2）解析   （1）设子弹穿过物块后物块的速度为v,由动量守恒定律得mv0=m+Mv                                                                                                                                           ①解得v=v0                                                                                                                                                                                                                                                                                      ②系统的机械能损失为ΔE=mv02 -[m（）2 +Mv2]                                                                                                       ③由②③式得ΔE=（3-）mv02                                                                                                                                                                                 ④（2）设物块下落到地面所需时间为t,落地点距桌面边缘的水平距离为s,则h=gt2                                                                                                                                                                                                               ⑤s=vt                                                                                                                                      ⑥由②⑤⑥式得s=                    ⑦

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