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

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．

（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                                                                                                                                                                                                               ⑤

s=vt                                                                                                                                      ⑥

由②⑤⑥式得s=                    ⑦