Question

在沙堆上放一木块（质量为M=5 kg),木块上放一爆竹（质量为m=0.10 kg).点燃爆竹后由于爆炸木块陷于沙中5 cm.若沙对运动木块的阻力为58 N，不计爆竹中火药的质量和空气阻力.求爆竹上升的高度.（g=10 m/s²)??????????

Answer 1

答案：
解析：

解析：爆竹爆炸时爆竹和木块动量守恒.取向上为正方向，由动量守恒定律得：? mv+(-MV)=0                                                                                                           ① (式中v、V分别是爆炸后爆竹和木块速度）? 木块陷于沙中，做匀减速运动，由动能定理得：? 0-MV2=-（F-Mg）s                                                                                             ② 爆炸后爆竹以v做竖直上抛运动，设达到的高度为h，则： 0-v2=-2gh                                                                                                                ③ 由②得V=0.4 m/s                                                                                                    ④ ④代入①得v=20 m/s                                                                                              ⑤ ⑤代入③得h=20 m?

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