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191问答库 > 质量为m=3kg的小球从5m高处自由下落，与地面碰撞后反弹到1.8m高处，已知小球和地面的接触时间为0.5s，不

质量为m=3kg的小球从5m高处自由下落，与地面碰撞后反弹到1.8m高处，已知小球和地面的接触时间为0.5s，不

2025-03-15 11:15:25

推荐回答（1个）

回答1：

解答：解：在整个过程中，规定竖直向下为正方向．
①钢球与地面碰撞前做自由落体运动，根据机械能守恒有：
mgh₁=

1

2

mv₁²
所以球落地时的速度为：v₁=
2gh1
=

2×10×5
=10m/s
②钢球碰撞后向上做竖直上抛运动，根据机械能守恒定律有：mgh₂=

1

2

mv₂²
所以钢球弹起，刚离地时的速度为：v₂=-
2gh2
=-

2×10×1.8
=-6m/s.8s
③在钢球与地面碰撞过程中，所用的时间：△t=0.5s
运用动量定理有：（mg-F_N）?△t=mv₂-mv₁
代入数据得：（3×10-F_N）×0.5=3×（-6-10）
所以有：F_N=126N
故选C．

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