Question

如图所示,水平桌面离地面高h=1.25 m.小物块A静止在桌面上,距桌右边缘l=1.00 m,小物块B从桌面的左边缘向A运动,并与之发生正碰(碰撞时间极短).碰后A从桌面的右边缘以垂直边缘的速度飞出,其落地点距桌面右边缘的水平距离s=0.75 m,B刚好到达桌面右边缘.A和B质量均为m=0.1 kg,与桌面间的动摩擦因数均为μ=0.2.重力加速度为g=10 m/s².试求:

(1)小物块A从桌面边缘飞出时的速度大小;

(2)小物块B与A碰撞过程中B与A组成的系统损失的机械能.

Answer 1

解:(1)物块A做平抛运动的时间t==0.5 s                                   ①

从桌面边缘飞出时的速度v==1.5 m/s                                           ②

(2)碰撞前A的速度为v₀,碰撞后,A的速度为v₁,B的速度为v₂,

据动能定理,有-μmgl=mv²-mv₁²                                              ③

-μmgl=0-mv₂²                                                               ④

代入数据,解得v₁=2.5 m/s                                                        ⑤

v₂=2 m/s                                                                      ⑥

据动量守恒定律,有mv₀=mv₁+mv₂                                                 ⑦

解得v₀=4.5 m/s                                                                 ⑧

碰撞损失的机械能E_损=mv₀²-mv₁²-mv₂²=0.5 J                               ⑨