Question

如圖所示，兩個(gè)質(zhì)量m₁=20 g、m₂=80 g的小球，用等長(zhǎng)的細(xì)線懸掛在O點(diǎn)。懸掛m₂的細(xì)線處于豎直狀態(tài)，懸掛m₁的細(xì)線處于伸直狀態(tài)且與豎直方向成37°角�，F(xiàn)將m₁由靜止釋放，m₁與m₂碰撞后粘在一起。若線長(zhǎng)L=1 m,重力加速度g取10 m/s²，取sin37°=0.6,cos37°=0.8，求：

(1)碰撞前瞬間m₁的速度v₀;

(2)碰撞中損失的機(jī)械能ΔE。

Answer 1

解：（1）由機(jī)械能守恒，得

m₁gL(1-cos37°)=                                                        ①

v₀=2 m/s。                                                                    ②

（2）設(shè)碰后速度為v，由動(dòng)量守恒得

m₁v₀=(m₁+m₂)v                                                                ③

ΔE=m₁gL(1-cos37°)-(m₁+m₂)v²                                                 ④

解得：ΔE=0.032 J。                                                            ⑤

考查機(jī)械能守恒定律，動(dòng)量守恒定律，能量守恒定律。