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# (物理)普通物理的問題

1.

two point particles,each of mass 100kg,are initially at rest 1m apart

in outer space. what are thier speeds when their separation is 0.5m?

(A)0.817m/s (B)8.17*10^-2m/s (C)8.17*10^-5m/s (D)8.17*10^-7

2.

a pendulum bob of mass m is released from a height H above the

lowest point with another pendulum of the same lenght but with

a bob of mass 2m initially at test. find the height to which the bobs

rise given that the collision is completely inelastic.

(A) 1/3H (B) 1/9H (C) 3H (D) 9H

要有計算和解釋喔

### 1 Answer

- Eric WangLv 71 decade agoFavorite Answer
1. 用能量不滅，兩物體間的重力位能為 -GMm/r(以相距無限遠處為0)

-Gm^2/1= -Gm^2/0.5+2*(1/2)mv^2→ 6.67*10^-11*100=v^2→ v=8.17*10^ -5 m /s …(C)

2. 完全非彈性碰撞後兩物黏在一起，質量m物體碰撞前速度可用能量不滅算出：

mgH=(1/2)mv^2 → v=√2gH

與質量2m物體碰撞後速度可用動量不滅算出：m√ 2g H= 3m v’→ v’=(1/3)√ 2g H

最後擺盪的高度用能量不滅算出：(1/2)* 3m *[(1/3)√ 2g H]^2= 3m gh→ h=H/9…(B)