A marble of mass 0.1kg is released from a height of 1m onto a hard floor. It rebounds to a height of 0.7m.
If the marble is in contact with the floor for 0.01s, what is the average force acting
(i) on the floor during impact ? (ans : 82.1N)
(ii) on the marble during impact ? (ans : 82.1N)
計數問題...
2007-01-14 8:20 am
回答 (1)
2007-01-14 8:58 am
✔ 最佳答案
let the downwad motion be the positivev^2=u^2+2as
v^2=2(10)1
v^2= 20------where is the velocity before the impact[postive magnitude]
=================
v^2=u^2+2as
v^2=2(10)(0.7)
v^2=14------where is the velocity after the impact[negative magnitude]
by F=mv-mu/t
0.1(20^1/2) - 0.1(-14^1/20
--------------------------------------
0.03
F=82.1N
by newton's third law
action and reacton pair so that the force acting on the marble is 82.1N
收錄日期: 2021-04-12 23:08:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070114000051KK00103