2009-3-3 17:52
qwert123
中四phy@就咁睇真係唔conserve,initial唔=final
以下呢幾題點做?
16(a)(i)
16(c)
16(d)
17(c)
18(c)
答案:
16(a)(i) : 2mv,0
16(c) : 否
16(d) : a:彈性碰撞 b:非彈性碰撞
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2009-3-4 17:30
qwert123
有沒有好心人?......
2009-3-5 21:17
碌仔
(a)i .. KE= 0.5mv^2-0.5mv^2
= 0
2009-3-6 00:16
昌昌
momentum係vector黎既
本身mv
之後-mv
掉過去=2mv
KE係scaler...
個collision係perfectly elastic...冇loss到能量
initail=final , change = 0
F=(mv)-(-mv)/t
F=-2mv/t (正負睇你點take方向)
formular就係咁,你冇比實數我
撞埋去停左
momentum係比左地球,即係車momentum變0
initail - final = mv-0 = mv
ke inital 1/2mv^2 後面v=,KE= 0
唔應該有負號...KE係SCALER講左
F=mv-m(0)/2t (同你講左時間係2t)
F=mv/2t
就咁睇真係唔conserve,initial唔=final wo... (計埋vector)
其實係因為比左牆,牆連左地球,即係momentum比左地球既
所以答案會話否,有d ans可能答conserve就要參考埋地球
講左,inital K.E= final K.E = perfectly elastic
initial KE. > final = inelastic (partly)
final KE = 0 (completely inelastic)
2009-3-6 00:32
昌昌
mgh=1/2mv^2
a)
m約左佢,下次計呢類數都係咁,方便
10h=0.5x 5^2
h=1.25m
contact time 1.2嘛
F=mv-mu/t
F=350/1.2
=....ANS
增加COLLISION TIME, 即係盡量將T(1.2)加大...咁F細,冇咁傷
F=mv-mu/t
80x0.1 =mv-mu (ch of momentum)
=8Ns
mv-mu=8
8=m(v-u)
32=v-15
17....(方向自己take)
t大d喎諗到啦
如果係0.2秒,就16Ns
咁佢rebounce speed都大d,個波反彈with higher speed