phy問題=)

小霞的fans · 發表於 3/9/2009 04:22 PM

本帖最後由小霞的fans 於 2/9/2009 08:48 PM 編輯

assume air resistance is negligable.a ball is thrown veically upward with velocity v and it returns to its original position after time t.velocity and acceleration pointing upwards is taken to be positive.which of following is/are correct?
1.its velocity is zero at 0.5t.
2.the acceleration is maxium at t=0
3.at t seconds,its velocity is -v
我知2 一定錯..但係唔明3點解岩亦都唔知1岩唔岩..望指教..
另外..我有相似既問題..

assume air resistance is negligable.a ball is thrown veically upward with velocity 30ms^-1 . acceleration pointing downwards is taken to be positive.find the distance travelled after 1 second.
咁姐係u=30ms^-1 ,,a=10ms^-2,,t=1 s=?
s=ut+1/2at^2
s=30+1/2(-10)
=25m

但係佢既velocity唔係應該係20ms^-1(比acceleration每秒鍾-10米)架咩?
咁應該果秒行20米架喎,,點解會係25既-0- thx!!

P=Fv
'F'is the applied force that opposes resisting force acting on the object moving at a constant speed.
呢句野姐係話個F 係MOTION既相反方句!? opposes 同 resisting都解妨礙-0- , , 所以有d唔明

sy · 發表於 3/9/2009 04:48 PM

但係佢既velocity唔係應該係20ms^-1(比acceleration每秒鍾-10米)架咩?
>>這些公式好用於不用考慮這種東西,代入數字即可

超夢夢的夢境 · 發表於 3/9/2009 09:00 PM

assume air resistance is negligable.a ball is thrown veically upward with velocity v and it returns to its original position after time t.velocity and acceleration pointing upwards is taken to be positive.which of following is/are correct?
小霞的fans 發表於 3/9/2009 04:22 PM

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清楚起見,下面v改用u
代式
v^2-u^2=2as
it returns to its original position after time t
=> s=0

v^2=u^2
v=u or v=-u

v=u 係 initial velocity
所以v=-u

清楚D就係
v=u+at
a=/=0 and t=/=0
=> v=/=u
所以v=-u

v=-u and v=u+at
-u=u+at
-2u=at
-u=a(t/2)

(v at time=t/2)=u+a(t/2)=0

1,3

assume air resistance is negligable.a ball is thrown veically upward with velocity 30ms^-1 .acceleration pointing downwards is taken to be positive.find the distance travelled after 1 second.
小霞的fans 發表於 3/9/2009 04:22 PM

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u=-30ms^-1, a=10ms^-2, t=1
s=ut+1/2*at^2
s=-30+1/2*(10)
=-25m
(上行25米)

但係佢既velocity唔係應該係20ms^-1(比acceleration每秒鍾-10米)架咩?
咁應該果秒行20米架喎,,點解會係25既-0- thx!!
小霞的fans 發表於 3/9/2009 04:22 PM

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velocity at time =1s 係 20ms^-1...
但係佢唔係uniform velocity...
initial velocity = 30 ms^-1
因為 acceleration 係 uniform,
average velocity = (v+u)/2 =25

P=Fv
'F'is the applied force that opposes resisting force acting on the object moving at a constant speed.
呢句野姐係話個F 係MOTION既相反方句!? opposes 同 resisting都解妨礙-0- , , 所以有d唔明
小霞的fans 發表於 3/9/2009 04:22 PM

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applied force thatopposesresisting force即係同resisting force相反方向
In order words,in the same direction as motion

p.s."moving at a constant speed" implies applied force = resisting force

小霞的fans · 發表於 5/9/2009 03:58 PM

本帖最後由小霞的fans 於 4/9/2009 08:01 PM 編輯

清楚起見,下面v改用u
代式
v^2-u^2=2as
it returns to its original position after time t
=> s=0

v^2=u^2
v=u or v=-u

v=u 係 initial velocity
所以v=-u

清楚D就係
v=u+at
a=/=0 and t=/=0
= ...
超夢夢的夢境發表於 3/9/2009 01:00 AM

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我明啦,, 好多謝你幫左我咁多^^ 仲有1個關於action and reaction pair既問題..

two block A and B are in contact.a force of 20N pushes the block fron the left

大約幅圖係咁..

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LET F(A) be the force on B by A. LET F(B) be the force on A by B

20=3+5(a)
a=2.5ms^-2

force acting on A by B
條式唔係
20+F(A)-F(B)=3a咩!?
20同F(A)係同方向既force (向右,,而F(B)係向左既,,
唔係右加右減左咩 .. .
點解會係20-F(A)=3a架!?同方向都相減!? 定係我抄錯,,唔係減F(A)係減F(B)..咁F(A)唔駛理!?個物件 exert 既force唔會計!?因為我睇返本書之後唔明 thx ><

捕風雪影 · 發表於 6/9/2009 01:22 PM

本帖最後由捕風雪影於 6/9/2009 01:23 PM 編輯

我明啦,, 好多謝你幫左我咁多^^ 仲有1個關於action and reaction pair既問題..

two block A and B are in contact.a force of 20N pushes the block fron the left

大約幅圖係咁..

http://i554.photobu ...
小霞的fans 發表於 5/9/2009 03:58 PM

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其實佢地係Action and Reaction Pair,所以Acting on A by B同埋Acting on B by A既force係相同的

所以A受到既Net Force=7.5N,咁Acting on B by A=7.5N,反之亦然

小霞的fans · 發表於 6/9/2009 02:46 PM

其實佢地係Action and Reaction Pair,所以Acting on A by B同埋Acting on B by A既force係相同的

所以A受到既Net Force=7.5N,咁Acting on B by A=7.5N,反之亦然
捕風雪影發表於 5/9/2009 05:22 PM

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但係acting on A by B同acting on B by A 既force都係12.5N.
我唔係好明呢d數點計- - "
阿sir 將block 既mass(5)x a (2.5)=12.5N咁就計完. .
咁個20N 唔駛理架咩-,- .. 佢有份受力>< " 唔明呀!!!!!!

fish · 發表於 6/9/2009 03:03 PM

本帖最後由 fish 於 6/9/2009 03:10 PM 編輯

但係acting on A by B同acting on B by A 既force都係12.5N.
我唔係好明呢d數點計- - "
阿sir 將block 既mass(5)x a (2.5)=12.5N咁就計完. .
咁個20N 唔駛理架咩-,- .. 佢有份受力>< " 唔明呀!!!!!!
小霞的fans 發表於 6/9/2009 14:46

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用free body diagram 來解釋不知會否好些呢?(看下圖, 只show horizontal component, no friction, right = positive, left = negative)

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即使兩個物體是放在一起時,那20N也只會畫在3kg上, 因為那20N是直接加於3kg上

Force on A:
F(x net) = 20N + F(B)

ma = 20N + F(B)

3kg * 2.5 m/s^2  = 20N + F(B)

F(B) = 7.5N - 20N = -12.5N

當然....直接點

Force on B:
F(x net) = F(A)

ma = F(A)    (and we know F(A) = - F(B) from the third law)

5kg *  2.5m/s^2 = - F(B)

F(B) = -12.5N

小霞的fans · 發表於 6/9/2009 03:30 PM

用free body diagram 來解釋不知會否好些呢?(看下圖, 只show horizontal component, no friction, right = positive, left = negative)

登錄/註冊後可看大圖

即使 ...
fish 發表於 5/9/2009 07:03 PM

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喔,,我大致都明了^^
Force on A:
F(x net) = 20N + F(B)

20N 同F(B)係相反方向點解係加既?
20N-F(B)=F(x net) 都得!?

捕風雪影 · 發表於 6/9/2009 03:34 PM

喔,,我大致都明了^^
Force on A:
F(x net) = 20N + F(B)

20N 同F(B)係相反方向點解係加既?
20N-F(B)=F(x net) 都得!?
小霞的fans 發表於 6/9/2009 03:30 PM

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正確寫法係

20N+(-F(B))=F(x net)

20N-F(B)=F(x net)

因為相反方向的

fish · 發表於 6/9/2009 03:40 PM

喔,,我大致都明了^^
Force on A:
F(x net) = 20N + F(B)

20N 同F(B)係相反方向點解係加既?
20N-F(B)=F(x net) 都得!?
小霞的fans 發表於 6/9/2009 15:30

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習慣問題....
我如果直接加的話在答案會出了個-12.5N (即是向左)

如果是 20N -F(B) =ma =7.5 的話
會出 F(B) = 12.5N (正數)
然後在report answer 時要寫 -12.5N 而已

不過我不清楚CE 是否都准許兩個方法了

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