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什麽?你的嚴謹性?天哪,嗬嗬,看來你是死活非要給我送個笑料啊,嗬嗬嗬嗬 :-))

(2012-06-12 13:30:51) 下一個

“嚴謹性”這個詞兒你選的非常好!

我從來都不想跟你抬扛,可惜你總是這麽認為,今天我也沒想跟你抬扛,可是你是死活非要給我送個笑料啊

那麽,就讓大家來看看你的嚴謹性吧,嗬嗬

我說你沒算彈性碰撞:“ 你沒算!因為你忘了彈性碰撞不僅動量守恒還有能量守恒 :-))

你回答我說:“ 你去校核一下那個0.6m/s怎麽來的:) 順便算算這個時候彈芯的反彈速度是多少,再順便算算各自的動能,加一起看看。 ”

這些都是事實吧?

我本來可沒想給你挖這個坑兒,雖然我老老實實的也有一定貢獻,但是這個坑兒主要是你挖滴!

我都不得不佩服我自己了,人家是“無心栽柳柳成蔭”,我是“無心挖坑坑挖成”,嗬嗬嗬嗬

那好,我如果不“去校核一下那個0.6m/s怎麽來的”都對不起你啦,嗬嗬嗬嗬

你說的你算的彈性碰撞是這一段吧:“坦克能獲得最大動能的情況是穿甲彈彈芯以接近初始速度的速度被彈回,坦克獲得速度0.6m/s,動能為9000焦耳,為彈芯初始動能的萬分之8。這種情況下,沒有動能轉化成熱能。也就是說坦克能接受到的動能應該在22.5焦和9000焦之間。”

你給了一個表格說明你會算數,還會套公式,雖然你這裏套錯了,因為你對彈性碰撞沒完全理解,所以你對彈性碰撞基本上沒概念。

總之,你背公式算數字還是會滴。恭喜你還贏得了一個粉絲,嘿嘿

可惜啊,你對力學沒感覺,你對彈性碰撞基本上沒概念!

請看證據:

你說的“穿甲彈彈芯以接近初始速度的速度被彈回”這個確實是彈性碰撞的特征,但是你說的“坦克能獲得最大動能的情況”是這個特征就大錯特錯了!

而且我都告訴你啦!

請回憶我對你說的你沒算彈性碰撞的話:“ 你沒算!因為你忘了彈性碰撞不僅動量守恒還有能量守恒 :-))

看到我說的“ 彈性碰撞不僅動量守恒還有能量守恒 :-)) ”沒有?

看到我說的“ 能量守恒 :-)) ”沒有?

你說的“穿甲彈彈芯以接近初始速度的速度被彈回”這個情況是穿甲彈彈芯打擊坦克前的速度與打擊坦克後被彈回的速度接近,對吧?

也就是說你說的“穿甲彈彈芯以接近初始速度的速度被彈回”這個情況是穿甲彈彈芯打擊坦克前的動能與打擊坦克後被彈回的動能接近,對吧?

如果你看到了我說的“ 能量守恒 :-)) ”的話,如果你對彈性碰撞有概念的話,如果你對力學有感覺的話,那麽,你就應該知道彈性碰撞不僅不是你說的“坦克能獲得最大動能的情況”,恰恰相反,彈性碰撞是坦克能獲得最小動能的情況!

因為理想情況的彈性碰撞,比如說小質量物體與大質量物體的彈性碰撞,大質量物體獲得〇動能!

你如果不相信我的話,請看“自由的百科全書”維基百科,我不僅給你鏈接:http://en.wikipedia.org/wiki/Momentum

我還不編輯給你抄整段。我還把有關的部分用黑體幫你標出來了,希望你能很容易地找到它們,嗬嗬嗬嗬

我真心實意滴祝你讀的跟我貼這個回複一樣愉快,嗬嗬嗬嗬

Elastic collisions

A collision between two pool balls is a good example of an almost totally elastic collision, due to their high rigidity; a totally elastic collision exists only in theory, occurring between bodies with mathematically infinite rigidity. In addition to momentum being conserved when the two balls collide, the sum of kinetic energy before a collision must equal the sum of kinetic energy after:
tfrac{1}{2} m_1 u_{1}^2 + tfrac{1}{2} m_2 u_{2}^2 = tfrac{1}{2} m_1 v_{1}^2 + tfrac{1}{2} m_2 v_{2}^2,.

In one dimension

When the initial velocities are known, the final velocities for a head-on collision are given by

 mathbf{v}_{1} = left( frac{m_1 - m_2}{m_1 + m_2} right) mathbf{u}_{1} + left( frac{2 m_2}{m_1 + m_2} right) mathbf{u}_{2},
 mathbf{v}_{2} = left( frac{m_2 - m_1}{m_1 + m_2} right) mathbf{u}_{2} + left( frac{2 m_1}{m_1 + m_2} right) mathbf{u}_{1},.
This example m1=1000kg, u1=5m/s, m2=0.1kg, u2=0m/s, the final velocities are approximately given v1=4.999m/s, v2=9.999m/s

When the first body is much more massive than the other (that is, m1 » m2), the final velocities are approximately given by

 mathbf{v}_{1} = mathbf{u}_{1},
 mathbf{v}_{2} = 2mathbf{u}_{1} - mathbf{u}_{2},.

Thus the more massive body does not change its velocity, and the less massive body travels at twice the velocity of the more massive body less its own original velocity. Assuming both masses were heading towards each other on impact, the less massive body is now therefore moving in the opposite direction at twice the speed of the more massive body plus its own original speed.

A Newton's cradle demonstrates conservation of momentum.

In a head-on collision between two bodies of equal mass (that is, m1 = m2), the final velocities are given by

mathbf{v}_1 = mathbf{u}_2,
mathbf{v}_2 = mathbf{u}_1,.

Thus the bodies simply exchange velocities. If the first body has nonzero initial velocity u1 and the second body is at rest, then after collision the first body will be at rest and the second body will travel with velocity u1. This phenomenon is demonstrated by Newton's cradle.

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