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什么?你的严谨性?天哪,呵呵,看来你是死活非要给我送个笑料啊,呵呵呵呵 :-))

(2012-06-12 13:30:51) 下一个

“严谨性”这个词儿你选的非常好!

我从来都不想跟你抬扛,可惜你总是这么认为,今天我也没想跟你抬扛,可是你是死活非要给我送个笑料啊

那么,就让大家来看看你的严谨性吧,呵呵

我说你没算弹性碰撞:“ 你没算!因为你忘了弹性碰撞不仅动量守恒还有能量守恒 :-))

你回答我说:“ 你去校核一下那个0.6m/s怎么来的:) 顺便算算这个时候弹芯的反弹速度是多少,再顺便算算各自的动能,加一起看看。 ”

这些都是事实吧?

我本来可没想给你挖这个坑儿,虽然我老老实实的也有一定贡献,但是这个坑儿主要是你挖滴!

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

那好,我如果不“去校核一下那个0.6m/s怎么来的”都对不起你啦,呵呵呵呵

你说的你算的弹性碰撞是这一段吧:“坦克能获得最大动能的情况是穿甲弹弹芯以接近初始速度的速度被弹回,坦克获得速度0.6m/s,动能为9000焦耳,为弹芯初始动能的万分之8。这种情况下,没有动能转化成热能。也就是说坦克能接受到的动能应该在22.5焦和9000焦之间。”

你给了一个表格说明你会算数,还会套公式,虽然你这里套错了,因为你对弹性碰撞没完全理解,所以你对弹性碰撞基本上没概念。

总之,你背公式算数字还是会滴。恭喜你还赢得了一个粉丝,嘿嘿

可惜啊,你对力学没感觉,你对弹性碰撞基本上没概念!

请看证据:

你说的“穿甲弹弹芯以接近初始速度的速度被弹回”这个确实是弹性碰撞的特征,但是你说的“坦克能获得最大动能的情况”是这个特征就大错特错了!

而且我都告诉你啦!

请回忆我对你说的你没算弹性碰撞的话:“ 你没算!因为你忘了弹性碰撞不仅动量守恒还有能量守恒 :-))

看到我说的“ 弹性碰撞不仅动量守恒还有能量守恒 :-)) ”没有?

看到我说的“ 能量守恒 :-)) ”没有?

你说的“穿甲弹弹芯以接近初始速度的速度被弹回”这个情况是穿甲弹弹芯打击坦克前的速度与打击坦克后被弹回的速度接近,对吧?

也就是说你说的“穿甲弹弹芯以接近初始速度的速度被弹回”这个情况是穿甲弹弹芯打击坦克前的动能与打击坦克后被弹回的动能接近,对吧?

如果你看到了我说的“ 能量守恒 :-)) ”的话,如果你对弹性碰撞有概念的话,如果你对力学有感觉的话,那么,你就应该知道弹性碰撞不仅不是你说的“坦克能获得最大动能的情况”,恰恰相反,弹性碰撞是坦克能获得最小动能的情况!

因为理想情况的弹性碰撞,比如说小质量物体与大质量物体的弹性碰撞,大质量物体获得〇动能!

你如果不相信我的话,请看“自由的百科全书”维基百科,我不仅给你链接:http://en.wikipedia.org/wiki/Momentum

我还不编辑给你抄整段。我还把有关的部分用黑体帮你标出来了,希望你能很容易地找到它们,呵呵呵呵

我真心实意滴祝你读的跟我贴这个回复一样愉快,呵呵呵呵

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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