徐陵 asked in 科學其他:科學 · 8 years ago

物理-動量守恆題(英文)

Q:

A spaceship of mass M is traveling in deep space with velocity V= 20 km/s relative to the Sun.It ejects a rear stage of mass 0.2M with a relative speed u= 5 km/s .

What then is the velocity of the spaceship?

A: 21km/s

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原題與答案如上

應該能以動量守恆表示式寫出MV=0.2M*V1+0.8M*V2

(V1為分離後rear stage的速度 V2為分離後宇宙船的速度)

但是我搞不太清楚

[It ejects a rear stage of mass 0.2M with a relative speed u= 5 km/s]

這一段的意思

意思是這個u是分離後對太陽的速度嗎?還是對船的速度呢?

想請知道的大大詳述一下想法以及算式,謝謝.

1 Answer

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  • 8 years ago
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    It ejects a rear stage of mass 0.2M with a " relative speed" u= 5 km/s .

    並沒有說" relative to the Sun",所以是對船的速度。

    依假設太空船速度為V2(方向取正),

    則u=-5 km/s = V1-V2 => V1= -5+V2

    代入版大提供的動量守恆表示式MV=0.2M*V1+0.8M*V2

    => M*20 = 0.2M*(-5+V2) +0.8M*V2

    => 20 =0.2*(-5)+0.2V2 +0.8V2

    => V2= 21 km/s

    2012-11-02 18:15:39 補充:

    給版大建議:

    如果是問太空船速率的話,我個人會以太空船為主設成V1

    其他才V2,V3....

    Source(s): 應作如是觀
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