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2012美国US F=MA物理竞赛 (id: 90d8758a8)

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admin 发表于 2025-12-20 23:13:50 | 显示全部楼层 |阅读模式
本题目来源于试卷: 2012美国US F=MA物理竞赛,类别为 美国F=MA物理竞赛

[单选题]
A particle at rest explodes in1w-z55d*d/ n hftzl vrto three particles of equal mass in the absencesg :4 *y;1wfx+er/n wpui0e,c 0nl tah of external forces. Two part/f0r*u:+0ep n cnwg4hyis,e alt x; w1icles emerge at a right angle to each other with equal speed $v$. What is the speed of the third particle?

A. $v$
B. $\sqrt{2}v$
C. $2v$
D. $2\sqrt{2}v$
E. The third particle can have a range of different speeds.


参考答案:  B


本题详细解析:
The initial momentum of the szh )cf4i13 sszn7y (bgystem is zero. By conservation of momentum, the total final momentum must also be zero. Let the two particles travel along the x and y axes. Their momentay4 hb13gniszc7zf ( )s are $\vec{p}_1 = mv \hat{i}$ and $\vec{p}_2 = mv \hat{j}$. The momentum of the third particle $\vec{p}_3$ must be $\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) = -mv \hat{i} - mv \hat{j}$. The magnitude of this momentum is $|\vec{p}_3| = \sqrt{(-mv)^2 + (-mv)^2} = \sqrt{m^2v^2 + m^2v^2} = \sqrt{2m^2v^2} = mv\sqrt{2}$. Since $p_3 = mv_3$, the speed of the third particle is $v_3 = \sqrt{2}v$.

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