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2007美国US F=MA物理竞赛 (id: 25901bad3)

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admin 发表于 2025-12-20 23:13:49 | 显示全部楼层 |阅读模式
本题目来源于试卷: 2007美国US F=MA物理竞赛,类别为 美国F=MA物理竞赛

[单选题]
A point-like mass moves horizontak s7d(dpd4 6x w7qkc ; rkr2sc09v.mrilly between cxd/szg fj6uo::/y /,mir8aw two walls on a frictionless surface with iraw 6yz8 /:co/ugdsi/:,jx mfnitial kinetic energy $E$. With every collision with the walls, the mass loses $\frac{1}{2}$ its kinetic energy to thermal energy. How many collisions with the walls are necessary before the speed of the mass is reduced by a factor of 8?

A. 3
B. 4
C. 6
D. 8
E. 16


参考答案:  C


本题详细解析:
Initial kinetic energjpez z+5s rb+uhz6tul0i : x:yc3x )2oy $E_0 = \frac{1}{2}mv_0^2$. The final speed is $v_f = v_0 / 8$. The final kinetic energy $E_f = \frac{1}{2}m(v_f)^2 = \frac{1}{2}m(v_0/8)^2 = \frac{1}{64} \left( \frac{1}{2}mv_0^2 \right) = E_0 / 64$. If the mass *loses* $\frac{1}{2}$ its KE on each collision, it *retains* $\frac{1}{2}$ its KE. Let $E_n$ be the kinetic energy after $n$ collisions. This forms a geometric progression: $E_n = E_0 \times (\frac{1}{2})^n$. We want to find $n$ such that $E_n = E_f = E_0 / 64$. $E_0 (\frac{1}{2})^n = E_0 / 64$ $(\frac{1}{2})^n = 1/64$ $2^n = 64$ Since $2^5 = 32$ and $2^6 = 64$, the number of collisions is $n=6$.

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