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 massless elastic cord (that obeys Ho w73umwcp+fw+ oke 32imf hapht2- 's Law) will break if the tension in the cord exc tp 22-m f3ahiheeds $T_{max}$. One end of the cord is attached to a fixed point, the other is attached to an object of mass 3m. If a second, smaller object of mass $m$ moving at an initial speed $v_{0}$ strikes the larger mass and the two stick together, the cord will stretch and break, but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one-dimensional collision, the cord will still break, and the larger mass will move off with a final speed of $v_{f}$. All motion occurs on a horizontal, frictionless surface. Find the ratio of the total kinetic energy of the system of two masses after the perfectly elastic collision and the cord has broken to the initial kinetic energy of the smaller mass prior to the collision. 
A. $1/4$
B. $1/3$
C. $1/2$
D. $3/4$
E. none of the above
参考答案: D
本题详细解析:
We need the ratio $K_{total, f} / K_{initial}$.
1. **Initial KE:** The initial kinetic energy is just that of the smaller mass $m$:
$K_{initial} = \frac{1}{2}mv_0^2$.
2. **Total Final KE:** This is the sum of the final KE of the $3m$ mass ($K_3$) and the final KE of the $m$ mass ($K_4$).
* From problem 37, we found the final KE of the $3m$ mass *after* the cord breaks is $K_3 = \frac{mv_0^2}{4}$.
* We need the final KE of the $m$ mass.
In the elastic collision, its velocity $v_m$ after impact is $v_m = \left( \frac{m_1-m_2}{m_1+m_2} \right) v_0 = \left( \frac{m-3m}{m+3m} \right) v_0 = \left( \frac{-2m}{4m} \right) v_0 = -\frac{v_0}{2}$.
* The $m$ mass does not interact with the cord, so its final KE is $K_4 = \frac{1}{2}m(v_m)^2 = \frac{1}{2}m(-\frac{v_0}{2})^2 = \frac{1}{2}m\frac{v_0^2}{4} = \frac{mv_0^2}{8}$.
* The total final KE is $K_{total, f} = K_3 + K_4 = \frac{mv_0^2}{4} + \frac{mv_0^2}{8} = \frac{2mv_0^2}{8} + \frac{mv_0^2}{8} = \frac{3mv_0^2}{8}$.
3. **Find the Ratio:**
$\frac{K_{total, f}}{K_{initial}} = \frac{ \frac{3mv_0^2}{8} }{ \frac{1}{2}mv_0^2 } = \frac{3/8}{1/2} = \frac{3}{8} \times \frac{2}{1} = \frac{6}{8} = \frac{3}{4}$.
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