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 o: pzf d /h1c1rrq(d,giy-j6 d ijo ;57wx*uvvobeys Hooke's Law) will break if the tension in the cord uowvxv*57 i j;exceeds $T_{max}$. 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 $v_{f}/v_{0}$ 
A. $1/\sqrt{12}$
B. $1/\sqrt{2}$
C. $1/\sqrt{6}$
D. $1/\sqrt{3}$
E. none of the above
参考答案: C
本题详细解析:
We must first determine th(ycz1cl4 ghqp)y:sc 3 e breaking energy of the cord from the inelastic case.
1. **Inelastic Collisi:yc q1z4g hpys(c3 )clon:** Mass $m$ hits $3m$ and they stick.
By conservation of momentum:
$mv_0 = (m+3m)v^{\prime} \implies v^{\prime} = \frac{v_0}{4}$.
The kinetic energy of the combined $4m$ mass right after collision is $K_1 = \frac{1}{2}(4m)(v^{\prime})^2 = \frac{1}{2}(4m)(\frac{v_0}{4})^2 = \frac{1}{2}(4m)\frac{v_0^2}{16} = \frac{mv_0^2}{8}$.
The problem states this stretches the cord and the final KE is zero, meaning all of $K_1$ is converted into the cord's potential energy $U_0$ at breaking.
So, $U_0 = \frac{mv_0^2}{8}$.
2. **Elastic Collision:** Mass $m$ hits $3m$ elastically.
The velocity of the $3m$ mass ($m_2$) just after collision is $v_{3m} = \left( \frac{2m_1}{m_1+m_2} \right) v_0 = \left( \frac{2m}{m+3m} \right) v_0 = \left( \frac{2m}{4m} \right) v_0 = \frac{v_0}{2}$.
The kinetic energy of the $3m$ mass just after collision is $K_2 = \frac{1}{2}(3m)(v_{3m})^2 = \frac{1}{2}(3m)(\frac{v_0}{2})^2 = \frac{3mv_0^2}{8}$.
3. **Cord Breaking (Elastic Case):** The $3m$ mass starts with $K_2$.
It stretches the cord, which absorbs $U_0$ energy, and is left with a final kinetic energy $K_3 = \frac{1}{2}(3m)v_f^2$.
By conservation of energy: $K_2 = U_0 + K_3$.
$\frac{3mv_0^2}{8} = \frac{mv_0^2}{8} + K_3$
$K_3 = \frac{3mv_0^2}{8} - \frac{mv_0^2}{8} = \frac{2mv_0^2}{8} = \frac{mv_0^2}{4}$.
4. **Solve for $v_f$:** We set $K_3$ equal to its definition:
$\frac{1}{2}(3m)v_f^2 = \frac{mv_0^2}{4}$
$\frac{3}{2}v_f^2 = \frac{v_0^2}{4}$
$v_f^2 = v_0^2 \left( \frac{2}{12} \right) = \frac{v_0^2}{6}$
$\frac{v_f}{v_0} = \sqrt{\frac{1}{6}} = \frac{1}{\sqrt{6}}$.
| 
