2 Two objects are connected by a light string passing over a light,
3 frictionless pulley as shown in Figure~P8.7. The object of mass $m_1$
4 is released from rest at a height $h$ above the table. Using the
5 isolated system model, \Part{a} determine the speed of $m_2$ just as
6 $m_1$ hits the table and \Part{b} find the maximum height above the
7 table to which $m_2$ rises.
19 Conserving energy, we just have to worry about gravitational potential
20 energy and kinetic energy (there are no springs, friction, external
21 forces, etc.). While they are attached by a taught string (i.e. while
22 $m_1$ is falling), the speeds of the two masses must match. Labeling
23 the release point $a$ and $m_1$ just hitting the floor $b$, we have
26 K_{1a} + U_{1a} + K_{2a} + U_{2a} &= K_{1b} + U_{1b} + K_{2b} + U_{2b} \\
28 &= \frac{1}{2} m_1 v_b^2 + 0 + \frac{1}{2} m_2 v_b^2 + m_2 gh \\
29 (m_1-m_2) gh &= \frac{1}{2} (m_1+m_2) v_b^2 \\
30 v_b^2 &= 2gh\frac{m_1-m_2}{m_1+m_2} \\
31 v_b &= \pm\sqrt{2gh\frac{m_1-m_2}{m_1+m_2}}
32 = \ans{\sqrt{2gh\frac{m_1-m_2}{m_1+m_2}}} \;,
34 where we dropped the $\pm$ because we only want the magnitude of the
35 velocity, not its direction.
38 After $m_1$ hits the table, the string goes slack as $m_2$ sails up in
39 a parabola $h(t)$ and peaks at some point $c$. Conserving energy (now
40 just for $m_2$, because $m_1$ just sits there on the table being
44 K_{2b} + U_{2b} &= K_{2c} + U_{2c} \\
45 \frac{1}{2} m_2 v_b^2 + m_2 gh &= 0 + m_2 g h_c \\
46 h_c &= \frac{v_b^2}{2g} + h
47 = \frac{2gh\frac{m_1-m_2}{m_1+m_2}}{2g} + h
48 = h\p({\frac{m_1-m_2}{m_1+m_2} + 1}) \\
49 &= h\p({\frac{m_1-m_2}{m_1+m_2}+\frac{m_1+m_2}{m_1+m_2}})
50 = h\frac{m_1-m_2+m_1+m_2}{m_1+m_2}
51 = \ans{\frac{2m_1 h}{m_1+m_2}}