2 The system shown in Figure~P8.11 consists of a light, inextensible
3 cord, light, frictionless pulleys, and blocks of equal mass. Notice
4 that block $B$ is attached to one of the pulleys. The system is
5 initially held at rest so that the blocks are at the same height above
6 the ground. The blocks are then released. Find the speed of block
7 $A$ at the moment the vertical separation of the blocks is $h$.
16 Because the tension on both sides of a light pulley must match, block
17 $B$ has twice the upwards force (from the strings) as block $A$, so it
18 will rise as block $A$ drops.
20 Block $A$ will move twice as fast and far as block $B$ (draw a few
21 snapshots or build a little model to prove that to yourself if you
22 need to). Therefore, when the blocks are $h$ appart, block $A$ will
23 be down $2h/3$ and block $B$ will be up $h/3$. Conserving energy, we
27 K_{Ai} + U_{Ai} + K_{Bi} + U_{Bi} &= K_{Af} + U_{Af} + K_{Bf} + U_{Bf} \\
29 &= \frac{1}{2} m v_{Af}^2 + mg\frac{-2h}{3}
30 \frac{1}{2} m \p({\frac{v_{Af}}{2}})^2 + mg\frac{h}{3} \\
31 0 &= v_{Af}^2 \p({\frac{1}{2} + \frac{1}{8}}) - g\frac{h}{3} \\
32 v_{Af}^2 \frac{5}{8} &= \frac{gh}{3} \\
33 v_{Af} &= \pm\sqrt{\frac{8gh}{15}}
34 = \ans{\sqrt{\frac{8gh}{15}}}
36 where we dropped the $\pm$ because we only want the magnitude of the
37 velocity, not its direction.