Question 9.86: Speed amplifier In Fig. 9-75, block 1 of mass m1 slides alon......

(a) We use Eq. 9-68 twice:

\nu_{2f}=\frac{2m_{1}}{m_{1}+\ m_{2}}\,\nu_{1i}.                                            (9-68)

\nu_{2}=\,{\frac{2m_{1}}{m_{1}+m_{2}}}\,\nu_{1i}\,=\,{\frac{2m_{1}}{1.5m_{1}}}\,(4.00\;\mathrm{m/s})=\,{\frac{16}{3}}\,\mathrm{m/s}

\nu_{3}=\frac{2m_{2}}{m_{2}+m_{3}}\,\nu_{2}\,=\,\frac{2m_{2}}{1.5m_{2}}\,(16/3\,\mathrm{m/s})=\frac{64}{9}m/s=\,7.11\,\mathrm{m/s}\;.

(b) Clearly, the speed of block 3 is greater than the (initial) speed of block 1.

(c) The kinetic energy of block 3 is

K_{3f}\!=\!\frac{1}{2}m_{3}\nu_{3}^{2}\!=\!\left(\frac{1}{2}\right)^{3}\!m_{1}\left(\frac{16}{9}\right)^{2}\!\nu_{1i}{}^{2}=\frac{64}{81}\,K_{1i}\,.

We see the kinetic energy of block 3 is less than the (initial) K of block 1. In the final situation, the initial K is being shared among the three blocks (which are all in motion), so this is not a surprising conclusion.

(d) The momentum of block 3 is

p_{3f}\!=\!m_{3}\nu_{3}\,=\!\left({\frac{1}{2}}\right)^{2}\!m_{1}\!\left({\frac{16}{9}}\right)\!\nu_{1i}\!=\!{\frac{4}{9}}p_{1i}

and is therefore less than the initial momentum (both of these being considered in magnitude, so questions about ± sign do not enter the discussion).