Question 9.85: Speed deamplifier. In Fig. 9-74, block 1 of mass m1 slides a......

Using Eq. 9-67 and Eq. 9-68, we have after the first collision

v_{2 f}=\frac{2 m_1}{m_1+m_2}v_{1 i}\text{.}       (9-68)
v_{1 f}=\frac{m_1-m_2}{m_1+m_2}v_{1 i}       (9-67)

\begin{aligned}& v_{1 f}=\frac{m_1-m_2}{m_1+m_2}v_{1 i}=\frac{m_1-2 m_1}{m_1+2 m_1}v_{1 i}=-\frac{1}{3}v_{1 i}\\& v_{2 f}=\frac{2 m_1}{m_1+m_2}v_{1 i}=\frac{2 m_1}{m_1+2 m_1}v_{1 i}=\frac{2}{3}v_{1 i}.\end{aligned}

After the second collision, the velocities are

\begin{aligned}& v_{2, f f}=\frac{m_2-m_3}{m_2+m_3}v_{2 f}=\frac{-m_2}{3 m_2}\frac{2}{3}v_{1 i}=-\frac{2}{9}v_{1 i}\\& v_{3, f f}=\frac{2 m_2}{m_2+m_3}v_{2 f}=\frac{2 m_2}{3 m_2}\frac{2}{3}v_{1 i}=\frac{4}{9}v_{1 i}.\end{aligned}

(a) Setting v_{1i} = 4 m/s, we find v_{3 ff} ≈ 1.78 m/s.

(b) We see that v_{3 ff} is less than v_{1i} .

(c) The final kinetic energy of block 3 (expressed in terms of the initial kinetic energy of block 1) is

K_{3, f f}=\frac{1}{2}m_3 v_3^2=\frac{1}{2}\left(4 m_1\right)\left(\frac{4}{9}\right)^2 v_{1 i}^2=\frac{64}{81}K_{1 i}

We see that this is less than K_{1i} .

(d) The final momentum of block 3 is p_{3 f f}=m_3 v_{3 f f}=\left(4 m_1\right)\left(\frac{16}{9}\right) v_1>m_1 v_1 .