Question 9.119: In Fig. 9-83, block 1 slides along an x axis on a frictionle......
In Fig. 9-83, block 1 slides along an x axis on a frictionless floor with a speed of 0.75 m/s. When it reaches stationary block 2, the two blocks undergo an elastic collision. The following table gives the mass and length of the (uniform) blocks and also the locations of their centers at time t = 0. Where is the center of mass of the two-block system located (a) at t = 0, (b) when the two blocks first touch, and (c) at t = 4.0 s?
\begin{array}{cccl}\hline \text{Block}& \text{Mass}( kg ) & \text{Length}( cm ) & \text{Center at}t=0 \\\hline 1 & 0.25 & 5.0 & x=-1.50 \,m \\2 & 0.50 & 6.0 & x=0 \\\hline\end{array}
(a) Each block is assumed to have uniform density, so that the center of mass of each block is at its geometric center (the positions of which are given in the table [see problem statement] at t = 0). Plugging these positions (and the block masses) into Eq. 9-29 readily gives x_{com} = –0.50 m (at t = 0).
\int_{t_i}^{t_f}d \vec{p}=\int_{t_i}^{t_f}\vec{F}(t) d t . (9-29)
(b) Note that the left edge of block 2 (the middle of which is still at x = 0) is at x = –2.5 cm, so that at the moment they touch the right edge of block 1 is at x = –2.5 cm and thus the middle of block 1 is at x = –5.5 cm. Putting these positions (for the middles) and the block masses into Eq. 9-29 leads to x_{com} = –1.83 cm or –0.018 m (at t = (1.445 m)/(0.75 m/s) = 1.93 s).
(c) We could figure where the blocks are at t = 4.0 s and use Eq. 9-29 again, but it is easier (and provides more insight) to note that in the absence of external forces on the system the center of mass should move at constant velocity:
\vec{v}_{\text{com}}=\frac{m_1 \vec{v}_1+m_2 \vec{v}_2}{m_1+m_2}=0.25 \,m / s \hat{i}
as can be easily verified by putting in the values at t = 0. Thus,
x_{com}=x_{com\ \text{initial}}+\vec{v}_{com}t=(-0.50 \,m )+(0.25\, m / s )(4.0\, s )=+0.50 \,m.