Question 13.198: Blocks A and B are connected by a cord which passes over pul...

(a) Velocity of A just before it hits C :

Conservations of energy:

Datum at ①:

Position ①:

\begin{aligned}\left(\nu_{A}\right)_{1} & =\left(\nu_{B}\right)_{1}=0 \\T_{1} & =0 \\\nu_{1} & =0\end{aligned}

Position ②: \quad T_{2}=\frac{1}{2} m_{A}\left(\nu_{A}\right)^{2}+\frac{1}{2} m_{B} \nu_{B}^{2}

\nu_{A}=\nu_{B} \quad (kinematics)

T_{2}=\frac{1}{2}(5+6) \nu_{A}^{2}=\frac{11}{2} \nu_{A}^{2}

V_{2}=m_{A} g(1.7)-m_{B} g(1.7)

=(5-6)(g)(1.7)

V_{2}=-1.7 g

T_{1}+V_{1}=T_{2}+V_{2}

0+0=\frac{11}{2} \nu_{A}^{2}-1.7 g

\nu_{A}^{2}=\left\lgroup\frac{3.4}{11}\right\rgroup(9.81)

=3.032 \mathrm{~m}^{2} / \mathrm{s}^{2}

\nu_{A}=1.741 \mathrm{~m} / \mathrm{s}

Velocity of A and C after A hits C:

Impulse-momentum A and C:

B alone:

Adding Equations (1) and (2), 

\begin{aligned}& 11(1.741) =14 \nu_{A}^{\prime} \\& \quad\quad\quad\quad \nu_{A}^{\prime}=1.3679 \mathrm{~m} / \mathrm{s}\end{aligned}

\nu_{A}^{\prime}=\nu_{B}^{\prime}=\nu_{C}^{\prime}=1.368 \mathrm{~m} / \mathrm{s}\blacktriangleleft

(b) Distance A and C move before stopping:

Conservations of energy:

Datum at ②:

Position ②:

\begin{aligned}& T_{2}=\frac{1}{2}\left(m_{A}+m_{B}+m_{C}\right)\left(\nu_{A}^{\prime}\right) \\& T_{2}=\left\lgroup\frac{14}{2}\right\rgroup(1.3681)^{2} \\& T_{2}=13.103 \mathrm{~J} \\& V_{2}=0\end{aligned}

Position ③:

T_{3}=0

V_{3}=\left(m_{A}+m_{C}\right) g d-m_{B} g d

V_{3}=(8-6) g d=2 g d

T_{2}+V_{2}=T_{3}+V_{3}

13.103+0=0+2 g d

d=(13.103) /(2)(9.81)=0.6679 \mathrm{~m}

d=0.668 \mathrm{~m}\blacktriangleleft

(c) As the system returns to position ② after stopping in position ③, energy is conserved, and the velocities of A, B, and C before the collar at C is removed are the same as they were in Part (a) above with the directions reversed. Thus, \nu_{A}^{\prime}=\nu_{C}^{\prime}=\nu_{B}^{\prime}=1.3679 \mathrm{~m} / \mathrm{s}. After the collar C is removed, the velocities of A and B remain the same since there is no impulsive force acting on either.

Conversation of energy:

Datum at ②:

\begin{aligned}& T_{2}=\frac{1}{2}\left(m_{A}+m_{B}\right)\left(\nu_{A}^{\prime}\right)^{2} \\& T_{2}=\frac{1}{2}(5+6)(1.3679)^{2} \\& T_{2}=10.291 \mathrm{~J}\end{aligned}

V_{2}=0

\begin{aligned}T_{4}=0 \quad & V_{4}=m_{B} g x-m_{A} g x \\V_{4} & =(6-5) g x\end{aligned}

T_{2}+V_{2}=T_{4}+V_{4}

10.291+0=(1)(9.81) x

x=1.049 \mathrm{~m}\blacktriangleleft