Blocks A and B are connected by a cord which passes over pulleys and through a collar C. The system is released from rest when x = 1.7 m. As block A rises, it strikes collar C with perfectly plastic impact (e = 0). After impact, the two blocks and the collar keep moving until they come to a stop

Question

Blocks A and B are connected by a cord which passes over pulleys and through a collar C. The system is released from rest when x = 1.7 m. As block A rises, it strikes collar C with perfectly plastic impact (e = 0). After impact, the two blocks and the collar keep moving until they come to a stop and reverse their motion. As A and C move down, C hits the ledge and blocks A and B keep moving until they come to another stop. Determine (a) the velocity of the blocks and collar immediately after A hits C, (b) the distance the blocks and collar move after the impact before coming to a stop, (c) the value of x at the end of one compete cycle.

Accepted Answer

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

Conservations of energy:

Datum at ①:

Position ①:

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

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

[latex]\nu_{A}=\nu_{B} \quad[/latex] (kinematics)

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

[latex]V_{2}=m_{A} g(1.7)-m_{B} g(1.7)[/latex]

[latex]=(5-6)(g)(1.7)[/latex]

[latex]V_{2}=-1.7 g[/latex]

[latex]T_{1}+V_{1}=T_{2}+V_{2}[/latex]

[latex]0+0=\frac{11}{2} \nu_{A}^{2}-1.7 g[/latex]

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

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

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

Velocity of A and C after A hits C:

[latex]\nu_{A}^{\prime}=\nu_{C}^{\prime} \quad(\text { plastic impact })[/latex]

Impulse-momentum A and C:

[latex]\begin{gathered}+↑ m_{A} \nu_{A}+T \Delta t=\left(m_{A}+m_{C}\right) \nu_{A}^{\prime} \(5)(1.741)+T \Delta t=8 \nu_{A}^{\prime} \quad\quad \quad\quad \text{(1)}\\left.\nu_{B}=\nu_{A} ; \nu_{B}^{\prime}=\nu_{A}^{\prime} \text { (cord remains taut }\right)\end{gathered}[/latex]

B alone:

[latex]\begin{aligned}& m_{B} \nu_{A}-T \Delta t=m_{B} \nu_{A}^{\prime} \& \text { (6)(1.741) }-T \Delta t=6 \nu_{A}^{\prime} \quad\quad\quad\quad \text{(2)} \\end{aligned}[/latex]

Adding Equations (1) and (2),

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

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

(b) Distance A and C move before stopping:

Conservations of energy:

Datum at ②:

Position ②:

[latex]\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}[/latex]

Position ③:

[latex]T_{3}=0[/latex]

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

[latex]V_{3}=(8-6) g d=2 g d[/latex]

[latex]T_{2}+V_{2}=T_{3}+V_{3}[/latex]

[latex]13.103+0=0+2 g d[/latex]

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

[latex]d=0.668 \mathrm{~m}\blacktriangleleft[/latex]

(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, [latex]\nu_{A}^{\prime}=\nu_{C}^{\prime}=\nu_{B}^{\prime}=1.3679 \mathrm{~m} / \mathrm{s}[/latex]. 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:

[latex]\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}[/latex]

[latex]V_{2}=0[/latex]

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

[latex]T_{2}+V_{2}=T_{4}+V_{4}[/latex]

[latex]10.291+0=(1)(9.81) x[/latex]

[latex]x=1.049 \mathrm{~m}\blacktriangleleft[/latex]

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

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