A 300-g collar A is released from rest, slides down a frictionless rod, and strikes a 900-g collar B which is at rest and supported by a spring of constant 500 N/m. Knowing that the coefficient of restitution between the two collars is 0.9, determine (a) the maximum distance collar A moves up the

Question

A 300-g collar A is released from rest, slides down a frictionless rod, and strikes a 900-g collar B which is at rest and supported by a spring of constant 500 N/m. Knowing that the coefficient of restitution between the two collars is 0.9, determine (a) the maximum distance collar A moves up the rod after impact, (b) the maximum distance collar B moves down the rod after impact.

Accepted Answer

After impact Velocity of A just before impact, [latex] u_{0}[/latex]

[latex]\begin{aligned} u_{0}=\sqrt{2 g h} & =\sqrt{2\left(9.81 \mathrm{~m} / \mathrm{s}^{2} ight)(1.2 \mathrm{~m}) \sin 30^{\circ}} \& =\sqrt{2(9.81)(1.2)(0.5)}=3.431 \mathrm{~m} / \mathrm{s}\end{aligned}[/latex]

Conservation of momentum

[latex]+\swarrow_{\underline{30^°}}m_A u_0=m_B u_B-m_A u_A: \quad 0.3 u_0=0.9 u_B-0.3 u_A \qquad (1)[/latex]

Restitution

[latex]\left( u_{A}+ u_{B} ight)=e\left( u_{0}+0 ight)=0.9 u_{0}\quad (2)[/latex]

Substituting for [latex] u_{B}[/latex] from (2) in (1)

[latex]\begin{gathered}0.3 u_{0}=0.9\left(0.9 u_{0}- u_{A} ight)-0.3 u_{A} \quad 1.2 u_{A}=0.51 u_{0} \ u_{A}=1.4582 \mathrm{~m} / \mathrm{s}, u_{B}=1.6297 \mathrm{~m} / \mathrm{s}\end{gathered}[/latex]

(a) A moves up the distance d where:

[latex]\frac{1}{2} ot m_{A} u_{A}^{2}= ot m_{A} g d \sin 30^{\circ} ; \quad \frac{1}{2}(1.4582 \mathrm{~m} / \mathrm{s})^{2}=\left(9.81 \mathrm{~m} / \mathrm{s}^{2} ight) d(0.5)[/latex]

[latex]\swarrow_{\underline{30^°}}d_{A}=0.21675\,\mathrm{m}=217\,\mathrm{mm}\,\blacktriangleleft[/latex]

(b) Static deflection = [latex]x_0 [/latex], B moves down [latex]\swarrow_{\underline{30^°}}^{d_B}[/latex]

Conservation of energy (1) to (2)

Position (1) – spring deflected, [latex]x_{0}[/latex]

[latex]\begin{gathered}k x_{0}=m_{B} g \sin 30^{\circ} \T_{1}+V_{1}=T_{2}+V_{2}: \quad T_{1}=\frac{1}{2} m_{B} u_{B}^{2}, T_{2}=0\end{gathered}[/latex]

[latex]\begin{gathered}V_{1}=V_{e}+V_{g}=\frac{1}{2} k x_{0}^{2}+m_{B} g d_{B} \sin 30^{\circ} \V_{2}=V_{e}^{\prime}+V_{g}^{\prime}=\int_{0}^{x_{0}+d_{B}} k x \ d x=\frac{1}{2} k\left(d_{B}^{2}+2 d_{B} x_{0}+x_{0}^{2} ight) \\frac{1}{2} k x_{0}^{2}+m g \ d_{B} \sin 30^{\circ}+\frac{1}{2} m_{B} u_{B}^{2}=\frac{1}{2} k\left(d_{B}^{2}+2 d_{B} x_{0}+x_{0}^{2} ight)+0+0 \ herefore k \ d_{B}^{2}=m_{B} u_{B}^{2} ; \quad 500 d_{B}^{2}=0.9(1.6297)^{2} \quad d_{B}=0.0691 \mathrm{~m}\end{gathered}[/latex]

[latex]d_{B}=69.1 \mathrm{~mm}\blacktriangleleft[/latex]

Question 13.197: A 300-g collar A is released from rest, slides down a fricti...

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