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$\begin{array}{l}{\text { A } 0.150-\mathrm{kg} \text { glider is moving to the right on a frictionless, }} \\ {\text { horizontal air track with a speed of } 0.80 \mathrm{m} / \mathrm{s} \text { . It has a head-on col- }} \\ {\text { lision with a } 0.300-\mathrm{kg} \text { glider that is moving to the left with a speed }} \\ {\text { of } 2.20 \mathrm{m} / \mathrm{s} \text { . Find the final velocity (magnitude and direction) of }} \\ {\text { each glider if the collision is elastic. }}\end{array}$

$m_{A}=0.15 \mathrm{kg} \quad \longrightarrow$

$V_{A{1}}=0.8 \mathrm{m} /s$

$m_{B}=0.3 k g$

$V_{B 1}=2.2 m / s$

Elastic

$m_{A} v_{A1x}+ m_{B} v_{B 1x}=m_{A} v_{A 2 x}+m_{B} v_{B 2x}$

$(0.15)(0.8)+(0.3)(-2.2)=(0.15)\left(v_{A2x}\right)+(0 .3) v_{B2x}$

$-0.54=0.15 \mathrm{V_{A2x}}+0.3 \mathrm{V}_{B2x}$

$\div 0.3$        $-1.8=0.5 \mathrm{V_{A 2 x}}+\mathrm{V}_{B2 x}$

$\times 2$        $-3.6=V_{A 2 x}+2 V_{B 2 x} \longrightarrow \mathbb(1)$

$V_{B2 x}-V _{A 2 x}=-\left(V_{B 1 x}-V_{A 1 x}\right)=-(-2.2-0.8)=3$

$V_{B2 x}-V_{A 2 x}=3 \rightarrow(2)$

$(1)+(2)$

$-0.6=3 \mathrm{V}_{B 2x} \Rightarrow \mathrm{V}_{B 2x}=-0.2 \mathrm{m/s}$

$V_{A 2 x}=V_{BAx}-3=-3.2 \mathrm{m} / \mathrm{s}$

$\begin{array}{l}{\text { Blocks } A \text { (mass } 2.00 \mathrm{kg}) \text { and } B \text { (mass } 10.00 \mathrm{kg}) \text { move on }} \\ {\text { a frictionless, horizontal surface. Initially, block } B \text { is at rest and }} \\ {\text { block } A \text { is moving toward it at } 2.00 \mathrm{m} / \mathrm{s} \text { . The blocks are equipped }} \\ {\text { with ideal spring bumpers, as in Example } 8.10 \text { (Section } 8.4) . \text { The }}\end{array}$

$\begin{array}{l}{\text { collision is head-on, so all motion before and after the collision is }} \\ {\text { along a straight line. (a) Find the maximum energy stored in the }} \\ {\text { spring bumpers and the velocity of each block at that time. (b) Find }} \\ {\text { the velocity of each block after they have moved apart. }}\end{array}$

$P_{x{1}}=P_{x{2}}$

$m_{A} v_{A x1}+m_B v_{B x{1}}=m_{A} v+m_B v$

$m_{A} v_{A x{1}}=v\left(m_{A}+m_{B}\right)$

$∴ V=\frac{m_{A} V_{A X1}}{m_{A}+m_{B}}=\frac{2(2)}{2+10}=0.333 \mathrm{m} / \mathrm{s}$

$k_{1}=U+k_{2}$

$\frac{1}{2} m_{A} {v_{A x1}}^{2}=U+\frac{1}{2} v^{2}\left(m_{A}+m_{B}\right)$

$\frac{1}{2}(2)(2)^{2}=U+\frac{1}{2}(0.333)^{2}(2+10)\rightarrow 4=U+0.666 \rightarrow U=4-0.666=3.333 \mathrm{J}$

$V_{A 2x} =\left(\frac{m_{A}-m_{B}}{m_{A}+m_{B}}\right) V_{A1} x=\frac{2-10}{2+10}(2)=-1.33 \mathrm{m/s} \quad \leftarrow$

$V_{B 2 x}=\frac{2m_A}{m_{A}+m _B} V_{A1x}=\frac{2(2)}{2+10}(2)=0.667 \mathrm{m} / \mathrm{s} \rightarrow$

$\begin{array}{l}{\text { Three odd-shaped blocks of chocolate have the following }} \\ {\text { masses and center-of-mass coordinates: }(1) 0.300 \mathrm{kg},(0.200 \mathrm{m},} \\ {0.300 \mathrm{m}) ;(2) 0.400 \mathrm{kg},(0.100 \mathrm{m},-0.400 \mathrm{m}) ; \text { (3) } 0.200 \mathrm{kg},} \\ {(-0.300 \mathrm{m}, 0.600 \mathrm{m}) . \text { Find the coordinates of the center of mass }} \\ {\text { of the system of three chocolate blocks. }}\end{array}$

$k_{1}=k_{2}$

$\frac{1}{2} m_A {v_{B1}}^{2}+0=\frac{1}{2}m _{A} {v_{A2}}^{2}+\frac{1}{2}m_B {v_{B2}}^{2}$

${V_{B2}}^{2}=\frac{m_{A} {V_{ A1}}^{2}-m_{A} {V_ {A2}}^{2}}{m_{B}}$

$=\frac{0.5(4)^{2}-(0.5)(2)^{2}}{0.3}=4.47 \mathrm{m} / \mathrm{s}$

$\rightarrow V_{B2}=4.47 \mathrm{m/s}$

(x)

$P_{1x}=P_{2x} \ , \quad V_{B 1x} =0$

$m_{A} v_{Ax}=m_{A}v_{A2x}+m_{B} v_{B 2x}$

$(0.5)(4)=0.5(2) \cos \alpha-0.3(4.47) \cos \beta$

$2=\cos \alpha+1.341 \cos \beta \rightarrow (1)$

(y)

$P_{1y}=P_{2y} \ , V_{B1y}=V_{A 1y}=0$

$P_{1y}=0$

$0=m_{A} V_{A 2 y}+m_{B} V_{B 2 y}$

$0=0.5(2) \sin \alpha-0.3(4.47) \sin \beta$

$0=\sin \alpha-1.341 \sin \beta \rightarrow(2)$

$(1) \rightarrow \quad \cos \beta=\frac{2- \cos \alpha}{1.341}$

$(2) \rightarrow \sin \beta=\frac{\sin \alpha}{1.341}$

تربيع $\cos ^{2} \beta=\left(\frac{2-\cos \alpha}{1.341}\right)^{2}$        $\sin ^{2} \beta=\frac{\sin ^{2} \alpha}{(1.34 1)^{2}}$

$\sin ^{2} \beta+\cos ^{2}\beta =1 \rightarrow\left(\frac{(2- \cos \alpha)}{(1.341)}\right)^2+\frac{\sin \alpha}{(1.341)^{2}}=1$

$∴ \frac{(2-\cos \alpha)^{2}}{1.341^{2}}+\frac{\sin ^{2} \alpha}{1.341^{2}}=1$

$(2-\cos \alpha)^{2}=\cos ^{2} \alpha-2 \cos \alpha+4\rightarrow 1-\sin ^{2} \alpha-4 \cos \alpha+4$

$\longrightarrow=-\sin ^{2} \alpha-4 \cos \alpha+5$

$\frac{-\sin ^{2} \alpha-4 \cos \alpha+5}{(1.341)^{2}}+\frac{\sin ^{2} \alpha}{(1.341)^{2}}=1$

$\rightarrow \frac{-\sin^{2} \alpha-4 \cos \alpha+5+\sin ^{2} \alpha}{(1.341)^{2}}=1 \rightarrow \frac{-4 \cos \alpha+5}{(1.341)^{2}}=1$

$-4 \cos \alpha+5=(1.341)^{2} \rightarrow 4 \cos \alpha=5-(1.341)^{2}=3.201$

$\cos \alpha=\frac{3.201}{4}=0.8 \rightarrow \alpha=\cos ^{-1}(0.8)=36.9^{\circ}$

$\sin \beta=\frac{\sin (\alpha)}{1.341} \rightarrow \sin \beta=\frac{\sin (36.9)}{1.341}=0.448$

$∴ \beta=\sin ^{-1} 0.448=26.6^{\circ}$