At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg, and 35 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity vB = 1.5 m/s to the left, but car B is initially at rest. The coefficient of

Question

At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg, and 35 kg respectively. Car A is moving to the right with a velocity [latex]\mathbf{ v }_{ A }=2 \ m/s[/latex] and car C has a velocity [latex]\mathbf{ v }_{ B }=1.5 \ m/s[/latex] to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming (a) cars A and C hit car B at the same time, (b) car A hits car B before car C does.

Accepted Answer

Assume that each car with its rider may be treated as a particle. The masses are:

[latex]\begin{aligned}& m_{A}=200+40=240 \mathrm{~kg}, \& m_{B}=200+60=260 \mathrm{~kg}, \& m_{C}=200+35=235 \mathrm{~kg} .\end{aligned}[/latex]

Assume velocities are positive to the right. The initial velocities are:

[latex]\nu_{A}=2 \mathrm{~m} / \mathrm{s} \quad \nu_{B}=0 \quad \nu_{C}=-1.5 \mathrm{~m} / \mathrm{s}[/latex]

Let [latex]\nu_{A}^{\prime}, \nu_{B}^{\prime}[/latex], and [latex]\nu_{C}^{\prime}[/latex] be the final velocities.

(a) Cars A and C hit B at the same time. Conservation of momentum for all three cars.

[latex]\begin{gathered}m_{A} \nu_{A}+m_{B} \nu_{B}+m_{C} \nu_{C}=m_{A} \nu_{A}^{\prime}+m_{B} \nu_{B}^{\prime}+m_{C} \nu_{C}^{\prime} \(240)(2)+0+(235)(-1.5)=240 \nu_{A}^{\prime}+260 \nu_{B}^{\prime}+235 \nu_{C}^{\prime}\quad\quad \text{(1)}\end{gathered}[/latex]

Coefficient of restition for cars A and B.

[latex]\nu_{B}^{\prime}-\nu_{A}^{\prime}=e\left(\nu_{A}-\nu_{B}\right)=(0.8)(2-0)=1.6[/latex] (2)

Coefficient of restitution for cars B and C.

[latex]\nu_{C}^{\prime}-\nu_{B}^{\prime}=e\left(\nu_{B}-\nu_{C}\right)=(0.8)[0-(-1.5)]=1.2[/latex] (3)

Solving Eqs. (1), (2), and (3) simultaneously,

[latex]\begin{array}{r}\nu_{A}^{\prime}=-1.288 \mathrm{~m} / \mathrm{s} \quad \nu_{B}^{\prime}=0.312 \mathrm{~m} / \mathrm{s} \quad \nu_{C}^{\prime}=1.512 \mathrm{~m} / \mathrm{s} \\mathbf{v}_{A}^{\prime}=1.288 \mathrm{~m} / \mathrm{s} \longleftarrow\blacktriangleleft \\mathbf{v}_{B}^{\prime}=0.312 \mathrm{~m} / \mathrm{s} \longrightarrow\blacktriangleleft \\mathbf{v}_{C}^{\prime}=1.512 \mathrm{~m} / \mathrm{s} \longrightarrow\blacktriangleleft\end{array}[/latex]

(b) Car A hits car B before C does.

First impact. Car A hits car B. Let [latex]\nu_{A}^{\prime}[/latex] and [latex]\nu_{B}^{\prime}[/latex] be the velocities after this impact. Conservation of momentum for cars A and B.

[latex]\begin{aligned}m_{A} \nu_{A}+m_{B} \nu_{B} & =m_{A} \nu_{A}^{\prime}+m_{B} \nu_{B}^{\prime} \(240)(2)+0 & =240 \nu_{A}^{\prime}+260 \nu_{B}^{\prime}\quad\quad\text{(4)}\end{aligned}[/latex]

Coefficient of restitution for cars A and B.

[latex]\nu_{B}^{\prime}-\nu_{A}^{\prime}=e\left(\nu_{A}-\nu_{B}\right)=(0.8)(2-0)=1.6[/latex] (5)

Solving Eqs. (4) and (5) simultaneously,

[latex]\begin{aligned}& \nu_{A}^{\prime}=0.128 \mathrm{~m} / \mathrm{s}, \quad \nu_{B}^{\prime}=1.728 \mathrm{~m} / \mathrm{s} \& \mathbf{v}_{A}^{\prime}=0.128 \mathrm{~m} / \mathrm{s} \longrightarrow \& \mathbf{v}_{B}^{\prime}=1.728 \mathrm{~m} / \mathrm{s} \longrightarrow\end{aligned}[/latex]

Second impact. Cars B and C hit. Let [latex]\nu_{B}^{\prime \prime}[/latex] and [latex]\nu_{C}^{\prime \prime}[/latex] be the velocities after this impact. Conservation of momentum for cars B and C.

[latex]m_{B} \nu_{B}^{\prime}+m_{C} \nu_{C}=m_{B} \nu_{B}^{\prime \prime}+m_{C} \nu_{C}^{\prime \prime}[/latex]

[latex](260)(1.728)+(235)(-1.5)=260 \nu_{B}^{\prime \prime}+235 \nu_{C}^{\prime \prime}[/latex] (6)

Coefficient of restitution for cars B and C.

[latex]\nu_{C}^{\prime \prime}-\nu_{B}^{\prime \prime}=e\left(\nu_{B}^{\prime}-\nu_{C}\right)=(0.8)[1.728-(-1.5)]=2.5824[/latex] (7)

Solving Eqs. (6) and (7) simultaneously,

[latex]\begin{aligned}\nu_{B}^{\prime \prime} & =-1.03047 \mathrm{~m} / \mathrm{s} \quad \nu_{C}^{\prime \prime}=1.55193 \mathrm{~m} / \mathrm{s} \\mathbf{v}_{B}^{\prime \prime} & =1.03047 \mathrm{~m} / \mathrm{s} \longleftarrow \\mathbf{v}_{C}^{\prime \prime} & =1.55193 \mathrm{~m} / \mathrm{s} \longrightarrow\end{aligned}[/latex]

Third impact. Cars A and B hit again. Let [latex]\nu_{A}^{\prime \prime \prime}[/latex] and [latex]\nu_{B}^{\prime \prime \prime}[/latex] be the velocities after this impact. Conservation of momentum for cars A and B.

[latex]m_{A} \nu_{A}^{\prime}+m_{B} \nu_{B}^{\prime \prime}=m_{A} \nu_{A}^{\prime \prime \prime}+m_{B} \nu_{B}^{\prime \prime \prime}[/latex]

[latex](240)(0.128)+(260)(-1.03047)=240 \nu_{A}^{\prime \prime \prime}+260 \nu_{B}^{\prime \prime \prime}[/latex] (8)

Coefficient of restitution for cars A and B.

[latex]\nu_{B}^{\prime \prime \prime}-\nu_{A}^{\prime \prime \prime}=e\left(\nu_{A}^{\prime}-\nu_{B}^{\prime \prime}\right)=(0.8)[0.128-(-1.03047)]=0.926776[/latex] (9)

Solving Eqs. (8) and (9) simultaneously,

[latex]\begin{aligned}\nu_{A}^{\prime \prime \prime} & =-0.95633 \mathrm{~m} / \mathrm{s} \\nu_{B}^{\prime \prime \prime} & =-0.02955 \mathrm{~m} / \mathrm{s} \\mathbf{v}_{A}^{\prime \prime \prime} & =0.95633 \mathrm{~m} / \mathrm{s}\longleftarrow \\mathbf{v}_{B}^{\prime \prime \prime} & =0.02955 \mathrm{~m} / \mathrm{s}\longleftarrow \end{aligned}[/latex]

There are no more impacts. The final velocities are:

[latex]\begin{gathered}\mathbf{v}_{A}^{\prime \prime \prime}=0.956 \mathrm{~m} / \mathrm{s} \longleftarrow\blacktriangleleft \\mathbf{v}_{B}^{\prime \prime \prime}=0.0296 \mathrm{~m} / \mathrm{s} \longleftarrow\blacktriangleleft \\mathbf{v}_{C}^{\prime \prime}=1.552 \mathrm{~m} / \mathrm{s} \longrightarrow\blacktriangleleft\end{gathered}[/latex]

We may check our results by considering conservation of momentum of all three cars over all three impacts.

[latex]\begin{aligned}m_{A} \nu_{A}+m_{B} \nu_{B}+m_{C} \nu_{C} & =(240)(2)+0+(235)(-1.5) \& =127.5 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \m_{A} \nu_{A}^{\prime \prime \prime}+m_{B} \nu_{B}^{\prime \prime \prime}+m_{C} \nu_{C}^{\prime \prime} & =(240)(-0.95633)+(260)(-0.02955)+(235)(1.55193) \& =127.50 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} .\end{aligned}[/latex]

Question 13.162: At an amusement park there are 200-kg bumper cars A, B, and ...

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