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A 0.143 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.900 m/s. It has a head-on collision with a 0.298 kg glider that is moving to the left with a speed of 2.25 m/s. Suppose the collision is elastic.Find the magnitude of the final velocity of the 0.143kg glider.Find the magnitude of the final velocity of the 0.298kg glider.

Respuesta :

Answer:

[tex]v_{2f} = -0.207 m/s[/tex]

[tex]v_{1f} = -3.36 m/s[/tex]

Explanation:

As we know that system of two gliders are moving on frictionless track

so momentum conservation is applied on it

So we will have

[tex]m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}[/tex]

so we will have

[tex]0.143(0.900) + 0.298(-2.25) = 0.143 v_{1f} + 0.298v_{2f}[/tex]

[tex]-0.5418 = 0.143 v_{1f} + 0.298 v_{2f}[/tex]

now we know that collision is elastic so we have

[tex]e = \frac{v_{2f} - v_{1f}}{0.900 + 2.25}[/tex]

[tex]v_{2f} - v_{1f} = 3.15[/tex]

so by solving above equations we have

[tex]v_{2f} = -0.207 m/s[/tex]

[tex]v_{1f} = -3.36 m/s[/tex]

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