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A cart with mass 320 g moving on a frictionless linear air track at an initial speed of 1.8 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.78 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact? (c) What is the speed of the two-cart center of mass?

Respuesta :

Answer:

Part a)

m = 126.5 g

Part b)

v = 2.58 m/s

Part c)

v = 1.29 m/s

Explanation:

Part a)

By momentum conservation we will have

[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]

here we have

[tex]m_1 = 320 g[/tex]

[tex]u_1 = 1.8 m/s[/tex]

[tex]u_2 = 0[/tex]

[tex]v_1 = 0.78 m/s[/tex]

also since collision is elastic collision so we have

[tex]v_2 = 2.58 m/s[/tex]

so now we have

[tex]320(1.8) + m_2(0) = 320(0.78) + m_2(2.58)[/tex]

[tex]m_2 = 126.5 g[/tex]

Part b)

As we know that in perfect elastic collision we will have

[tex]e = \frac{v_2 - v_1}{u_1 - u_2}[/tex]

now we will have

[tex]1 = \frac{v_2 - 0.78}{1.8 - 0}[/tex]

now we have

[tex]1.8 = v_2 - 0.78[/tex]

[tex]v_2 = 2.58 m/s[/tex]

Part c)

Since there is no external force on it

so here velocity of center of mass will remain the same

[tex]v_{cm} = \frac{m_1v_1 + m_2 v_2}{m_1 + m_2}[/tex]

[tex]v_{cm} = \frac{320(1.8) + 126.5(0)}{320 + 126.5}[/tex]

[tex]v_{cm} = 1.29 m/s[/tex]

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