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engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collision between two train cars traveling toward each other. Car one is traveling south at 18 M/S and has a full load, giving it a total mass of 14,650 kg. Kartoo is traveling north at 11 m/s and has a mass of 3825 kg. After the collision car one has a final velocity of 6 M/S south what is the final velocity of car 2?

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Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

[tex]m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f[/tex]

where ;

[tex]m_1=mass of object 1\\v_1i=initial velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2[/tex]

Given in the question that;

[tex]m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?[/tex]

Apply the formula as;

[tex]m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f[/tex]

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

57 m/s = v₂f { nearest whole number}

Answer:

35 m/s south

Explanation:

took the test

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