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A 1210 kg car is driving NE

(at 45.0°) at 15.2 m/s when it is

struck by a moving 1540 kg car.

Afterward, they stick together and

move directly east (at 0°) at 23.3

m/s. What was the x-component

of the second car's initial velocity?

Please help

Respuesta :

okpalawalter8

Answer:

[tex]V_2_X=33.16m/s[/tex]

Explanation:

From the question we are told that

Mass of car        [tex]M_1=1210kg \\Angle1=\theta _1 45\textdegree NE[/tex]

Velocity of car   [tex]v_1= 15m/s[/tex]

Mass of Truck  [tex]M_2= 1540kg \\Angle 2=\theta_2 0\textdegree E[/tex]

Final velocity    [tex]v_2= 23.3m/s[/tex]

Generally the the equation of the law of conservation of momentum is mathematically given by

Given the x direction

  [tex]m_1v_1cos\theta+m_2v_2=(m_1+m_2)v[/tex]

  [tex]1210*15.2cos45+1540*V_2=(1210+1540)*23.3[/tex]

  [tex]V_2=\frac{(1210+1540)*23.3}{210*15.2cos45+1540}[/tex]

  [tex]V_2=33.16m/s[/tex]

The x-component  of the second car's initial velocity is

[tex]V_2_X=33.16m/s[/tex]

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