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Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and was approaching at 6.00 m/s due south. The second car has a mass of 900 kg and was approaching at 24.0 m/s due west.

A. Calculate the final velocity of the cars. (Note that since both cars have an initial velocity, you cannot use the equations for conservation of momentum along the x-axis and y-axis; instead, you must look for other simplifying aspects..)
Magnitude incorrect: Your answer is incorrect. m/s Direction ° (counterclockwise from west is positive)

B. How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.)

Respuesta :

cjmejiab

To resolve point A and B we need the concepts related to conservation of momentum (By collision) and Kinetic Energy. Conservation of momentum is given by the equation,

[tex]m_1\vec{v_1}+m_2\vec{v_2} = (m_1+m_2)\vec{v}[/tex]

Our values in the statment are:

[tex]m_1 = 1200kg[/tex]

[tex]v_1 = 6m/s[/tex]

[tex]m_2 = 900kg[/tex]

[tex]v_2 = 24m/s[/tex]

Part A) As it is in an icy intersection, there is two different components (x,y) then,

[tex]1200(-6\hat{j})+900(-24\hat{i}) = (1200+900)\vec{v}[/tex]

[tex]2100\vec{v} = -21600\hat{i}-7200\hat{j}[/tex]

[tex]\vec{v} = -72/7\hat{i}-24/7\hat{j}[/tex]

Then the magnitude is,

[tex]|\vec{v}| = 9.6525m/s[/tex]

Part B) To obtain the Kinetic Energy Loss we need to use its equation, which is given by,

[tex]KE_i = \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]

[tex]KE_i = \frac{1}{2}(1200)(6)^2+\frac{1}{2}(900)(24)^2[/tex]

[tex]KE_i = 280.8kJ[/tex]

The final energy is given by,

[tex]KE_f = \frac{1}{2}(m_1+m_2)v_f^2[/tex]

[tex]KE_f = \frac{1}{2} (1200+900)(9.65)[/tex]

[tex]KE_f =97778.625J[/tex]

Then the change in Kinetic Energy is

[tex]\Delta KE = KE_f-KE_i = 97.778kJ- 280.8kJ[/tex]

[tex]\Delta KE = -183.02kJ[/tex]

There was a loss of KE of 183.02kJ

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