2D Car Collision 1 m2 Figure 721 2 (c7p50) A 900- kg car collides with a 1300- k
ID: 1776852 • Letter: 2
Question
2D Car Collision 1 m2 Figure 721 2 (c7p50) A 900- kg car collides with a 1300- kg car that was initially at rest at the origin of an x-y coordinate system. After the collision, the lighter car moves at 25.0 km/h in a direction of 35 o with respect to the positive x axis. The heavier car moves at 28 km/h at -45° with respect to the positive x axis. What was the initial speed of the lighter car (in km/h)? 48.36 km/h The car is not coming in along the x-axis in the picture. Calulate the x and y values of the initial monenta and compute the speed and direction of the velocity. Submit Answer Incorrect. Tries 1/7Previous Tries What was the initial direction (as measured counterclockwise from the x-axis)? Submit Answer Tries 0/7Explanation / Answer
x - component of 1300 kg car after collision
= (1300 kg)(28.0 km/h)cos(-45) = 25738.68 kg km/.h
x-component of 900 kg car after collision
= (900 kg)(25.0 km/h)cos(35) 18430.92= kg km/h
Total x-component = 25738.68 kg km/h + 18430.92 kg km/h = 44169.60 kg km/h.
y-component of 1300 kg car after collision
= (1300 kg)(28.0 km/h) sin(-45) = - 25738.68 kg km/h
y-component of 900 kg car after collision
= (900 kg)(25.0 km/h)sin(35) = 12905.469 kg km/h
Total y-component = -25738.68 kg +12905.469 kg km/h = - 12833.21 kg km/h.
Initial magnitude of momentum must be (since x- and y- axes are at right angles to each other)
[(44169.60 kg km/h)^2 + (-12833.21 kg km/h)^2] = 45996.139 kg km/h
initial sped of lighter car v = 45996.139/900 = 51.10 km/h
And the initial direction must be
tan^-1[(-12833.21 kg km/h)/(44169.60 kg km/h)] = - 16.20 degrees.
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