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1 October, 03:30

A 1100 - 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 20.0 km/h in a direction of 30 o with respect to the positive x axis. The heavier car moves at 23 km/h at - 46 o with respect to the positive x axis. What was the initial speed of the lighter car (in km/h)

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  1. 1 October, 07:02
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    37.45 km/hr

    Explanation:

    To solve this, we use the law of conservation of momentum in two directions (x, and y).

    in x direction

    1100 * v * cosθ = 1100 * 20 * Cos30 + 1300 * 23 * cos46

    1100 * vcosθ = 22000 * 0.866 + 29900 * 0.695

    1100 * vcosθ = 19052 + 20780.5

    1100 * vcosθ = 39832.5

    vcosθ = 39832.5 / 1100

    vcosθ = 36.21

    In the y direction

    1100 * v * Sinθ = 1100 * 20 sin30 - 1300 * 23 sin46

    1100 * vsinθ = 22000 * 0.5 - 29900 * 0.719

    1100 * vsinθ = 11000 - 21498.1

    1100 * vsinθ = - 10498.1

    vsinθ = - 10498.1 / 1100

    vsinθ = - 9.54

    Since we are looking for v, then

    v² = vcos²θ + vsin²θ

    v² = 36.21² + (-9.54²)

    v² = 1311.16 + 91.01

    v² = 1402.17

    v = √1402.17

    v = 37.45 km/hr

    Thus, the initial speed of the lighter car is 37.45 km/hr
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