A 800- kg car collides with a 1200- kg car that was initiallyat rest at the orig

Question

A 800- kg car collides with a 1200- kg car that was initiallyat rest at the origin of an x-y coordinate system. After thecollision, the lighter car moves at 25.0 km/h in a direction of 40o with respect to the positive x axis. The heavier carmoves at 28 km/h at -50 o with respect to the positive xaxis.
What was the initial speed of the lighter car (inkm/h)? What was the initial direction (as measured counterclockwisefrom the x-axis)? A 800- kg car collides with a 1200- kg car that was initiallyat rest at the origin of an x-y coordinate system. After thecollision, the lighter car moves at 25.0 km/h in a direction of 40o with respect to the positive x axis. The heavier carmoves at 28 km/h at -50 o with respect to the positive xaxis.
What was the initial speed of the lighter car (inkm/h)? What was the initial direction (as measured counterclockwisefrom the x-axis)?

Explanation / Answer

here we have, momentum conservation so , let the initial speed of 800 kg is ux km/hr in x direction then 800u + 0 =800x 25 x cos 40   + 1200x28 cos -50         so ux = 46.14 km/hr now let the initial speed of 800 kg be uy in y direction then we have, 800uy = 800x 25 x sin 40   + 1200x28 sin -50 uy = -16.1 km/hr so total magnitude of u = (ux2 + uy2)1/2                                    =( 46.142 + 16.12 )1/2                                   = 48.86 km/hr and the direction is :- tan - ( uy/ux)    = -19.23 deg

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