A 800- kg car collides with a 1500- kg car that was initially at rest at the ori

Question

A 800- kg car collides with a 1500- 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 25 o with respect to the positive x axis. The heavier car moves at 28 km/h at -43 o with respect to the positive x axis. The initial speed of the lighter car is 66.1 km/h. What was the initial direction (as measured counterclockwise from the x-axis)?

The correct answer is 337.5 degrees. Please explain step-by-step how to arrive at this answer.

Explanation / Answer

The key concept here is conservation of momentum. Momentum is mass times velocity. Note that as the 1500 kg car was initially at rest, the total momentum must be equal to the momentum of the 800 kg car before the collision

x - component of 1500 kg car after collision
= (1500 kg)(28.0 km/h)cos(-43) = 30716.85 kg km/.h

x-component of 800 kg car after collision
= (800 kg)(25.0 km/h)cos(25) = 18126.15 kg km/h

Total x-component = 30716.85 kg km/h +18126.15 kg km/h = 48843.00 kg km/h.

y - component of 1500 kg car after collision
= (1500 kg)(28.0 km/h)sin(-43) = -28643.93 kg km/.h

y-component of 800 kg car after collision
= (800 kg)(25.0 km/h)sin(25) = 8,452.36 kg km/h

Total y-component = -28643.93 kg +8,452.36 kg km/h = -20191.56kg km/h.

Initial magnitude of momentum must be (since x- and y- axes are at right angles to each other)
[( 48843.00 kg km/h)² + (-20191.56 kg km/h)²]
= 52852.03 kg km/h, and dividing by the mass of the lighter car, 800 kg, gives an initial speed of
66.06 km/h.

And the initial direction must be tan-1[(-20191.56 kg km/h)/(48843.00 kg km/h)] = -22.46o

= 360 - 22.46 = 337.5o

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