(c7p50) A 1000- kg car collides with a 1300- kg car that was initially at rest...

(c7p50) A 1000- 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 25 ^o with respect to the positive x axis. The heavier car moves at 28 km/h at -50 ^o with respect to the positive x axis.
What was the initial speed of the lighter car (in km/h)?

Also, What was the initial direction (as measured counterclockwise from the x-axis)?

Solutions

Expert Solution

The key concept here is conservation of momentum. Momentum is mass times velocity. Note that as the 1300 kg car was initially at rest, the total momentum must be equal to the momentum of the 1000 kg car before the collision

x - component of 1300 kg car after collision
= (1300 kg)(28.0 km/h)cos(-50) = 23397.46 kg km/.h

x-component of 1000 kg car after collision
= (1000 kg)(25.0 km/h)cos(25) = 22657.69 kg km/h

Total x-component = 23397.46 kg km/.h + 22657.69 kg km/h = 46055.15 kg km/h.

y - component of 1300 kg car after collision
= (1300 kg)(28.0 km/h)sin(-50) = -27884.0 kg km/.h

y-component of 1000 kg car after collision
= (1000 kg)(25.0 km/h)sin(25) = 10565.45 kg km/h

Total y-component = -27884.0 kg km/.h + 10565.45 kg km/h = -17318.54 kg km/h.

Initial magnitude of momentum must be (since x- and y- axes are at right angles to each other)
√[(46055.15 kg km/h)² + (-17318.54 kg km/h)²]
= 49204.74 kg km/h, and dividing by the mass of the lighter car, 1000 kg, gives an initial speed of
49.20 km/h.

And the initial direction must be tan^-1[(-17318.54 kg km/h)/(46055.15 kg km/h)] = -20.6 degrees

genius_generous answered 11 months ago

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