A 700- kg car collides with a 1500- kg car that was initially at rest at the ori

Question

A 700- 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 30.0 km/h in a direction of 25 o with respect to the positive x axis. The heavier car moves at 33 km/h at -50 o with respect to the positive x axis. What was the initial speed of the lighter car (in 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. What was the initial direction (as measured counterclockwise from the x-axis)?

Explanation / Answer

We solve this using the conservation of momentum in both the x an y directions. Let's start with x.

o = f

m1v1xo + m2v2xo = m1v1xf + m2v2xf

The momentum of m2 in the left side of the equation goes away, since the velocity is zero. Then plug in the values given, being sure the take account for the angles:

700v1xo = 700(30cos(25)) + 1500(33cos(310)) //we can use cosine of 310o or -50o to get the same value  

Divide by 700

v1xo = 30cos(25) + (1500(33cos(310)) / 700)

The same idea is applied for the y directions, using sine instead of cosine:

700v1yo = 700(30sin(25)) + 1500(33sin(310))

v1yo = 30sin(25) + (1500(33sin(310)) / 700)

Once you have the two values for the velocities, you can use Pythagoream Theorem to find the initial velocity:

v1xo2 +v1yo2 = v12

Using the same values for the x and y values of the velocity, you can find the initial angle, using the inverse tangent:

= tan-1(v1xo / v1yo)

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