A 0.75 kg model car is moving west at a speed of 9.0 m/s when it collides head-on with a 2.00 kg model truck that is traveling east at a speed of 10.0 m/s. After the collision, the 0.75 kg model car is now moving east at 11 m/s. What is the speed and direction of the model truck after the collision?

  2.5 m/s east

Explanation:

Let east be the positive direction for velocity.

The change in momentum of the 0.75 kg model car is ...

  m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)

  = (0.75 kg)(20 m/s) = 15 kg·m/s

The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.

  m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s

The required relation is ...

  15 kg·m/s = -2(v2 -10) kg·m/s

  -7.5 = v2 -10 . . . . divide by -2

  2.5 = v2 . . . . . . . add 10

The velocity of the model truck after the collision is 2.5 m/s east.