A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume that any other unbalanced forces are negligible. What is the speed of the wreckage just after the collision?

12.23m/s

Explanation:

Let's recall that

The Initial momentum in (East)axis  = 900 x 15 which is = 13500 N-s

Initial momentum in (North) axis = 750 x 20 which is = 15000 N-s

Total initial momentum  becomes = √[(13500)^2 + (15000)^2] = 20180.44 N-s

The  initial momentum direction is  at θ* East of North

Therefore, tanθ = 13500/15000 = 0.9 = tan42*

 tanθ=42

The speed of the wreckage after collection is,

Recall that The law of momentum conservation states that

Momentum initial = Momentum final

20180.44 = (900+750) x v

v = 12.23 m/s

6/12=1/2

step-by-step explanation: