Suppose that, in the x-y plane, the first car is moving to the right so that its velocity is given by the vector
v₁ = (14 m/s) i
and the second car is moving upward so that its velocity vector is
v₂ = (20 m/s) j
Then the total momentum of two cars before their collision is
m₁v₁ + m₂v₂ = (1400 kg) (14 m/s) i + (2300 kg) (20 m/s) j
= (19,600 i + 46,000 j) kg•m/s
Their momentum after the collision is
(1400 kg + 2300 kg) v = (3700 kg) v
where v is the velocity vector of the wreckage.
By conservation of momentum,
(19,600 i + 46,000 j) kg•m/s = (3700 kg) v
Let a and b be the horizontal and vertical components of v, respectively. Then
19,600 kg•m/s = (3700 kg) a ⇒ a ≈ 5.2973 m/s ≈ 5.3 m/s
46,000 kg•m/s = (3700 kg) b ⇒ b ≈ 12.4324 m/s ≈ 12 m/s
so that the final speed of the wreckage is
||v|| = √(a² + b²) ≈ 13.5139 m/s ≈ 14 m/s
