Question

Two cars, one of mass 1400 kg, and the

Answer 1

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