Question

Two cars, one of mass 1400 kg, and the

second of mass 2300 kg, are moving at right
angles to each other when they collide and
stick together. The initial velocity of the first
car is 14 m/s in the positive r direction and
that of the second car is 20 m/s in the positive
y direction.
What is the magnitude of the velocity of the
wreckage of the two cars immediately after the
collision?
Answer in units of m/s.
es

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