A1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the same road at 12.0 m/s . the two vehicles remain locked together after the collision.

The questions are:
(a)  What is the velocity (magnitude and direction) of the two vehicles just after the collision?  
(b) At what speed should the truck have been moving so that it and car are both stopped in the collision?  
(c) Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b).  For which situation is the change in kinetic energy greater in magnitude?

Answers:
A) Since the two cars stick together, this is a totally inelastic collision, therefore you have conservation of momentum:
m₁·v₁₁ + m₂·v₁₂ = (m₁ + m₂) · v₂

Therefore:
v₂ = (m₁·v₁₁ + m₂·v₁₂) / (m₁ + m₂)
For convention, velocities towards the right (estbound) are positive while towards the left (westbound) are negative.
v₂ = (1050 · (-13) + 6320 · 12) / (1050 + 6320)
    = 8.438 m/s
Hence, the two cars have a speed of 8.4 m/s eastbound.

B) The conservation of momentum is the same as in part A):
m₁·v₁₁ + m₂·v₁₂ = (m₁ + m₂) · v₂

If v₂ = 0, the formula becomes:
m₁·v₁₁ + m₂·v₁₂ = 0

Now, solve for v₁₂:
v₁₂ = -m₁·v₁₁ / m₂
     = -1050 · (-13) / 6320
     = 2.159 m/s

Therefore, the truck should have moved at a speed of 2.2 m/s eastbound.

C) The change in kinetic energy is given by:
ΔK = K₂ - (K₁₁ + K₁₂) 
      = 1/2·(m₁+m₂)·v₂² - (1/2·m₁·v₁₁² + 1/2·m₂·v₁₂²)

For part A) we have:
ΔK = 1/2·(1050+6320)·(8.4)² - (1/2·1050·(-13)² + 1/2·6320·12²)
      = -283751 J

For part B) we have:
ΔK = 1/2·(1050+6320)·(0)² - (1/2·1050·(-13)² + 1/2·6320·2.2²)
      = -104019 J

The change in kinetic energy is bigger in magnitude for part A).
      

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