Question

A 2.0-kg object moving at 7.0 m/s collides with an 8.0-kg object initially at rest. The two objects stick together after the collision. If the two objects system is isolated from environment, calculate the kinetic energy change of the system after the collision.

A. -5.0J

B. -13J

C. -29J

D. -20J

E. -39J

Answer 1

Mass of the first object = m₁ = 2 kg

Mass of the second object = m₂ = 8 kg

Initial velocity of the first object = V₁ = 7 m/s

Initial velocity of the second object = V₂ = 0 m/s (At rest)

Velocity of the objects after the collision = V₃

By conservation of linear momentum,

m₁V₁ + m₂V₂ = (m₁ + m₂)V₃

(2)(7) + (8)(0) = (2 + 8)V₃

V₃ = 1.4 m/s

Initial kinetic energy of the system = E₁

E₁ = 49 J

Final kinetic energy of the system = E₂

E₂ = 9.8 J

Change in kinetic energy of the system = E

E = -39 J

Change in kinetic energy of the system = -39 J