35. an object of mass m moving at speed v0 strikes an object of mass 2m which had been at rest. the first object bounces backward along its initial path at speed v0. is this collision elastic, and if not, what is the change in kinetic energy of the system?

The collision is not elastic. The system increases his kinetic energy m*v₀² times.

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved.

Considering the information provided, we can write the momentum conservation equation as follows:

m*v₀ = -m*v₀ + 2*m*vf

Solving for vf, we arrive to this somehow surprising result:

vf = v₀ (in the same direction that m was moving before the collision).

In order to determine if the collision was elastic, or not, we need to calculate the kinetic energy of the system before and after the collision:

K₀ = 1/2*m*v₀²

Kf = 1/2*m*v₀² (due to the object of mass m, as the kinetic energy is always positive) + 1/2 (2m) * v₀²

⇒Kf = 1/2*m*v₀² + 1/2 (2m) * v₀² = 3/2*m*v₀²

ΔK = Kf - K₀ = 3/2*m*v₀² - 1/2*m*v₀² = m*v₀²

As there is a net difference between the final and initial kinetic energies, and the total kinetic energy must be conserved in an elastic collision (by definition) we conclude that the collision is not elastic, and the change in the kinetic energy of the system is equal to m*v₀².

chemical properties of thing