A diatomic ideal gas acts as a heat engine following a process described by the diagram below: Step 1. Isothermal expansion at ThotThot from volume VaVa to a final volume VbVb. The final volume is a factor kk times larger than the initial volume (ie. VbVb=kVakVa).
Step 2. Isobaric contraction from volume VbVb back to original volume VaVa, and cooling from ThotThot to TcTc
Step 3. Isochoric heating returning the system to its original state.

If there are n=82.4mols of the gas, the expansion factor kk=3.2, and initial temperature ThotThot=217.8K, what is the efficiency ee of this cyclic heat engine?

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s