Joseph remembers reading that for a particular pressure, p1, there is a maximum height to which a fluid can rise. he reasons that in this case the pressure (p2) must drop to zero so that the water can no longer flow to increased heights. he decides to check if the simulation can be made to demonstrate such conditions. he decides to make a2 as small as possible in the simulation (2.2 cm2) so that he increases the speed of the fluid in tube 2 as high as is possible (he decides to do this after examining the continuity equation). he then calculates the height of the fluid under these conditions where the pressure would fall to zero so that the fluid could no longer be made to flow under the conditions set up in the simulation. he then runs the simulation under the largest δy allowed to check his calculation. what does joseph calculate for the maximum height?

    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

45 is the correct answer