One cubic centimeter (1.0 cm3) of water has a mass of 1.0 10-3 kg.
(a) determine the mass of 1.0 m3 of water

One meter has 100cm. One cubic meter is therefore 100cm x 100cm x 100cm, multiplied are 1,000,000 cubic centimeters.Water density of 1.0x10^-3 Kg per cubic centimeter MULTIPLIED BY 1.00x10^6 (one million) cubic centimeters equals 1.0x10^3Kg or 1000Kg or one metric ton.I multiplied because I wanted to cancel the cubic centimeter from "the denominator of the density units with the numerator of the volume units"

    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