SOMEBODY ONCE TOLD ME THE WORLD WAS GONNA ROLL ME I AIN'T THE SHARPEST TOOL IN THE SHED~
SHE WAS LOOKING KINDA DUM WITH HER FINGER AND HER THUMB
IN THE SHAPE OF AN "L" ON HER FOREHEAD~
WELL THE YEARS START COMING AND THEY DON'T STOP COMING
FED TO THE RULES AND I HIT THE GROUND RUNNING
DIDN'T MAKE SENSE NOT TO LIVE FOR FUN
YOUR BRAIN GETS SMART BUT YOUR HEAD GETS DUM
SO MUCH TO DO, SO MUCH TO SEE--
SO WHATS WRONG WITH TAKING THE BACKSTREETS?
YOU'LL NEVER KNOW IF YOU DON'T GO
YOU'LL NEVER SHINE IF YOU DON'T GLOW.

    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