Ablock slides across a frictionless horizontal surface with constant acceleration 4.15 m/s2 and over a time interval reaches a final velocity of 12.2 m/s.
(a) if its initial velocity is 6.10 m/s, what is its displacement (in m) during the time interval? (indicate the direction with the sign of your answer.) m
(b) what is the distance it travels (in m) during this interval? m
(c) if its initial velocity is −6.10 m/s, what is its displacement (in m) during the time interval? (indicate the direction with the sign of your answer.) m
(d) what is the total distance it travels (in m) during the interval in part (c)?

    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