Recall that the blocks can only move along the x axis. The x components of their velocities at a certain moment are v1x and v2x. Find the x component of the velocity of the center of mass (vcm)x at that moment. Keep in mind that, in general: vx=dx/dt.

Vcm = m1v1 + m2v2 / m1 + m2

Explanation:

The distance of centre of mass Xcm can be taken by taking mass moments about the origin.

(m1 + m2)Xcg = m1x1 + m2x2

Xcm = m1x1 + m2x2 / m1 + m2

Differentiating Xcm = m1x1 + m2x2 / m1 + m2 with respect to time t, will give us the velocity of the centre of mass.

I have attached a diagram showing said differentiation.

    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