Ablock (m=5.00 kg) is attached to a spring (k=315 n/m) and placed on a frictionless surface. the block is displaced 15.0 cm to the left of its equilibrium position and released at time = 0. where is the block at t = 0.555 sec?

    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

Iam on that question know can you tell me the answer i am realy strruglingasap

The answer is 4f======since electrostatic force is given by    we let f be the electrostatic force when  r = 1 m    if we let  denote the electrostatic force when the distance is decreased so that r = 0.5m, then    we established earlier that f = kq₁q₂. therefore    the force will now be 4f.