A block mass m = 34 kg and a speed of v is behind a block mass M = 81 kg and a speed of .5m/s, as shown. The surface is frictionless and the blocks collide and a couple. After the collision, the blocks have a common speed of .9 m/s. What is the magnitude of the impulse on 34 kg block due to collisions? A. 32Ns
B. 14Ns
C. 41Ns
D. 52Ns
E. 73Ns

    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

Two 10¢  coins  (dimes)  carrying identical charges  are  lying 2.5 m apart  on a  table.  if each  of  these coins experiences  an  electrostatic force  of  magnitude 2.0 n due  to the  other coin.  the charge on each coin is :     52.2 uc