Auniform brick of length 25 m is placed overthe edge of a horizontal surface with a maximum overhang of 12.5 m attained withouttipping. now two identical uniform bricks of length25 m are stacked over the edge of a horizontalsurface. what maximum overhang is possible forthe two bricks (without tipping)? answer in units of m.

The system of two rods will lie on the table as show in the figure

The center of first rod will lie exactly at the edge of first rod and then the center of mass of two combined rods will lie at the edge of the table.

So now the whole system will rest on the rod and it will not tipping off.

Since both rods are identical so we can say that the system will have its center of mass at the mid point on the line joining the two centers

So the value of x will be at mid point of line joining the two points on rod

So total length over the edge will be given as

    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