Acommon practical joke is to pull a seat out from under someone as they attempt to sit down. the person falls to the floor and nobody has fun. assume this happens to a 65 kg person, they fall 0.48 m, and it takes the floor 0.078 s to bring them to rest. (a) what was the impulse exerted on the person by the floor? (b) what was the average force exerted on the person by the floor?

a)  I = 199.36 N s

,  b) F = 2555.9 N

Explanation:

a) For this problem we will use the relationship between momentum and momentum, let's start by finding the speed with which the person reaches the floor, using energy

Initial

    Em₀ = U = m h y

Final

    = ½ m v²

    Em₀ =

    mg y = ½ m v²

    v = √ 2gy

    v = √ 2 9.8 0.48 m

    v = 3,067 m / s

Let's calculate the momentum

    I = Δp = m - m v₀

When it hits the floor it stops so its final speed is zero

    I = 0 - m v₀

    I = - 65 3,067

    I = - 199.36 N s

The momentum of the floor is of equal magnitude but opposite direction

    I = 199.36 N s

b) we write the other impulse relationship

    I = F t

   F = I / t

   F = 199.36 / 0.078

   F = 2555.9 N

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.