Question

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by 0.50 m, the mass that moves downward is 80 kg, and the collision on the floor lasts 0.081 s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

Answer 1

Impulse, J = 250.4 kgm/s, Avg Force F=3091.4 N

Step-by-step explanation:

Since we know that impulse is the change in momentum i.e. Δp and Δp
`=mv`, therefore, to calculate the velocity we perform:

As the person has fallen by a 0.50m height, its potential energy changes into kinetic energy, therefore,

K.E.=P.E.

`\frac{1}2}mv^(2)`=
`mgh`

`v=√(2gh)`

`v=√(2*9.8*0.50)`

`v=3.13ms^(-1)`

(a) Impulse
`J` = Δp
`=mv`

`J= 80*3.13`

`J = 250.4 kgms^(-1)`

(b) Avg Force F = Δp/Δt

`F=(250.4)/(0.081)`

`F=3091.4` N