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 0.40 m, the mass that moves downward is 61.0 kg, and the collision on the floor lasts 0.0820 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

SOL:

(a)

falling from height h =0.4 m

velocity with which he can hits the floor =v m/s

from kine matic relation ,

v^2 - u^2 = 2g h

u = 0 m/s     (since , falls freely down word)

thus, v = sqrt ( 2 g h )

           = sqrt  ( 2 * 9.8 * 0.4 )   = 2.8 m/s

as the floor is hard , it will not re bounce back ,

thus , rebounce velocity = 0 m/s

thus, change of velocity = 2.8 - 0 = 2.8 m/s

so, change of momentum   = mv - 0 kgm/s

                                       = 61 (2.8) kgm/s         ( mass m = 61 kg given)

                                       = 170.8 kgm/s

we know that , change of momentum = impulse (I) = 170.8 kgm/s  

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(b) collision on the floor lasting time dt = 0.082 s

        impulse (I )= average force (F) . lasting time (dt)

Thus, the average force acting on the victim (F) = impulse (I ) / lasting time (dt)

                                                                       = 170.8 / 0.082

                                                                       = 2082.9    or 2083 N       (appx.)