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 79.0 kg, and the collision on the floor lasts 0.0750 s. what are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

A)247.4kgm/s b) 3299.0N

Explanation:

The potential energy of the victim = mgh where m is mass, g is acceleration due to gravity and h is height from where the victim fell

Pe = 79*9.81*0.5 = 1/2mv2 (kinetic energy of the motion) v is speed in m/s

v = √9.81 = 3.132m/s

Change in momentum of the victim = m(vi-vf)

Vi is zero and vf is the velocity of impact

Change in momentum = 79*(3.132-0) = 247kgm/s which also the impulse

B) average force x time = impulse = 247

Average force = 247/0.075 = 3299.04N

f is prop to m1xm2 double m1 double force

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west