Question

E. O-8.237 kg m/s4. An object of mass 0.2 kg falling vertically downwardhits the ground with a speed of 13 m/s and bouncesback vertically upward with a speed 4 m/s. If the object was incontact with the ground for 0.5 seconds, calculate theaverage force exerted by the ground on the object. (1 point)A. 06.8 NB. 10.563 NC. 09.048 ND. 12.856 NE. 03.149 N5. An object was acted upon by a force of 24 for 0.8seconds. Calculate the change in its momentum. (1 point)

Answer 1

4)

We woud apply the formula for calculating rate of change of momentum which is expressed as

F = m(v2 - v1)/t

where

F is the force

m is the mass of the object

v2 is the final velocity

v1 is the initial velocity

t is the time

From the information given,

m = 0.2

v1 = -13

v2 = 4

t = 0.5

F = 0.2(4 - -13)/0.5

F = 6.8 N