Question

0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before and just after impact with the floor is 10 m/s. Determine the magnitude of the impulse delivered to the floor by the steel ball. ​

Answer 1

Answer:

impulsive force is = F × TIME TAKEN

SO , F = m x g ( g = 10 m/s2)

thus the f= 0.2 × 10(kg m/ s2)

f = 2 N

finally. the impulsive force = 2 N × 1sec = 2

so , magnitude is (2)

This ans.may be correct , so please confirm this ans.