Question

I m getting my answer as (1.4 N) force.....plz check it.!​

Answer 1

Let us first take out the velocity when the marble hit the surface....

When the object is dropped from height of 2.5m

u=0

a=g

S=2.5

V^2=u^2+2gh

V^2=2×10×2.5

V=5√2

velocity after contact=initak velocity through which it reached to the height 2.5

In that case

S=2.5

a=-g (-showing upposite direction of accleration)

V=0

using third law

we get

0^2=u^2-2×10×2.5

u=5√2

we know,force= change in momentum/time taken

force=(Mv-Mu))T={.1×5√2-(-5√2×.1)}/.1

force=

force=10√2

FORCE=10×1.41=14.1N

{hope it helps you}