AP EAPCET 23-Aug-2021 Shift 2 Paper

Given, mass of body =M
Height above the sand floor =h
Distance penetrated into sand =x
Let v be the velocity with which body strikes the surface.
Then, by equation of motion, we get
v2=u2+2gh⇒v2=2gh ...(i)
[∵ Initial velocity at highest point, u=0]
With this velocity v, when body passes through the sand floor it comes to rest after travelling a distance x.
Let F be the resisting force acting on body.
Then, net downward force will be Fnet=Mg−F
By work-energy theorem, work done by all forces is equal to change in KE, which gives
W=ΔK
Fnet×x=Kf−Ki
(Mg−F)x=0−

1

2

Mv2 [∵Final‌KE,Kf=0]
(Mg−F)x=−

1

2

Mv2 ...(ii)
Substituting the value from Eq. (i) in Eq. (ii), we get
⇒Mg‌x−Fx=−

1

2

M(2gh)
⇒Mg‌x−Fx=−Mg‌h
⇒Fx=Mg‌x+Mg‌h
⇒F=Mg(

x+h

x

)