A body of mass M is dropped from a height h on a sand floor. If the body penetrates x cm into the sand, the average resistance offered by the sand to the body is

Question

A body of mass M is dropped from a height h on a sand floor. If the body penetrates x cm into the sand, the average resistance offered by the sand to the body is (A) Mg((h/x)) (B) Mg(1+(h/x)) (C) Mgh + Mgx (D) Mg(1-(h/x))

Tardigrade Admin · Accepted Answer

the body strikes the sand floor with a velocity

v

, then
Potential energy = Kinetic energy

M g h = \frac{1}{2} M υ^{2}

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 the body. Net force in downward direction

= M g - F

Work done by all the forces is equal to change in KE

(M g - F) x = 0 - \frac{1}{2} M υ^{2}

(M g - F) x = - M g h

or

F x = M g h + M gx

or

F = M g (1 + \frac{h}{x})

Q. A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x c m$ into the sand, the average resistance offered by the sand to the body is

$M g (\frac{h}{x})$

$M g (1 + \frac{h}{x})$

$M g h + M gx$

$M g (1 - \frac{h}{x})$

Q. A body of mass M is dropped from a height h on a sand floor. If the body penetrates xcm into the sand, the average resistance offered by the sand to the body is

Mg(xh​)

Mg(1+xh​)

Mgh+Mgx

Mg(1−xh​)

Q. A body of mass $M$ is dropped from a height $h$ on a sand floor. If the body penetrates $x c m$ into the sand, the average resistance offered by the sand to the body is

$M g (\frac{h}{x})$

$M g (1 + \frac{h}{x})$

$M g h + M gx$

$M g (1 - \frac{h}{x})$