Question

a body of mass 20kg at a height of 100m falls freely. calculate the kinetic energy when it strikes the ground is the linear momentum of the body when it strikes the ground.​

Answer 1

Given,

$m=20\ kg\\\\h=100\ m$

Since the body is falling from rest,

$u=0\ m\ s^{-1}$

Let the body have a final velocity $v\ m\ s^{-1}$ at the surface of the earth.

At the height of 100 metre from the surface of the earth, the total mechanical energy of the body is,

$E_1=\dfrac{1}{2}mu^2+mgh\\\\\\E_1=mgh\quad [u=0]\\\\\\E_1=20\times 10\times 100\\\\\\E_1=20\ kJ$

The total mechanical energy of the body at the surface of the earth is,

$E_2=\dfrac{1}{2}mv^2+mg(0)\\\\\\E_2=\dfrac{1}{2}mv^2$

By law of conservation of mechanical energy,

$E_1=E_2\\\\\\\boxed{\dfrac {1}{2}mv^2=20\ kJ}$

So this is the kinetic energy of the body at the surface of the earth.

$\dfrac {1}{2}mv^2=20000\ J\\\\\\10v^2=20000\\\\\\v^2=2000\\\\\\v=20\sqrt5\ m\ s^{-1}$

Then the linear momentum of the body when it strikes the ground is,

$p=mv\\\\\\p=20\times20\sqrt5\\\\\\\boxed{p=400\sqrt5\ kg\ m\ s^{-1}}$

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