Question

A body of mas $m=1kg$ falls from a height $h=20m$ to the ground level.
(a) What is the magnitude of total change in momentum of the body before it strikes the ground?
(b) What is the corresponding average force experienced by it? $(g=10m/se{c}^2)$

Answer 1

(a) Since the body falls from rest $(u=0)$ through a distance $h$ before striking the ground, the speed $v$ of the body is given by kinematical equation.
$v^2=u^2+2$ as; Putting $a=g$ and $s=h$
We obtain $v=\sqrt{2gh}$
$\Rightarrow$ The magnitude of total change in momentum of the body
$\Delta p=|mv-0|=mv$, where $v=\sqrt{2gh}$
$\Rightarrow$ $\Delta p=m\sqrt{2gh}=\left(1\right)\sqrt{(2\times 10\times 20)}kg-m/sec$
$\Rightarrow$ $\Delta p=20kg-m/sec$.
(b) The average force experienced by the body $=F_{av}=\frac{\Delta p}{\Delta t}$, where $\Delta t=$ times of motion of the body $=t$(say). We know $\Delta p=20kgm/sec$. Therefore we will have to find $t$ using the given data/ From kinematics,
$s=ut+\frac{1}{2}{at}^2\Rightarrow h=\frac{1}{2}g{t}^2(u=0)$
$\Rightarrow t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 20}{10}}=2s$
$\therefore F_{av}=\frac{\Delta p}{\Delta t}=\frac{\Delta p}{t}=\frac{20}{2}=10N$
Putting the general values of $\Delta p$ & $t$, we obtain
$F_{av}=\frac{m\sqrt{2gh}}{\sqrt{\frac{2h}{g}}}=mg$
$\Rightarrow {\overrightarrow{F}}_{av}=m\overrightarrow{g}=1\times 10=10N$
where $mg$ is the weight $\left(W\right)$ of the body and $g$ is always directed vertically downward force of magnitude $mg$.