Question

A 21.2 kg mass falls from a height of 4.000m. The momentum of the mass just before it hits the ground is
A. 144.2
B. 187.8
C. 320.0
D. 442.4
E. 502.1

Answer 1

Given :-

• Initial velocity of mass,u = 0
• Distance travelled by mass before hitting ground ,S = 4 m
• Mass of object, m = 21.2 kg

Solution:-

By third equation of motion -

$\green{ \underline { \boxed{ \sf{v^2-u^2=2aS}}}}$

where

• v= final velocity
• u = initial velocity
• a = acceleration
• S = distance travelled

Putting Values to find final velocity of mass before hitting the ground-

$\begin{gathered}\\\implies\quad \sf v^2-(0)^2=2\times g \times 4 \quad (g = acceleration \: due \:to \: gravity) \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf v^2=2\times 9.8 \times 4\quad (g= 9.8 \:m/s) \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf v=\sqrt{78.6} \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf v= 8.86 \:m/s \\\end{gathered}$

Now finding the momentum of the mass at that moment -

$\green{ \underline { \boxed{ \sf{Momentum= mass \times velocity}}}}$

$\begin{gathered}\\\implies\quad \sf Momentum= 21.2 \times 8.86 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf Momentum= 187.8 \:kgms^{-1} \\\end{gathered}$

$\longrightarrow$The momentum of the mass just before it hits the ground is 187.8 kgm/s

$\therefore \sf Option \: B) \: is \:correct$✔️