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* g * 4 \quad (g = acceleration \: due \:to \: gravity) \\\end{gathered}`

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

`\begin{gathered}\\\implies\quad \sf v=√(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 * velocity}}}}`

`\begin{gathered}\\\implies\quad \sf Momentum= 21.2 * 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`✔️