Question

In an hour-glass approximately $100$ grains of sand fall per second (starting from rest); and it takes $2s$ for each sand particle to reach the bottom of the hour-glass. If the average mass of each sand particle is $0.2g$, then the average force exerted by the falling sand on the bottom of the hour-glass is close to

• A. $0.4N$
• B. $0.8N$
• C. $1.2N$
• D. $1.6N$

Answer 1

Answer: (A)

Force = Rate of change of momentum



Velocity with which a sand particle strikes the bottom of hour-glass,

$v=u+gt$

$\Rightarrow v=0+10\times 2=20m{s}^{-1}$

Change in momentum of particle

$=p_f-P_i=0-mv$
0
$=-0.2\times 10^{-3}\times 20$

$=-4\times 10^{-3}kg-m{s}^{-1}$

Momentum imparted to base by the particle

$=4\times 10^{-3}kg-m{s}^{-1}$

Total change of momentum imparted per second by all $100$ particles

$=4\times 10^{-3}kg-m{s}^{-1}\times 100s^{-1}$

$=0.4kg-m{s}^{-1}$

So, force on bottom $=0.4N$