Question

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

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

Answer 1

Answer: (A)

Force = Rate of change of momentum

Velocity with which a sand particle strikes the bottom of hour-glass,
$v = u + gt$
$\Rightarrow v = 0 + 10 \times 2 = 20 \,ms^{-1}$
Change in momentum of particle
$= p_f - P_i = 0 - mv$
0
$= - 0.2 \times 10^{-3} \times 20$
$= - 4 \times 10^{-3}\, kg-ms^{-1}$
Momentum imparted to base by the particle
$ = 4 \times 10^{-3}\, kg-ms^{-1}$
Total change of momentum imparted per second by all $100$ particles
$= 4\times 10^{-3}\, kg-ms^{-1} \times 100\, s^{-1}$
$= 0.4 \,kg-ms^{-1}$
So, force on bottom $= 0.4 \,N$