Question

In an hour glass approximately $100$ grains of sand fall per second (starting from rest), and it takes $2\,\sec $ 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

Hint: The force exerted by the falling sand is determined by using the acceleration equation of the motion, momentum equation and the force equation. By using these three equations, the average force exerted by the falling sand on the bottom of the hour glass is determined.

Useful formula:
The acceleration equation of the motion is given by,
$v = u + at$
Where, $v$ is the final velocity of the sand, $u$ is the initial velocity of the sand, $a$ is the acceleration of the sand due to the gravitational force and $t$ in the time taken by the sand to reach the bottom.
The momentum is given by,
$P = mv$
Where, $P$ is the momentum of the sand, $m$ is the mass of the sand and $v$ is the final velocity of the sand.
Force exerted by the sand is given by,
$F = \dfrac{{\Delta P}}{{\Delta t}}$
Where, $F$ is the force exerted by the sand, $\Delta P$ is the change in momentum and $\Delta t$ is the change in time taken.

Complete step by step solution:
Given that,
Total number of grains, $100$ grains.
The time taken by each sand particle, $t = 2\,\sec $.
The mass of each sand, $m = 0.2\,g \Rightarrow 0.2 \times {10^{ - 3}}\,kg$.
Now, the final velocity of the grain is given by,
$v = u + at\,..............\left( 1 \right)$
Assume the initial velocity is zero, and substitute the acceleration due to gravity and the time in the above equation, then,
$v = 0 + \left( {10 \times 2} \right)$
On multiplying the above equation, then
$v = 20\,m{s^{ - 1}}$
The momentum of the one grain of sand is given as,
$P = mv\,...............\left( 2 \right)$
Substituting the mass of the one grain and velocity in the above equation, then
${P_i} = \left( {0.2 \times {{10}^{ - 3}}} \right) \times 20$
Where, ${P_i}$ is the initial momentum.
On multiplying the above equation, then
${P_i} = 4 \times {10^{ - 3}}\,kgm{s^{ - 1}}$
The momentum of the grain when the sand reaches the bottom,
$P = mv$
The velocity of the sand grain when the sand reaches the bottom is $v = 0\,m{s^{ - 1}}$, substituting this velocity in the above equation, then
${P_f} = \left( {0.2 \times {{10}^{ - 3}}} \right) \times 0$
On multiplying the above equation, then
${P_f} = 0\,kgm{s^{ - 1}}$
Now, the change in momentum is the difference of the two momentums, then
$\left| {\Delta P} \right| = \left| {{P_f} - {P_i}} \right|$
By substituting the momentum values, then
\[\left| {\Delta P} \right| = \left| {0 - 4 \times {{10}^{ - 3}}} \right|\]
On subtracting and using the modulus, then
$\Delta P = 4 \times {10^{ - 3}}{\kern 1pt} kgm{s^{ - 1}}$.
Force exerted by the sand is given by,
$F = \dfrac{{\Delta P}}{{\Delta t}}\,................\left( 3 \right)$
On substituting the momentum and time values, then
$F = \dfrac{{4 \times {{10}^{ - 3}}}}{1}$
On dividing the above equation, then
$F = 4 \times {10^{ - 3}}{\kern 1pt} N$
Force gained by $100$ grains is,
$F = 4 \times {10^{ - 3}} \times 100$
On multiplying, then
$F = 0.4\,N$

Hence, the option (A) is the correct answer.

Note: After the equation (3), the time is taken as $1\,\sec $, because in the question it is given that the $100$ grains are falling per second. So, the force of the $100$ grain per second is calculated. The change in momentum will become positive when the value is taken out from the modulus.