Question

\[10,000\] small balls, each weighing \[1\;{\rm{gm}}\], strike one square cm of area per second with a velocity \[100\;{\rm{m/s}}\] in a normal direction and rebound with the same velocity. The value of pressure on the surface will be.
A. \[2 \times {10^3}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\]
B. \[2 \times {10^5}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\]
C. \[{10^7}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\]
D. \[2 \times {10^7}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\]

Answer 1

Hint:The above problem can be resolved by using some mathematical formulations and concepts. The formulas include the expression for change in momentum, the force acting on the surface and the pressure acting on the surface. We are provided with some basic pieces of information like the number of balls, the magnitude of velocity and weight of each ball. These values can be made to utilise for the substitution in the formulas to obtain the final answer.

Complete step by step answer:
Given:
The number of small balls is, \[n = 10,000\].
The weight of each ball is, \[w = 1\;{\rm{gm}} = 1\;{\rm{gm}} \times \dfrac{{1\;{\rm{kg}}}}{{1000\;{\rm{gm}}}} = 0.001\;{\rm{kg}}\].
The velocity of the ball is, \[v = 100\;{\rm{m/s}}\].
Apply the mathematical formulation for the change in momentum of one ball during rebound is,
\[
\Rightarrow \Delta P = mv - \left( { - mv} \right)\\
\Rightarrow \Delta P = 2mv
\]
Then the change of momentum for 10,000 balls is,
\[
\Rightarrow \Delta {P_1} = n\left( {\Delta P} \right)\\
\Rightarrow \Delta {P_1} = n\left( {2mv} \right)
\]
Solve by substituting the values in above expression as,
\[
\Rightarrow \Delta {P_1} = n\left( {2mv} \right)\\
\Rightarrow \Delta {P_1} = 10,000 \times \left( {2 \times 0.001\;{\rm{kg}} \times 100\;{\rm{m/s}}} \right)\\
\Rightarrow \Delta {P_1} = 2000\;{\rm{kg - m/s}}
\]
Then the magnitude of force acting on the surface is,’
\[\Rightarrow F = \dfrac{{\Delta {P_1}}}{t}\]
Here, t denotes the time in seconds and here its value is 1 second.
Again, substitute the values as,
\[
\Rightarrow F = \dfrac{{\Delta {P_1}}}{t}\\
\Rightarrow F = \dfrac{{2000\;{\rm{kg - m/s}}}}{{1\;{\rm{s}}}}\\
\Rightarrow F = 2000\;{\rm{N}}
\]
 Now apply the formula for the pressure on the surface as,
\[\Rightarrow P = \dfrac{F}{A}\]
Here, A denotes the area of the surface and its value is unit square centimetres,
\[\Rightarrow A = 1\;{\rm{c}}{{\rm{m}}^{\rm{2}}} = 1\;{\rm{c}}{{\rm{m}}^{\rm{2}}} \times \dfrac{{1\;{{\rm{m}}^{\rm{2}}}}}{{{{10}^4}\;{\rm{c}}{{\rm{m}}^{\rm{2}}}}} = {10^{ - 4}}\;{{\rm{m}}^{\rm{2}}}\]
Substitute the values in the expression for pressure as,
 \[
\Rightarrow P = \dfrac{F}{A}\\
\Rightarrow P = \dfrac{{2000\;{\rm{N}}}}{{{{10}^{ - 4}}\;{{\rm{m}}^{\rm{2}}}}}\\
\Rightarrow P = 2 \times {10^7}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}
\]
Therefore, the value of pressure on the surface is \[2 \times {10^7}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\] and option (D) is correct.

Note: To solve the given problem, one must be aware of the concept and fundamentals of the momentum. Along with this, the basic knowledge for the applications of mathematical relations is required to understand deeply.