1000 small balls,each weighing 1 text{ g},str | vedclass

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(B) Given that:Number of balls $N = 1000 = 10^3$Mass of each ball $m = 1 	ext{ g} = 10^{-3} 	ext{ kg}$Area $A = 1 	ext{ cm}^2 = 10^{-4} 	ext{ m}^2

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$10^6 	ext{ N/m}^2$

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(B) Given that:Number of balls $N = 1000 = 10^3$Mass of each ball $m = 1 	ext{ g} = 10^{-3} 	ext{ kg}$Area $A = 1 	ext{ cm}^2 = 10^{-4} 	ext{ m}^2$Velocity $v = 50 	ext{ m/s}$The change in momentum for each collision is $\Delta p = m[v - (-v)] = 2mv$.Since there are $N$ collisions per second,the total force $F$ exerted on the surface is:$F = N 	imes \Delta p = N 	imes 2mv$$F = 10^3 	imes 2 	imes 10^{-3} 	ext{ kg} 	imes 50 	ext{ m/s} = 100 	ext{ N}$.Pressure $P$ is defined as force per unit area:$P = \frac{F}{A} = \frac{100 	ext{ N}}{10^{-4} 	ext{ m}^2} = 10^6 	ext{ N/m}^2$.

$1000$ small balls,each weighing $1 \text{ g}$,strike $1 \text{ cm}^2$ of area per second with a velocity of $50 \text{ m/s}$ in a normal direction and rebound with the same velocity. The value of pressure on the surface will be:

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