[tex]\large\bf{\underline{Answer:}}[/tex]
[tex]\large\bf{a) \triangle p_{1s} = - 120 \: kgm {s}^{ - 1} }[/tex]
[tex]\large\bf{b) F = -120N}[/tex]
[tex]\large\bf{c) Pressure=40.10\times 10^5 pa }[/tex]
__________________________________________
[tex]\large\bf{\underline{In\: this\: problem\:we\:have:}}[/tex]
β To find the change in momentum for bullets , we need to remember the momentum p of a bullet is equal to product of mass and speed
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌp_{1}= mv}[/tex]
β This means , that change in momentum for one bullet will be equal to
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle P_{1} = mv_{f} - mv_{i}}[/tex]
[tex]\large\bf{where\:v_{f}=0}[/tex]
Total change in momentum for the bullet in 10 sec is equal to product of change in momentum for one bullet and number of bullets hit the wall in 10 sec
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle P_{10s} = N\triangle P_{i}}[/tex]
Change in momentum given is the change of momentum in 10 sec is 10 times less
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p_{1s} = \frac{N\triangle p_{i}}{10}}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p_{1s}=\frac{200.(mv_{f}-mv_{i}}{10}}[/tex]
[tex]\large\bf{as\:said,v_{f}=0}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p_{1s}=\frac{-200.mv_{i}}{10}}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p_{1s}=\frac{200.5\times 10^{-3}kg.1200ms^{-1}}{10}}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p_{1s}=-1200\:Kgms^{-1}}[/tex]
__________________________________________
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌF =\frac{\triangle P}{\triangle t}}[/tex]
Total change of momentum of bullets in 10 sec
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p =N\triangle p_{i} }[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p= N(mv_{f}-mv_{i})}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p= -N mv_{i}}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p= -200.5\times 10^{-3}.1200ms^{-1}}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌ\triangle p= -1200kgms^{-1}}[/tex]
β We can find total force exerted in the wall in 10sec by dividing the momentum of bullet with 10 sec
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌF = \frac{\triangle p}{\triangle t}}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌF = \frac{-1200}{10}}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌF = -120N}[/tex]
__________________________________________
[tex]\large\bf{area = 3\times 10^{-4}}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌP=\frac{|F|}{A} }[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌP=\frac{-120}{3\times 10^{-4}}}[/tex]
β βββ β βββ β βββ β βββ β βββ β βββ β βββ β β[tex]\large\bf{βΌP=40\times 10^4 Pa}[/tex]
