A) 8 Ns
B) Zero
C) 0.9 Ns
D) 1.8 Ns
Correct Answer: C
Solution :
We have given, \[F=600-2\times {{10}^{5}}\,J\] At the bullet leaves the barrel, the force on the bullet becomes zero. So, \[600-2\times {{10}^{5}}\,t=0\] \[\Rightarrow t=\frac{600}{2\times {{10}^{5}}}=3\times {{10}^{-3}}\,s\] Then, average impulse imparted to the bullet \[I=\int\limits_{0}^{t}{\,Fdt}\] \[\int_{0}^{3\times {{10}^{-3}}}{(600-2\times {{10}^{5}}\,t)\,dt}\] \[=\left[ 600\,t-\frac{2\times {{10}^{5}}\,{{t}^{2}}}{2} \right]_{0}^{3\times {{10}^{-3}}}\] \[=600\times 3\times {{10}^{-3}}-{{10}^{5}}\times {{(3\times {{10}^{-3}})}^{2}}\] = 1.8 - 0.9 = 0.9 Ns Alternative: As obtained in previous method, the time taken by bullet when it leaves the barrel \[t=3\times {{10}^{-3}}\,s\] Let \[{{F}_{1}}\] and \[{{F}_{2}}\] denote the force at time of firing of bullets i.e., at \[t=0\] and at the time of leaving the bullet i.e., at \[t=3\times {{10}^{-3}}\,s\]. \[{{F}_{1}}=600-2\times {{10}^{5}}\times 0=600\,N\] \[{{F}_{2}}=600-2\times {{10}^{5}}\times 3\times {{10}^{-3}}=0\] Mean value of force \[F=\frac{1}{2}({{F}_{1}}+{{F}_{2}})=\frac{600+0}{2}=300\,N\] Thus, impulse \[=F\times t\] \[=300\times 3\times {{10}^{-3}}\] = 0.9 Ns
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