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 |
| \[I=\int\limits_{0}^{t}{\,Fdt}\] |
| \[\int_{0}^{3\times {{10}^{-3}}}{(600-2\times {{10}^{5}}\,t)\,dt}\] |
| \[=\left[ 600\,t-\frac{2\times {{10}^{5}}\,{{t}^{2}}}{2} \right]_{0}^{3\times {{10}^{-3}}}\] |
| \[=600\times 3\times {{10}^{-3}}-{{10}^{5}}\times {{(3\times {{10}^{-3}})}^{2}}\] |
| \[=1.8-0.9=0.9\text{ }Ns\] |
| 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 \[=F\times t\] |
| \[=300\times 3\times {{10}^{-3}}\] |
| \[=\text{ }0.9\text{ }Ns\] |
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