A) 9Ns
B) Zero
C) 0.9 Ns
D) 1.8 Ns
Correct Answer: C
Solution :
\[F=600-2\times {{10}^{5}}t\] At \[t=0,\] \[F=600N\] As\[F=0,\]on leaving the barrel, \[\therefore \] \[0=600-2\times {{10}^{5}}t\] This is the time spent by the bullet in the barrel. Average force\[=\frac{600+0}{2}=300\text{ }N\] \[\therefore \] Average impulse imparted \[=F\times t\] \[=300\times 3\times {{10}^{-3}}=0.9Ns\]You need to login to perform this action.
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