Laws Of Motion Question 297
Question: . A bullet is fired from a gun. The force on the bullet is given by $ F=600-2\times 10^{5}t $ where. F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?
Options:
A) 1.8 N-s
B) zero
C) 9 N-s
D) 0.9 N-s
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Given $ F=600-( 2\times 10^{5}t ) $
The force is zero at time t, given by 0 $ =600-2\times 10^{5}t $
$ \Rightarrow t=\frac{600}{2\times 10^{5}}=3\times {{10}^{-3}}\text{ seconds} $
$ \therefore Impulse=\int _{0}^{t}{Fdt=\int _{0}^{3\times {{10}^{-3}}}{( 600-2\times 10^{5}t )dt}} $
$ =[ 600t-\frac{2\times 10^{5}t^{2}}{2} ] _{0}^{3\times {{10}^{-3}}} $
$ =600\times 3\times {{10}^{-3}}-10^{5}{{( 3\times {{10}^{-3}} )}^{2}} $
$ =1.8-0.9=0.9Ns $
