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 $



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