Laws Of Motion Question 47
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 newtons and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet[CBSE PMT 1998]
Options:
A) 9 Ns
B) Zero
C) 0.9 Ns
D) 1.8 Ns
Show Answer
Answer:
Correct Answer: C
Solution:
$ F=600-2\times 10^{5}t=0 $
therefore $ t=3\times {{10}^{-3}}\sec $
Impulse $ I=\int _{0}^{t}{Fdt}=\int _{0}^{3\times {{10}^{-3}}}{(600-2\times 10^{3}t)dt} $
$ =[600t-10^{5}t^{2}] _{0}^{3\times {{10}^{-3}}}=0.9N\times \sec $
BETA
