Laws Of Motion Question 47

Question: A bullet is fired from a gun. The force on the bullet is given by F=6002×105t , 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=6002×105t=0

therefore t=3×103sec

Impulse I=0tFdt=03×103(6002×103t)dt

=[600t105t2]03×103=0.9N×sec



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