A bullet is fired from a gun, the force on the bullet is given by F=600−2×105t 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 a bullet?

 Force acting on the bullet is given as

⇒ F=600−2×105t

⇒ t=600/2×105 

t = 3×10−3s

Now impulse imparted by the bullet is I=

So, we get ⇒I=|600t−2×105t2/2|3×10−3

 ∴ I = 0.9Ns

Getting Info...
Cookie Consent
We serve cookies on this site to analyze traffic, remember your preferences, and optimize your experience.
Oops!
It seems there is something wrong with your internet connection. Please connect to the internet and start browsing again.
AdBlock Detected!
We have detected that you are using adblocking plugin in your browser.
The revenue we earn by the advertisements is used to manage this website, we request you to whitelist our website in your adblocking plugin.
Site is Blocked
Sorry! This site is not available in your country.