Question

A bullet is fired from a gun .The force on the bullet is given by F=600-2X105t,where F is in newton and t is in second.The force on the bullet becomes zero as soon as it leaves the barrel.What is the impulse imparted to the bullet?

Answer 1

Answer:

The force imparted on the bullet is 0.9 Ns

Explanation:

The force given is

$F=600-2\times 10^5t$

If the force is zero then

$0=600-2\times 10^5t$

$\implies 2\times 10^5t=600$

$\implies t=300\times 10^{-5}$

$\implies t=0.003$ seconds

The impulse due to force F is given by

$\boxed{I=\int Fdt}$

Therefore, Impulse imparted to the bullet

$=\int_0^{0.003} Fdt$

$=\int_0^{0.003} [600-2\times 10^5t]dt$

$=\int_0^{0.003} 600dt-2\times 10^5\int_0^{0.003}tdt$

$=600\int_0^{0.003}dt-2\times 10^5\int_0^{0.003}tdt$

$=600t\Bigr|_0^{0.003}-2\times 10^5(t^2/2)\Bigr|_0^{0.003}$

$=1.8-0.9$

$=0.9$ Ns

Answer 2

Answer:

I hope this is the correct answer ☺️