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4-1.Newton's Laws of Motion
hard
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 ........... $N-s$
A
$9$
B
$0$
C
$0.9$
D
$1.8$
(AIPMT-1998)
Solution
(c)$F = 600 – 2 \times {10^5}t = 0$
$⇒$ $t = 3 \times {10^{ – 3}}\,\sec $
Impulse $I = \int_0^t {F\,dt} = \int_0^{3 \times {{10}^{ – 3}}} {(600 – 2 \times {{10}^3}t)dt} $ $ = [600t – {10^5}{t^2}]_0^{3 \times {{10}^{ – 3}}} = 0.9\,N \sec $
Standard 11
Physics
