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4-1.Newton's Laws of Motion
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A gun applies a force $F$ on a bullet which is given by $F =\left(100-0.5 \times 10^{5} t \right) N$. The bullet emerges out with speed $400 \,m / s$. Then find out the impulse exerted till force on bullet becomes zero. (in $N - s$)
A
$0.2$
B
$0.3$
C
$0.1$
D
$0.4$
(AIIMS-2019)
Solution
Consider the given expression.
$F =\left(100-0.5 \times 10^{5} t \right) N$
Given that, $F =0$
$\left(100-0.5 \times 10^{5} t \right)=0$
$t =\frac{100}{0.5 \times 10^{5}}$
$=2 \times 10^{-3} sec$
Calculate the impulse as,
$I =\int Fdt$
$=\int\left(100-0.5 \times 10^{5} t \right) dt$
$=\left[100 t -\frac{10^{5}}{2} \frac{ t ^{2}}{2}\right]$
$=\left[100\left(2 \times 10^{-3}\right)-\frac{10^{5}}{2} \frac{\left(2 \times 10^{-3}\right)^{2}}{2}\right]$
$=0.1 Ns$
Standard 11
Physics
