Laws of Motion - Result Question 10
10. 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 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 the bullet?
[1998]
(a) $1.8 N-s$
(b) zero
(c) $9 N-s$
(d) $0.9 N-s$
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Answer:
Correct Answer: 10. (d)
Solution:
- (d) Given $F=600-(2 \times 10^{5} t)$
The force is zero at time $t$, given by $0=600-2 \times 10^{5} t$
$\Rightarrow t=\frac{600}{2 \times 10^{5}}=3 \times 10^{-3}$ seconds
$\therefore$ Impulse $=\int_0^{t} F d t=\int_0^{3 \times 10^{-3}}(600-2 \times 10^{5} t) d t$
$=[600 t-\frac{2 \times 10^{5} t^{2}}{2}]_0^{3 \times 10^{-3}}$
$ \begin{aligned} & =600 \times 3 \times 10^{-3}-10^{5}(3 \times 10^{-3})^{2} \\ & =1.8-0.9=0.9 Ns \end{aligned} $
In case of impulse or motion of a charged particle in an alternating electric field, force is time dependent.
