Centre Of Mass Ques 50
- The magnitude of the force (in Newtons) acting on a body varies with time $t$ (in microseconds) as shown in the figure. $A B, B C$ and $C D$ are straight line segments. The magnitude of the total impulse of the force on the body from $t=4 \mu s$ to $t=16 \mu s$ is .. N-s.
(1994, 2M)
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Answer:
Correct Answer: 50.$\frac{3}{2} m v^{2}$
Solution:
Formula:
- Impulse $=\int F d t=$ area under $F$ - $t$ graph
$\therefore$ Total impulse from $t=4 \mu s$ to $t=16 \mu s$
$=$ Area $EBCD$
$=$ Area of trapezium $E B C F+$ Area of triangle $F C D$
$=\frac{1}{2}(200+800) \times 2 \times 10^{-6}+\frac{1}{2} \times 800 \times 10 \times 10^{-6}$
$=5 \times 10^{-3} N-s$
BETA
