Centre of Mass 3 Question 26

29. 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: 29. $\frac{3}{2} m v^{2}$

Solution:

  1. Impulse $=\int F d t=$ area under $F$ - $t$ graph

$\therefore$ Total impulse from $t=4 \mu s$ to $t=16 \mu s$

$=$ Area $E B C D$

$=$ Area of trapezium $E B C F+$ Area of triangle $F C D$

$=\frac{1}{2}(200+800) 2 \times 10^{-6}+\frac{1}{2} \times 800 \times 10 \times 10^{-6}$

$=5 \times 10^{-3} N-s$



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