Question 102: Laws of Motion

Question

Figures (a), (b), (c) and (d) show variation of force with time.
The impulse is highest in figure.

Accepted Answer

As we know that impulse is given by
$$I = \Delta P = F 	imes \Delta t$$
or $ I=$ Area of $f-t$ graph
For fig (a)
$ightarrow \mathrm{I}=\frac{1}{2} 	imes$ base $	imes$ height
$$
=\frac{1}{2} 	imes 0.5 	imes 1=0.25 \mathrm{~N}-\mathrm{sec} .
$$
For fig (b),
$I=$ length $	imes$ width
$$
=2 	imes 0.5=1 \mathrm{~N} 	ext {-sec }
$$
For fig (c),
I $=\frac{1}{2} 	imes$ base $	imes$ height
$$
=\frac{1}{2} 	imes 1 	imes 0.75=0.375 \mathrm{~N} 	ext {-sec. }
$$
For fig (d),
$I=\frac{1}{2} 	imes$ base $	imes$ height
$$
=\frac{1}{2} 	imes 2 	imes 0.5=0.5 \mathrm{~N}-\mathrm{sec} 	ext {. }
$$
Impulse is highest for that figure, whose area under $F$-t is maximum and i.e. figure(b)
Option (D) is correct.

Answer

Fig (c)

Answer

Fig (d)

Answer

Fig (a)

Answer

Fig (b)

Laws of Motion

Figures (a), (b), (c) and (d) show variation of force with time. The impulse is highest in figure.

Choose the correct answer: