4-1.Newton's Laws of Motion
medium

The position-time graph of a body of mass $2\, kg$ is as given in figure. What is the impulse on the body at $t = 0\, s$ and $t = 4\, s$.

A$0.5\,N s$
 
B$2.5\,N s$
C$1.5\,N s$
D$5\,N s$

Solution

Mass of body $=2 \mathrm{~kg}$
From graph given at $x=0$ body is at rest $\mathrm{s}=0, v=0$
$\therefore$ impulse of force $=0$
In time interval $t=0$ to $t=4 \mathrm{sec}$
$x \rightarrow t$ graph is a straight line means body moves with constant velocity.
$v=\frac{\Delta x}{\Delta t}=\frac{3-0}{4-0}=\frac{3}{4} \mathrm{~m} / \mathrm{s}$
Impulse at $t=5$ second $=$ change in momentum
$=m(v-u)$
$=2\left(0-\frac{3}{4}\right)$
$=-\frac{3}{2} \mathrm{~kg}-\mathrm{m} / \mathrm{s}=-1.5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
Standard 11
Physics

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