Question

The force vs time graph of an object of mass $1kg$ which is initially at rest is shown below. What will be the graph of momentum vs time for the given graph?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

$\Delta \vec{P}=\int \vec{F}dt$
For $t\le 0.05s,F=\frac{5}{0.05}t\Rightarrow F=100t$
For $0.05s\le t\le 0.1s,F=-100t+10$
For $0$ to $0.05s$,
$P_2-P_1=\underset{0}{\overset{0.05}{\int }}100tdt\Rightarrow P_2=100\left(\frac{0.{\left(05\right)}^2}{2}\right)$
$\Rightarrow P_2=50\times {\left(\frac{1}{20}\right)}^2\Rightarrow P_2=\frac{50}{400}$
$\Rightarrow P_2=\frac{1}{8}\Rightarrow P_2=0.125Ns$
For $0.05s$ to $0.1s$,
$P_2-0.125=\underset{0.05}{\overset{0.1}{\int }}\left(-100t+10\right)dt$
$\Rightarrow P_2-0.125={\left[-100\left(\frac{t^2}{2}\right)+10t\right]}_{0.05}^{0.1}$
$\Rightarrow P_2-0.125=-50\left[\frac{1}{100}-\frac{1}{400}\right]+10\left[\frac{1}{10}-\frac{1}{20}\right]$
$\Rightarrow P_2-0.125=-50\left[\frac{4-1}{400}\right]+\frac{1}{2}$
$\Rightarrow P_2-0.125=-50\times \frac{3}{400}+\frac{1}{2}$
$\Rightarrow P_2-0.125=-\frac{3}{8}+\frac{1}{2}$
$\Rightarrow P_2-0.125=\frac{-3+4}{8}$
$\Rightarrow P_2-0.125=0.125\Rightarrow P_2=0.25kgm{s}^{-1}$