Question

The figure shows the position-time (x-t) graph of one-dimensional motion of a particle of mass 0.4 kg. Find the magnitude of impulse imparted to the particle at $t=2\text{s}\text{and}4\text{s}$ respectively.

• A. $0.2\text{Ns}$ and $0.4\text{Ns}$
• B. $0.4\text{Ns}$ and $0.4\text{Ns}$
• C. $0.8\text{Ns}$ and $0.4\text{Ns}$
• D. $0.8\text{Ns}$ and $0.8\text{Ns}$

Answer 1

The correct option is D $0.8\text{Ns}$ and $0.8\text{Ns}$


As we know velocity = slope of $(x-t)$ graph.
$\therefore$ From graph
For$(0<t<2)\text{s}$ - particle has uniform velocity of $v_i=1\text{m/s}$.
For$(2<t<4)\text{s}$ - particle has uniform velocity of $v_f=-1\text{m/s}$.
$\therefore$At $t=2\text{s}$ velocity of particle is changing.

Using
Impulse $\left(|J|\right)=|\Delta P|=m|(v_f-v_i)$|
$\therefore$ magnitude of impulse at $t=2\text{s}$$=|0.4(-1-1)|=0.8\text{Ns}$

Simillarly,
For$(2<t<4)\text{s}$ - particle has uniform velocity of $v_i=-1\text{m/s}$.
For$(4<t<6)\text{s}$ - particle has uniform velocity of $v_f=1\text{m/s}$.
$\therefore$At $t=4\text{s}$ velocity of particle is changing.
Using
Impulse $\left(|J|\right)=|\Delta P|=m|(v_f-v_i)$|
$\therefore$ magnitude of impulse at $t=4\text{s}$$=|0.4(1-(-1\left)\right)|=0.8\text{Ns}$