Question

A particle of mass $2\text{kg}$ initially at origin $(0,0)$ moves with an initial velocity $\overrightarrow{V}=4\hat{i}+4\hat{j}\text{m/s}$. If a constant force $\overrightarrow{F}=-20\hat{j}\text{N}$ is applied on the particle, the $x-$coordinate of the particle when its $y-$coordinate again becomes zero is

• A. $1.2\text{m}$
• B. $4.0\text{m}$
• C. $6\text{m}$
• D. $3.2\text{m}$

Answer 1

The correct option is D $3.2\text{m}$

We know that,
$\overrightarrow{F}=m\overrightarrow{a}$
$\Rightarrow \overrightarrow{a}=\frac{-20\hat{j}}{2}=-10\hat{j}{\text{m/s}}^2$
Since, force is in only $y-$direction, therefore acceleration will be in $y-$direction.
and we have, $V_y=4\text{m/s}$
$\therefore$ Applying equation of motion in $y-$ direction, we get
$y=ut+\frac{1}{2}a{t}^2$
$0=4\times t-\frac{1}{2}\times 10\times t^2$
$t=\frac{4}{5}\text{s}$
Now applying equation of motion in $x-$direction
$x=4\times t=4\times \frac{4}{5}=\frac{16}{5}$
$x=3.2\text{m}$