A force of $(2.6 \hat{\mathbf{i}}+1.6 \hat{\mathbf{j}}) \mathrm{N}$ acts on a body of mass $2 \mathrm{~kg}$.…
A force of $(2.6 \hat{\mathbf{i}}+1.6 \hat{\mathbf{j}}) \mathrm{N}$ acts on a body of mass $2 \mathrm{~kg}$. If the velocity of the body at time, $t=0$ is $(3.6 \hat{\mathbf{i}}-4.8 \hat{\mathbf{j}}) \mathrm{ms}^{-1}$, the time at which the body will just have a velocity along $x$-axis only is
1 s
2 s
3 s
6 s
Solution
Given,
force acts on a body,
$
\mathbf{F}=(2.6 \hat{\mathbf{i}}+1.6 \hat{\mathbf{j}}) \mathrm{N}
$
mass of a body, $m=2 \mathrm{~kg}$
At $t=0$, velocity of the body, $v=(3.6 \hat{\mathbf{i}}-4.8 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$
We know that,
Force $=$ Mass $\times$ Acceleration
$
\therefore
$
or
$
\begin{aligned}
& \mathbf{F}=m \times \mathbf{a} \\
& \mathbf{a}=\frac{\mathbf{F}}{m}
\end{aligned}
$
or
$
\mathbf{a}=\frac{(2.6 \hat{\mathbf{i}}+1.6 \hat{\mathbf{j}})}{2}
$
or
$
\mathbf{a}=(1.3 \hat{\mathbf{i}}+0.8 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^2
$
Now, the velocity vector, $\frac{d \mathbf{v}}{d t}=\mathbf{a}$
$
\int d \mathbf{v}=\int \mathbf{a} d t
$
$
\begin{aligned}
& \text { or } \mathbf{v}=(1.3 \hat{\mathbf{i}}+0.8 \hat{\mathbf{j}}) t+c \\
& \text { at } t=0, c=\mathbf{v}=(3.6 \hat{\mathbf{i}}-4.8 \hat{\mathbf{j}}) \mathrm{ms}^{-1} \\
& \therefore \mathbf{v}=(3.6+1.3 t) \hat{\mathbf{i}}+{ }^{-}(-4.8+0.8 t) \hat{\mathbf{j}} \mathrm{m} / \mathrm{s} \\
& \therefore v_y=0 \text { (because body will just have a velocity } \\
& \text { along } x \text {-axis.) } \\
& -4.8+0.8 t=0 \\
& \Rightarrow \quad 0.8 t=4.8 \\
& \Rightarrow \quad t=\frac{4.8}{0.8}=6 \mathrm{~s} \\
&
\end{aligned}
$
This is the time at which the body will just have velocity along $x$-axis
$
t=6 \mathrm{~s}
$