Search any question & find its solution
Question:
Answered & Verified by Expert
A body of mass $5 \mathrm{~kg}$ starts from the origin with an initial velocity $\mathbf{u}=30 \hat{\mathbf{i}}+40 \hat{\mathbf{j}} \mathrm{ms}^{-1}$. When a constant force $\mathbf{F}=-(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{N}$ acts on the body, the time in which the $y$-component of the velocity becomes zero is
Options:
Solution:
1928 Upvotes
Verified Answer
The correct answer is:
$40 \mathrm{~s}$
Given, $\mathbf{u}=30 \hat{\mathrm{i}}+40 \hat{\mathrm{j}} \mathrm{ms}^{-1}$
Force, $\mathbf{F}=-(\hat{\mathbf{i}}+5 \hat{\mathrm{j}}) \mathrm{N}$
$$
\begin{aligned}
m & =5 \mathrm{~kg} \\
u_y & =40 \mathrm{~ms}^{-1}, F_y=-5 \mathrm{~N}
\end{aligned}
$$
$\therefore$ Acceleration of body along $y$-direction,
$$
a_y=\frac{F_y}{m}=\frac{-5}{5}=-1 \mathrm{~ms}^{-2}
$$
Using equation of straight line motion,
$$
\begin{aligned}
v & =u+a t \\
v_y & =u_y+a t \\
0 & =40+(-1) t \Rightarrow t=40 \mathrm{~s}
\end{aligned}
$$
Force, $\mathbf{F}=-(\hat{\mathbf{i}}+5 \hat{\mathrm{j}}) \mathrm{N}$
$$
\begin{aligned}
m & =5 \mathrm{~kg} \\
u_y & =40 \mathrm{~ms}^{-1}, F_y=-5 \mathrm{~N}
\end{aligned}
$$
$\therefore$ Acceleration of body along $y$-direction,
$$
a_y=\frac{F_y}{m}=\frac{-5}{5}=-1 \mathrm{~ms}^{-2}
$$
Using equation of straight line motion,
$$
\begin{aligned}
v & =u+a t \\
v_y & =u_y+a t \\
0 & =40+(-1) t \Rightarrow t=40 \mathrm{~s}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.