Question

An object with mass 5 kg is acted upon by a force, $\vec{F}=\left(-3\widehat{i}+4\widehat{j}\right)N$. If its initial velocity at t = 0 is $\vec{v}=\left(3\widehat{i}+12\widehat{j}\right)m/s$ , the time at which it will just have a velocity along y-axis is

• A. 2 s
• B. 5 s
• C. 15 s
• D. 10 s

Answer 1

Answer: (D)

10 s

Answer 2

$\vec{F}=\left(-3 \hat{i}+4\hat{j}\right)$
$M \vec{a}=\left(-3 \hat{i}+4\hat{j}\right)$
$M=5 kg$
$\vec{a}=\left(\frac{-3}{5}\hat{i}+\frac{4}{5} \hat{j}\right)\Rightarrow \vec{a}=a_{x} \hat{i}+a_{y} \hat{j}$
$a_{x}=\frac{-3}{5}$
$at \, t=0$
$\vec{V}_{0}=\left(6 \hat{i}-12 \hat{j}\right)\, \Rightarrow\, V_{0 x} \hat{i}+V_{0 y} \hat{j}$
$V_{0 x}=6$
For the body to have velocity along y-axis only, x-component of velocity should be zero
$i.e., V_{x}=0$
$V_{x}=V_{0x}+a_{x t}$
$0=6+\left(\frac{-3}{5}\right)t$
$6=\frac{3}{5} t$
$t=\frac{6\times5}{3}=10 s$
$OR$
$A=\frac{F}{m}=\frac{-3}{5} \hat{i}+\frac{4}{5} \hat{j}$
$u=6 \hat{i}-12 \hat{j} \, m s^{-1}$
$v=u+a t.$
$1 \hat{j}=6\hat{i}-12\hat{j}+\left(\frac{-3}{5} \hat{i}+\frac{4}{5} \hat{j}\right)t$
$=\left(6-\frac{3}{5}t\right)\hat{i}\left(-12+\frac{4}{5}t\right) \hat{j}$
$-12+\frac{4}{5}t \hat{j}=1 \hat{j}$
$-\frac{56}{5} t \hat{j}=1 \hat{j}$
$t=10\, s$