Q134 / 180 · KCET 2019 · Physics

An object with mass \( 5 \mathrm{~kg} \) is acted upon by a force, \( \vec{F}=(-3 \hat{i}+4 \hat{j}) \mathrm{N} \). If its initial velocity at
\( \mathrm{t}=0 \) is \( \vec{v}=(6 \hat{i}-12 \hat{j}) \mathrm{ms}^{-1} \), the time at which it will just have a velocity along \( \mathrm{y} \)-axis is

  • A
    \( 02 s \)
  • B
    \( 05 s \)
  • C
    \( 15 s \)
  • Correct answer
    D
    \( 10 s \)

Solution

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

Hence, the correct answer is option: 'D'.

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