A particle of mass 2 kg moves with an initial velocity of (4i^+2j^) m s−1 on the x-y plane. A force F→=(2i^−8j^)N acts on the particle. The initial position of the particle is 2 m, 3 m. Then for y = 3 m,

u→=4i^+2j^, a→=F→m=i^−4j^Let at any time, the coordinates be (x, y). x−2=uxt+12axt2⇒ x−2=4t+12t2 and y−3=2t−124t2⇒ y−3=2t−2t2When y=3 m, t=0, 1 s; when t=0, x=2 mWhen t=1 s, x=6.5 m

A particle of mass 2 kg moves with an initial velocity of (4i^+2j^) m s−1 on the x-y plane. A force F→=(2i^−8j^)N acts on the particle. The initial position of the particle is 2 m, 3 m. Then for y = 3 m,

u→=4i^+2j^, a→=F→m=i^−4j^Let at any time, the coordinates be (x, y). x−2=uxt+12axt2⇒ x−2=4t+12t2 and y−3=2t−124t2⇒ y−3=2t−2t2When y=3 m, t=0, 1 s; when t=0, x=2 mWhen t=1 s, x=6.5 m

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A particle of mass 2 kg moves with an initial velocity of $(4 \hat{i} + 2 \hat{j}) m s^{- 1}$ on the x-y plane. A force $\vec{F} = (2 \hat{i} - 8 \hat{j}) N$ acts on the particle. The initial position of the particle is $(2 m, 3 m)$ . Then for y = 3 m,

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Possible value of x is only $x = 2 m$

Time taken is $2 s$

Possible value of x is not only $x = 2 m$ , but there exists some other value of x also

All of the above

answer is B.

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Detailed Solution

$\vec{u} = 4 \hat{i} + 2 \hat{j}, \vec{a} = \frac{\vec{F}}{m} = \hat{i} - 4 \hat{j}$

Let at any time, the coordinates be (x, y).

$\begin{array}{l} x - 2 = u_{x} t + \frac{1}{2} a_{x} t^{2} \\ \Rightarrow x - 2 = 4 t + \frac{1}{2} t^{2} and y - 3 = 2 t - \frac{1}{2} 4 t^{2} \\ \Rightarrow y - 3 = 2 t - 2 t^{2} \end{array}$