Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A body of mass $2 \text{ kg}$ has an initial velocity of $3 \text{ m/s}$ along $OE$ and it is subjected to a force of $4 \text{ N}$ in a direction perpendicular to $OE$ . The distance of the body from $O$ after $4 \text{ seconds}$ will be:

12 m

20 m

8 m

48 m

Step-by-Step Solution

Analyze the Motion: The body has an initial velocity along one direction ( $x$ -axis, along $OE$ ) and is subjected to a constant force in the perpendicular direction ( $y$ -axis). This results in a 2D motion with constant velocity in the $x$ -direction and constant acceleration in the $y$ -direction.
Calculate Acceleration: Using Newton's Second Law ( $F = ma$ ): $a_y = \frac{F}{m} = \frac{4 \text{ N}}{2 \text{ kg}} = 2 \text{ m/s}^2$ The acceleration along $OE$ ( $a_x$ ) is zero since there is no force component in that direction [Source 35].
Calculate Displacements:

Along OE ( $x$ -axis): Uniform velocity motion. $x = u_x t + \frac{1}{2}a_x t^2$ $x = 3 \text{ m/s} \times 4 \text{ s} + 0 = 12 \text{ m}$
Perpendicular to OE ( $y$ -axis): Uniformly accelerated motion (initial velocity $u_y = 0$ ). $y = u_y t + \frac{1}{2}a_y t^2$ $y = 0 + \frac{1}{2} \times 2 \text{ m/s}^2 \times (4 \text{ s})^2 = 16 \text{ m}$ [Source 30]

Calculate Resultant Distance: The distance from the origin $O$ is the magnitude of the resultant displacement vector. $s = \sqrt{x^2 + y^2}$ $s = \sqrt{(12)^2 + (16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \text{ m}$

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