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NEET PHYSICSLAWS OF MOTIONMedium

Question

A body of mass 2 kg2 \text{ kg} has an initial velocity of 3 m/s3 \text{ m/s} along OEOE and it is subjected to a force of 4 N4 \text{ N} in a direction perpendicular to OEOE. The distance of the body from OO after 4 seconds4 \text{ seconds} will be:

A

12 m

B

20 m

C

8 m

D

48 m

Step-by-Step Solution

  1. Analyze the Motion: The body has an initial velocity along one direction (xx-axis, along OEOE) and is subjected to a constant force in the perpendicular direction (yy-axis). This results in a 2D motion with constant velocity in the xx-direction and constant acceleration in the yy-direction.

  2. Calculate Acceleration: Using Newton's Second Law (F=maF = ma): ay=Fm=4 N2 kg=2 m/s2a_y = \frac{F}{m} = \frac{4 \text{ N}}{2 \text{ kg}} = 2 \text{ m/s}^2 The acceleration along OEOE (axa_x) is zero since there is no force component in that direction [Source 35].

  3. Calculate Displacements:

  • Along OE (xx-axis): Uniform velocity motion. x=uxt+12axt2x = u_x t + \frac{1}{2}a_x t^2 x=3 m/s×4 s+0=12 mx = 3 \text{ m/s} \times 4 \text{ s} + 0 = 12 \text{ m}
  • Perpendicular to OE (yy-axis): Uniformly accelerated motion (initial velocity uy=0u_y = 0). y=uyt+12ayt2y = u_y t + \frac{1}{2}a_y t^2 y=0+12×2 m/s2×(4 s)2=16 my = 0 + \frac{1}{2} \times 2 \text{ m/s}^2 \times (4 \text{ s})^2 = 16 \text{ m} [Source 30]
  1. Calculate Resultant Distance: The distance from the origin OO is the magnitude of the resultant displacement vector. s=x2+y2s = \sqrt{x^2 + y^2} s=(12)2+(16)2=144+256=400=20 ms = \sqrt{(12)^2 + (16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \text{ m}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONinitialvelocitysubjecteddirectionperpendicular

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