A particle of mass 1 kg has a velocity of 2 m/sec. A constant force of 2 N acts on the particle for 1 sec in a direction perpendicular to its initial velocity. Find the angle between velocity vector and displacement vector of the particle at the end of 1 sec:
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Question

A particle of mass 1 kg has a velocity of 2 m/sec. A constant force of 2 N acts on the particle for 1 sec in a direction perpendicular to its initial velocity. Find the angle between velocity vector and displacement vector of the particle at the end of 1 sec:

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G1 Physics · Accepted Answer

Take the initial velocity along +x: v⃗₀ = 2î m/s. Since F = 2 N on a 1 kg mass and is perpendicular to v⃗₀, take a⃗ = 2ĵ m/s². After 1 s, v⃗ = 2î + 2ĵ and displacement s⃗ = 2î + 1ĵ. Thus cosφ = (v⃗·s⃗)/(|v⃗||s⃗|) = 6/√40 = 3/√10, so φ ≈ 18.4°, which is not 45°, 90° or 30°. Hence option D.