Physics 1st Ch 02 Vectors

Physics - First Paper Chapter 2: Vectors Selected Questions and Answers (HSC English Version) Chapter 2: Vectors B. Creative Questions 1. Rafi goes to college everyday by a cycle. While cycling he thinks that he applies force on the paddle vertically downward, but the cycle moves forward. Rafi’s friend Al Amin makes the matter simple. He says that it is possible due to the division of force. The downward force acts by dividing itself into two components. One of the components of the force makes the cycle move in the forward direction. F F cos θ F sin θ R θ (a) What is unit vector? (b) How velocity and acceleration are obtained from position vector? (c) If the force of magnitude of F is divided into two components, one component is equal to the magnitude of the force and makes a right angle with it, then find the magnitude and direction of the other component. Solution : Let the force be F at angle θ with one component. One component = F1 = F (equal to the force), and it is perpendicular to F. Using the parallelogram law: if the resultant is F, one component is F1 = F (at 90◦), then by Pythagoras: F 2 = F 2 1 + F 2 2 =⇒F 2 = F 2 + F 2 2 This gives F2 = 0 - which is impossible physic

(d) By resolving the force F of the stimulus into any two arbitrary directions, compare the two resolved components. Answer: (a) A unit vector is a vector whose magnitude is exactly 1 (dimensionless). It indicates direction only. Any vector⃗A can be expressed as⃗A = A ˆA, where ˆA =⃗A/|⃗A| is the unit vector in the direction of⃗A. (b) If⃗r(t) is the position vector of a particle, then: Velocity:⃗v = d⃗r dt ; Acceleration:⃗a = d⃗v dt = d2⃗r dt2 . Velocity is the first time-derivative of position and acceleration is the second derivative. Both⃗ v and⃗a are vector quantities. (c) (d) Let the force F be resolved at angle θ to the horizontal (along the pedal arm, as in the figure). The two rectangular components are: F∥= F cos θ (along the arm - effective component driving the crank) F⊥= F sin θ (perpendicular to the arm - ineffective, bends the axle) Comparison: F∥/F⊥= cos θ/ sin θ = cot θ. For θ = 30◦: F∥= F cos 30◦≈0.866F and F⊥= 0.5F, so the effective component is larger. As θ →0◦(force applied along the arm), F∥→F and F⊥→0; as θ →90◦, F∥→0. The productive component driving the cycle forward equals F cos θ - it is always less than or equal to the applied force, with equality only wh

(d) Will the normal unit vector on planes⃗OX,⃗OY be same? Give necessary reason with mathematical analysis. Answer: (a) Curl of a vector field⃗F is defined as⃗∇×⃗F. It is a vector quantity that measures the rotational tendency (circulation per unit area) of the field at a point. If⃗∇×⃗F = 0, the field is called irrotational. (b) When a body is projected upward (or has a vertical component of velocity), the gravitational force⃗g acts vertically downward. This force continuously decelerates the vertical component of motion. By Newton’s second law, a = −g in the vertical direction, so vy = vy0−gt decreases with time, becoming zero at maximum height. The horizontal component is unaffected (no horizontal force, neglecting air resistance). (c) (d) The unit normal to the plane of⃗OX and⃗OY can be found from⃗OX ×⃗OY and⃗OY ×⃗OX.⃗ OX ×⃗OY = ˆi ˆj ˆk 3 2 1 −2 1 4 = ˆi(8 −1) −ˆj(12 + 2) + ˆk(3 + 4) = 7ˆi −14ˆj + 7ˆk |⃗OX ×⃗OY | = √ 49 + 196 + 49 = √ 294 = 7 √ 6 Unit normal: ˆn1 = 7ˆi −14ˆj + 7ˆk 7 √ 6 = ˆi −2ˆj + ˆk √ 6 Since⃗OY ×⃗OX = −(⃗OX ×⃗OY ), the unit normal in the opposite direction is ˆn2 = −ˆn1. Conclusion: The normal unit vectors on planes⃗OX–⃗OY and⃗OY –⃗OX have the same magnitude