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Motion in a Lane MCQs for NEET — Physics Questions with Answers

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A vector $\vec{A}$ is directed along $30 °$ west of north direction and another vector $\vec{B}$ along $15 °$ south of east. Their resultant cannot be in ____________ direction.

North

East

North-East

South

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Explanation

The resultant of two vectors A and B will be in a direction between the directions of A and B. Since A is directed 30Â° west of north and B is directed 15Â° south of east, their resultant cannot be in the south direction.

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ABCD is a quadrilateral. Forces $\vec{BA}, \vec{BC}, \vec{CD} & \vec{DA}$ act at a point. Their resultant is

$2 \vec{AB}$

$2 \vec{DA}$

zero vector

$2 \vec{BA}$

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The resultant of the forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$ . If $\vec{Q}$ is double then the resultant also doubles in magnitude. Find the angle between $\vec{P}$ and $\vec{Q}$ .

$\cos θ = \frac{Q}{2 P}$

$\cos θ = \frac{- 4 Q}{3 P}$

$\cos θ = \frac{- 2 Q}{3 P}$

$\cos θ = \frac{- 3 P}{4 Q}$

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Explanation

(D)

$Resultant of \vec{P} & \vec{Q} is \vec{R} R^{2} = P^{2} + Q^{2} + 2 PQ cosθ . . . . . (i) If Q is doubled, resultant is 2 R {(2 R)}^{2} = P^{2} + {(2 Q)}^{2} + 2 P (2 Q) cosθ . . . . . . (ii) From (i) & (ii) 4 (P^{2} + Q^{2} + 2 PQ cosθ) = P^{2} + 4 Q^{2} + 4 PQ cosθ 4 P^{2} + 8 PQ cosθ = P^{2} + 4 QP cosθ 4 Q cosθ = - 3 P cosθ = \frac{- 3 P}{4 Q}$

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The maximum and minimum magnitude of the resultant of two vectors are 17 units and 7 units respectively. Then the magnitude of resultant of the vectors when they act perpendicular to each other is:

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Explanation

When two vectors are perpendicular, their resultant magnitude is given by the Pythagorean theorem: âˆš(a^2 + b^2), where a and b are the magnitudes of the vectors. If the maximum and minimum resultants are 17 and 7 units, then âˆš(17^2 + 7^2) = 13 units.

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A vector $\vec{A}$ makes an angle of $20 °$ and $\vec{B}$ makes an angle of $110 °$ with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the magnitude of the resultant.

$1$

$5 \sqrt{2}$

$5$

$7$

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Explanation

Angle between

$\vec{A} a n d \vec{B} = 90 ° R = \sqrt{A^{2} + B^{2}} = \sqrt{{(3)}^{2} + {(4)}^{2}} = 5 m$

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A particle is moving westward with a velocity ${\vec{v}}_{1} = 5 m / s .$ Its velocity changed to ${\vec{v}}_{2} = 5 m / s$ northward. The change in velocity vector $(∆ \vec{V} = {\vec{v}}_{2} - {\vec{v}}_{1})$ is:

$5 \sqrt{2} m / s$ towards north east

$5 m / s$ towards north west

zero

$5 \sqrt{2} m / s$ towards north west

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Consider east as positive x-axis, north as positive y-axis and vertically upward direction as z-axis. A helicopter first rises up to an altitude of 100 m than flies straight in north 500 m and then suddenly takes a turn towards east and travels 1000 m east. What is position vector of helicopter. (Take starting point as origin)

$1000 \hat{i} - 500 \hat{j} + 100 \hat{k}$

$1000 \hat{i} + 500 \hat{j} - 100 \hat{k}$

$1000 \hat{i} + 500 \hat{j} + 100 \hat{k}$

$- 1000 \hat{i} + 500 \hat{j} + 100 \hat{k}$

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Explanation

Position of helicopter= $100 \hat{k} + 500 \hat{j} + 1000 \hat{i}$

= $1000 \hat{i} + 500 \hat{j} + 100 \hat{k}$

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If vector $\vec{P}, \vec{Q}$ and $\vec{R}$ have magnitude 5, 12 and 13 units and $\vec{P} + \vec{Q} = \vec{R}$ the angle between $\vec{Q}$ and $\vec{R}$ is -

$\cos^{- 1} \frac{5}{12}$

$\cos^{- 1} \frac{5}{13}$

$\cos^{- 1} \frac{12}{13}$

$\cos^{- 1} \frac{2}{13}$

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Coordinates of a moving particle are given by $x = {ct}^{2}$ and $y = {bt}^{2}$ . The speed of the particle is given by

$2 t (c + b)$

$2 t \sqrt{c^{2} - b^{2}}$

$t \sqrt{c^{2} + b^{2}}$

$2 t \sqrt{c^{2} + b^{2}}$

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Explanation

$\vec{v} = \overset{}{v_{x} \hat{i} + v_{y} \hat{j}} S p e e d |\vec{v}| = \sqrt{{v^{2}}_{x} + {v_{y}}^{2}} \vec{v} = \frac{d x}{d t} \hat{i} + \frac{d y}{d t} \hat{j} = 2 c t \hat{i} + 2 b t \hat{j} ∴ |\vec{v}| = \sqrt{{(2 c t)}^{2} + {(2 b t)}^{2}} = 2 t \sqrt{c^{2} + b^{2}}$

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The x and y components of vector $\vec{A}$ are 4m and 6m respectively. The x, y components of vector $\vec{A} + \vec{B}$ are 10m and 9m respectively. The length of $\vec{B}$ is ______ and angle that $\vec{B}$ makes with the x axis is given by _______.

$\sqrt{0.45} m, \tan^{- 1} (1 / 2)$

$\sqrt{4.5} m, \tan^{- 1} (1 / 2)$

$\sqrt{45} m, \tan^{- 1} (1 / 2)$

$4 \sqrt{5} m, \tan^{- 1} (1 / 2)$

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