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Kinematics - Motion in a Plane

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Motion in a plane is two-dimensional motion, where position is defined by two coordinates (usually xx and yy) and requires vectors for complete description.

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A vector A⃗\vec{A} can be resolved into two rectangular components: Ax=Acos⁡θA_x = A \cos \theta along the x-axis and Ay=Asin⁡θA_y = A \sin \theta along the y-axis.

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Projectile motion is a form of motion where an object is thrown near the earth's surface and moves along a curved path under the action of gravity only. The horizontal component of velocity (ucos⁡θu \cos \theta) remains constant if air resistance is neglected.

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The trajectory of a projectile is a parabola, described by the equation relating the vertical displacement yy to the horizontal displacement xx.

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Uniform Circular Motion occurs when a particle moves in a circle with constant speed. Although speed is constant, velocity changes due to continuous change in direction, resulting in centripetal acceleration (aca_c) directed towards the center.

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Relative velocity in two dimensions is calculated using vector subtraction: v⃗AB=v⃗A−v⃗B\vec{v}_{AB} = \vec{v}_A - \vec{v}_B.

📐Formulae

R⃗=A2+B2+2ABcos⁡θ\vec{R} = \sqrt{A^2 + B^2 + 2AB \cos \theta}

tan⁡α=Bsin⁡θA+Bcos⁡θ\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}

T=2usin⁡θgT = \frac{2u \sin \theta}{g}

Hmax=u2sin⁡2θ2gH_{max} = \frac{u^2 \sin^2 \theta}{2g}

R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}

y=xtan⁡θ−gx22u2cos⁡2θy = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}

v⃗=dr⃗dt=dxdti^+dydtj^\vec{v} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}

ac=v2r=ω2ra_c = \frac{v^2}{r} = \omega^2 r

ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f

💡Examples

Problem 1:

A projectile is fired with an initial velocity of 40 m/s40 \, m/s at an angle of 30∘30^\circ with the horizontal. Calculate the Time of Flight and the Maximum Height reached. (Take g=10 m/s2g = 10 \, m/s^2)

Solution:

Given: u=40 m/su = 40 \, m/s, θ=30∘\theta = 30^\circ, g=10 m/s2g = 10 \, m/s^2.

  1. Time of Flight: T=2usin⁡θg=2×40×sin⁡30∘10=80×0.510=4 sT = \frac{2u \sin \theta}{g} = \frac{2 \times 40 \times \sin 30^\circ}{10} = \frac{80 \times 0.5}{10} = 4 \, s.
  2. Maximum Height: H=u2sin⁡2θ2g=402×(sin⁡30∘)22×10=1600×0.2520=40020=20 mH = \frac{u^2 \sin^2 \theta}{2g} = \frac{40^2 \times (\sin 30^\circ)^2}{2 \times 10} = \frac{1600 \times 0.25}{20} = \frac{400}{20} = 20 \, m.

Explanation:

The time of flight depends on the vertical component of the initial velocity, while the maximum height is reached when the vertical velocity component becomes zero.

Problem 2:

A cyclist is riding at 10 m/s10 \, m/s on a circular track of radius 40 m40 \, m. Find the magnitude of his centripetal acceleration.

Solution:

Given: v=10 m/sv = 10 \, m/s, r=40 mr = 40 \, m. Centripetal acceleration ac=v2r=10240=10040=2.5 m/s2a_c = \frac{v^2}{r} = \frac{10^2}{40} = \frac{100}{40} = 2.5 \, m/s^2.

Explanation:

In uniform circular motion, the acceleration is always directed towards the center and is proportional to the square of the speed and inversely proportional to the radius.