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JEE Mains Physics · Formula sheet

Motion in a Plane formulas

17 formulas and 35 common traps for JEE Mains Physics Motion in a Plane, grouped by subtopic.

Full notes with worked examples

Vectors: Resultant, Components and Products

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Resultant of two vectors at an angle

Resultant of two vectors

R=A2+B2+2ABcos⁡θtan⁡α=Bsin⁡θA+Bcos⁡θ∣A⃗+B⃗∣=2Acos⁡θ2 (A=B)R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta} \qquad \tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} \qquad |\vec{A} + \vec{B}| = 2A\cos\frac{\theta}{2}\ (A = B)

Resolving vectors into components

Components and magnitude

Ax=Acos⁡θ, Ay=Asin⁡θA=Ax2+Ay2+Az2A⃗ along B⃗=AB^A_x = A\cos\theta,\ A_y = A\sin\theta \qquad A = \sqrt{A_x^{2} + A_y^{2} + A_z^{2}} \qquad \vec{A}\ \text{along}\ \vec{B} = A\hat{B}

Dot product, cross product and projection

Products and projection

A⃗⋅B⃗=ABcos⁡θ∣A⃗×B⃗∣=ABsin⁡θprojBA⃗=A⃗⋅B⃗∣B⃗∣\vec{A}\cdot\vec{B} = AB\cos\theta \qquad |\vec{A}\times\vec{B}| = AB\sin\theta \qquad \text{proj}_{B}\vec{A} = \frac{\vec{A}\cdot\vec{B}}{|\vec{B}|}

Common traps

Adding magnitudes

A 3 N and a 4 N force give 7 N only when they point the same way. At right angles the resultant is 5 N; at any other angle use R² = A² + B² + 2AB cos θ.

Perpendicular to A, not to B

If the resultant is perpendicular to A, the condition is A + B cos θ = 0: the part of B along A cancels A. Writing B + A cos θ = 0 makes the resultant perpendicular to B instead.

Angle with the y-axis

When the angle θ is measured from the y-axis, the y component is A cos θ and the x component is A sin θ. Using A cos θ for x by habit swaps the two.

Losing the quadrant

tan α = A_y/A_x gives the same value for (3, 3) and (−3, −3). Look at the signs of the components to place the vector before you quote its angle.

Dividing the projection by the wrong length

The projection of A on B is A·B divided by |B|, the vector you project ON. Dividing by |A| gives the projection of B on A.

Order matters in a cross product

B × A = −(A × B). The size is the same, but the direction flips, which changes the answer whenever a question asks for a direction or a unit vector.

Velocity in a Plane and Relative Velocity

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Position, velocity and acceleration vectors

Motion in two dimensions

v⃗=dr⃗dt, a⃗=dv⃗dtr⃗=r⃗0+u⃗t+12a⃗t2tan⁡θ=vyvx\vec{v} = \frac{d\vec{r}}{dt},\ \vec{a} = \frac{d\vec{v}}{dt} \qquad \vec{r} = \vec{r}_0 + \vec{u}t + \tfrac{1}{2}\vec{a}t^{2} \qquad \tan\theta = \frac{v_y}{v_x}

Crossing a river

Crossing time and drift

tmin⁡=dvxdrift=udvsin⁡α=uv, t=dv2−u2t_{\min} = \frac{d}{v} \qquad x_{drift} = \frac{ud}{v} \qquad \sin\alpha = \frac{u}{v},\ t = \frac{d}{\sqrt{v^{2} - u^{2}}}

Relative velocity and change of frame

Relative velocity

v⃗AB=v⃗A−v⃗Btan⁡θumbrella=vmanvrain (vertical rain)\vec{v}_{AB} = \vec{v}_A - \vec{v}_B \qquad \tan\theta_{umbrella} = \frac{v_{man}}{v_{rain}}\ (\text{vertical rain})

Common traps

Quoting an angle without its axis

A velocity 4i − j m/s is at tan⁻¹(1/4) below the +x axis and at tan⁻¹ 4 from the −y axis. Options often give the right number against the wrong axis.

Force along the motion

The net force points along the acceleration. For r = 2t i + t² j the particle moves diagonally, but the force is along +y only.

Dividing the width by the resultant speed

The crossing time is the width divided by the ACROSS component of the boat's own velocity. The river's flow is along the banks, so it never shortens or lengthens the crossing.

Least time is not shortest path

Heading straight across gives the least time but lands downstream. Landing directly opposite needs an upstream heading and always takes longer.

Subtracting in the wrong order

The rain's velocity relative to the man is v_rain − v_man. Reversing it gives the man's velocity relative to the rain, which points the other way.

Taking a released object to start from rest

An object let go from a moving plane, boat or balloon starts with that vehicle's velocity. Taking it to start from rest gives the wrong range and the wrong path.

Projectile: Time of Flight, Maximum Height and Range

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Time of flight, maximum height and range

Projectile on level ground

T=2usin⁡θgH=u2sin⁡2θ2gR=u2sin⁡2θgT = \frac{2u\sin\theta}{g} \qquad H = \frac{u^{2}\sin^{2}\theta}{2g} \qquad R = \frac{u^{2}\sin 2\theta}{g}

Maximum range and the range-height relation

Greatest range; range against height

Rmax⁡=u2g (45∘)RH=4tan⁡θR_{\max} = \frac{u^{2}}{g}\ (45^{\circ}) \qquad \frac{R}{H} = \frac{4}{\tan\theta}

Complementary angles give the same range

Two angles, one range

R=4H1H2T1T2=2RgH1+H2=u22gR = 4\sqrt{H_1H_2} \qquad T_1T_2 = \frac{2R}{g} \qquad H_1 + H_2 = \frac{u^{2}}{2g}

Common traps

sin 2θ in the range, sin²θ in the height

R = u² sin 2θ/g and H = u² sin²θ/2g. Swapping them is the commonest slip; check with 90°, where the range must be zero and the height u²/2g.

Time of flight does not use sin 2θ

T = 2u sin θ/g comes from the vertical motion alone. Ranges compare as sin 2θ, but times compare as sin θ and heights as sin²θ.

Greatest height equal to greatest range

The greatest height (thrown straight up) is u²/2g; the greatest range (at 45°) is u²/g. The range is twice the height, not equal to it.

45° is only for level ground

Range is largest at 45° only when the ball lands at its launch height. Thrown from a tower or onto a slope, the best angle is different.

Equal range does not mean equal height

θ and 90° − θ land together, but the steeper launch rises higher and stays up longer. Only the range is shared.

Adding the angles to 180°

The two angles that give one range add to 90°, not 180°. It is the doubled angles, 2θ and 180° − 2θ, that add to 180°.

Projectile: Velocity, Trajectory and Projection from a Height

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Velocity and kinetic energy during the flight

Velocity during the flight

vx=ucos⁡θ, vy=usin⁡θ−gtKtop=Kcos⁡2θh=12gt1t2v_x = u\cos\theta,\ v_y = u\sin\theta - gt \qquad K_{top} = K\cos^{2}\theta \qquad h = \tfrac{1}{2}gt_1t_2

Equation of the trajectory

Path of a projectile

y=xtan⁡θ−gx22u2cos⁡2θy=αx−βx2: R=αβ, H=α24βy = x\tan\theta - \frac{gx^{2}}{2u^{2}\cos^{2}\theta} \qquad y = \alpha x - \beta x^{2}:\ R = \frac{\alpha}{\beta},\ H = \frac{\alpha^{2}}{4\beta}

Thrown horizontally from a height

Horizontal projection from height h

t=2hgx=u2hgv=u2+2ght = \sqrt{\frac{2h}{g}} \qquad x = u\sqrt{\frac{2h}{g}} \qquad v = \sqrt{u^{2} + 2gh}

Common traps

Zero speed at the top

Only the vertical velocity is zero at the top. The horizontal velocity u cos θ is still there, so the speed and the kinetic energy are not zero.

cos θ instead of cos²θ

Kinetic energy goes as speed squared. The speed at the top is u cos θ, so the kinetic energy there is K cos²θ: at 37° (cos 37° = 0.8) that is 0.64K, not 0.8K.

Dropping the cos²θ

The x² coefficient is g/(2u² cos²θ) = g/(2uₓ²): it uses the horizontal speed, not the launch speed. Using g/(2u²) gives the wrong speed.

The top is at x = R/2

The greatest height comes halfway across. Setting dy/dx = 0 gives x = α/2β, which is R/2; substituting gives H = α²/4β.

Giving the horizontally thrown body a vertical speed

Thrown horizontally means v_y = 0 at release. The time of fall is √(2h/g), the same as dropping it; a faster throw lands farther out but not sooner.

Horizontal distance is not displacement

The landing point is x out and h down from the release point. The displacement is √(x² + h²); the horizontal distance x alone is a common wrong option.

Uniform Circular Motion: Angular Speed and Acceleration

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Distance, displacement and average velocity on a circle

Arc and chord

s=rθ∣Δr⃗∣=2rsin⁡θ2vvavg=θ2sin⁡(θ/2)s = r\theta \qquad |\Delta\vec{r}| = 2r\sin\frac{\theta}{2} \qquad \frac{v}{v_{avg}} = \frac{\theta}{2\sin(\theta/2)}

Centripetal and tangential acceleration

Circular motion

v=ωrac=v2r=ω2ra=at2+ac2, at=dvdtv = \omega r \qquad a_c = \frac{v^{2}}{r} = \omega^{2}r \qquad a = \sqrt{a_t^{2} + a_c^{2}},\ a_t = \frac{dv}{dt}

Common traps

Average velocity from the arc

Average velocity is displacement over time, and the displacement is the chord 2r sin(θ/2). Dividing the arc length by the time gives the average speed instead.

Degrees in s = rθ

The arc length rθ needs θ in radians. Ninety degrees is π/2, so a quarter turn of radius r is πr/2 long, not 90r.

Constant speed, zero acceleration

In uniform circular motion the speed is constant but the direction is not, so the acceleration is v²/r towards the centre. Only straight-line motion at constant speed has zero acceleration.

Forgetting to convert rpm

ω in rad/s is 2π × (revolutions per second). For N rpm divide by 60 first: 120 rpm is 4π rad/s, not 240π.

Adding the two accelerations directly

Tangential and centripetal accelerations are at right angles. Their total is √(a_t² + a_c²), never a_t + a_c.

Dynamics of Circular Motion: Roads, Strings and Vertical Circles

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Friction and banking on a curve

Roads and turntables

vmax⁡=μrgtan⁡θ=v2rgvmax⁡2=rg tan⁡θ+μ1−μtan⁡θv_{\max} = \sqrt{\mu rg} \qquad \tan\theta = \frac{v^{2}}{rg} \qquad v_{\max}^{2} = rg\,\frac{\tan\theta + \mu}{1 - \mu\tan\theta}

String, spring and wall in a horizontal circle

Horizontal circles

T=mω2Lkx=mω2(l0+x)tan⁡θ=v2rg (conical pendulum)T = m\omega^{2}L \qquad kx = m\omega^{2}(l_0 + x) \qquad \tan\theta = \frac{v^{2}}{rg}\ (\text{conical pendulum})

Motion in a vertical circle

Vertical circle on a string

Tb=mvb2r+mg, Tt=mvt2r−mgvb2=vt2+4grvb≥5grT_b = \frac{mv_b^{2}}{r} + mg,\ T_t = \frac{mv_t^{2}}{r} - mg \qquad v_b^{2} = v_t^{2} + 4gr \qquad v_b \ge \sqrt{5gr}

Common traps

Putting the mass into the safe speed

v_max = √(μrg) has no mass in it: a heavier car needs more force but also gets more friction. A changed mass alone changes nothing.

Friction with the wrong sign on a banked road

At the greatest speed the car tends to slide outwards, so friction acts down the slope: (tan θ + μ)/(1 − μ tan θ). At the least speed it acts up the slope and both signs flip.

Using the natural length as the radius

A spring stretches to provide the pull, so the radius is l₀ + x. Writing kx = mω²l₀ drops the stretch from the radius and gives the wrong extension.

Radius or string length in a conical pendulum

The bob moves on a circle of radius L sin θ, but the tension comes out as mω²L, with the full string length. Mixing the two gives a tension too small by sin θ.

Zero speed at the top

On a string, 'just completes the circle' means the tension is zero at the top, which needs a speed of √(gr) there. At zero speed the string would have gone slack well before the top.

Change in velocity as a change in speed

Between the bottom and the point where the string is horizontal, the velocity turns through 90°. The change in velocity is √(v₁² + v₂²), not the difference of the speeds.

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