Chapter 3: Motion in a Plane Long Questions | physics Class 11 CBSE

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Chapter 3: Motion in a Plane Long Questions | physics Class 11 CBSE | ByteNotes.in

Long Answer

Medium difficulty • Structured explanation

Medium

Question 1

Long Form

Derive the expression for the magnitude and direction of the resultant of two vectors A and B making an angle θ, using the parallelogram law.

Place vectors A and B with tails at common origin O; complete the parallelogram OQSP. The resultant R = A + B is the diagonal OS.
Drop perpendicular SN from S to extended OP. Using geometry: ON = A + B cos θ and SN = B sin θ.
Applying Pythagoras: R² = OS² = ON² + SN² = (A + B cos θ)² + (B sin θ)² = A² + B² + 2AB cos θ.
Therefore R = √(A² + B² + 2AB cos θ), which is the law of cosines.
For direction angle α with A: tan α = SN/(OP + PN) = B sin θ/(A + B cos θ).
From the law of sines: R/sin θ = A/sin β = B/sin α, giving complete information about the resultant's direction.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 2

Long Form

Derive the equations of motion for a projectile launched with speed v0 at angle θ0, and obtain expressions for trajectory, maximum height, time of flight, and horizontal range.

With origin at launch point, ax = 0 and ay = −g; initial components are vox = v0 cos θ0, voy = v0 sin θ0.
Position equations: x = (v0 cos θ0)t and y = (v0 sin θ0)t − (1/2)gt².
Eliminating t: y = (tan θ0)x − gx²/[2(v0 cos θ0)²], showing the trajectory is a parabola.
Maximum height: set vy = 0 to get tm = v0 sin θ0/g; substituting gives hm = (v0 sin θ0)²/(2g).
Time of flight: set y = 0 to get Tf = 2v0 sin θ0/g = 2tm, confirming the symmetry of the path.
Horizontal range: R = (v0 cos θ0)Tf = v0² sin 2θ0/g, maximum at θ0 = 45° giving Rm = v0²/g.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 3

Long Form

Describe the resolution of a vector into components along two perpendicular axes. How can a vector be completely specified using its components?

Unit vectors i-hat and j-hat have magnitude 1 and point along the x- and y-axes respectively; they are mutually perpendicular.
For vector A making angle θ with the x-axis, dropping perpendiculars to the axes gives A1 = Ax i-hat and A2 = Ay j-hat such that A = A1 + A2.
The components are Ax = A cos θ and Ay = A sin θ, which can be positive, negative, or zero depending on θ.
Given components, the magnitude is recovered as A = √(Ax² + Ay²) using the Pythagorean theorem.
The direction is given by θ = tan⁻¹(Ay/Ax), fully specifying the vector.
A vector is therefore equivalently described either by (A, θ) or by its components (Ax, Ay); both representations are complete.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 4

Long Form

Analyse the velocity and acceleration of an object in uniform circular motion and show that the centripetal acceleration has magnitude v²/R directed towards the centre.

In uniform circular motion, speed v is constant but velocity direction changes continuously; velocity is always tangent to the circle.
For position vectors r and r′ separated by angle Δθ, the triangle formed by r, r′ and Δr is similar to the triangle formed by v, v′ and Δv.
By similarity: |Δv|/v = |Δr|/R, so |Δv| = v|Δr|/R.
Magnitude of acceleration: |a| = lim(Δt→0)|Δv|/Δt = (v/R) × lim(Δt→0)|Δr|/Δt = (v/R) × v = v²/R.
Since Δv is perpendicular to Δr and as Δt → 0 it points towards the centre, the acceleration is always centripetal.
Centripetal acceleration ac = v²/R = ω²R; its magnitude is constant but direction changes continuously, so it is not a constant vector.

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Long Answer

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Medium

Question 5

Long Form

Compare the graphical (head-to-tail) method and the analytical (component) method of vector addition, discussing their merits, limitations, and the laws they obey.

The head-to-tail (triangle) method places vectors in sequence with tail of each at head of previous; the resultant closes the figure from first tail to last head.
The parallelogram method places both vectors tail-to-tail; the diagonal from the common origin gives the resultant.
Both graphical methods confirm that vector addition is commutative (A + B = B + A) and associative ((A + B) + C = A + (B + C)).
The analytical method resolves all vectors into components, then adds corresponding components: Rx = Ax + Bx, Ry = Ay + By.
Graphical methods are useful for visualisation but are limited in accuracy and become tedious for many vectors.
The analytical method is exact, systematic, and easily extendable to three dimensions and any number of vectors.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 6

Long Form

Explain how two-dimensional motion with constant acceleration is treated as superposition of two independent one-dimensional motions. Illustrate with component equations.

For constant acceleration a in a plane, the vector equation r = r0 + v0t + (1/2)at² can be resolved into components.
x-component: x = x0 + voxt + (1/2)axt² operates entirely independently of the y-motion.
y-component: y = y0 + voyt + (1/2)ayt² is governed solely by initial y-conditions and ay.
The motions along x and y axes are completely independent — the x-motion is unaffected by ay and vice versa.
This is the principle of superposition for perpendicular directions, valid for constant acceleration.
In projectile motion, this reduces to uniform x-motion (ax = 0) and free-fall y-motion (ay = −g), each handled independently.

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Long Answer

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Hard

Question 7

Long Form

Derive the expression for maximum height attained by a projectile and explain why the horizontal component of velocity does not change throughout projectile motion.

After projection with v0 at angle θ0, horizontal acceleration ax = 0 because no force acts horizontally (air resistance neglected).
Since ax = 0, Newton's law gives dvx/dt = 0, so vx = v0 cos θ0 = constant throughout the flight.
The vertical velocity changes due to gravity: vy = v0 sin θ0 − gt.
At maximum height, vy = 0: 0 = v0 sin θ0 − g tm, giving tm = v0 sin θ0/g.
Substituting tm into y = (v0 sin θ0)t − (1/2)gt²: hm = (v0 sin θ0)²/g − (1/2)g(v0 sin θ0/g)² = (v0 sin θ0)²/(2g).
The horizontal velocity is constant because gravity acts perpendicular to the horizontal direction and cannot change the horizontal component of velocity.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 8

Long Form

Prove that for a given initial speed, projectiles launched at angles (45° + α) and (45° − α) have equal horizontal ranges.

Range formula: R = v0² sin 2θ0/g, where θ0 is the projection angle.
For θ0 = 45° + α: 2θ0 = 90° + 2α, so sin(90° + 2α) = cos 2α.
For θ0 = 45° − α: 2θ0 = 90° − 2α, so sin(90° − 2α) = cos 2α.
Both expressions equal cos 2α; therefore their ranges are identical: R1 = v0² cos 2α/g = R2.
This result, first stated by Galileo in Two New Sciences, shows range has a symmetry about 45°.
Physically, angles above and below 45° by the same amount α trade time of flight for horizontal speed such that the range remains unchanged.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 9

Long Form

Describe position vector, displacement, and velocity of a particle moving in a plane. Explain why instantaneous velocity is always tangent to the path.

Position vector r = x i-hat + y j-hat locates the particle relative to the origin of the x-y frame.
Displacement Δr = r′ − r = Δx i-hat + Δy j-hat; it is the straight-line vector from initial to final position and is independent of the path taken.
Average velocity v-bar = Δr/Δt has the same direction as Δr.
As Δt → 0, the displacement Δr becomes infinitesimally small and aligns with the tangent to the curve at the particle's position.
Instantaneous velocity v = dr/dt = (dx/dt) i-hat + (dy/dt) j-hat is therefore always tangential to the path.
The magnitude |v| = √(vx² + vy²) is the instantaneous speed, and its direction angle θ = tan⁻¹(vy/vx).

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 10

Long Form

Differentiate between scalars and vectors systematically, and explain why the rules of ordinary algebra cannot be applied to vector addition.

Scalars have magnitude only (e.g., mass, temperature) and combine via ordinary arithmetic; vectors have both magnitude and direction (e.g., displacement, force).
Ordinary algebraic addition ignores direction; for example, 3 + 4 = 7, but two forces of 3 N and 4 N may yield a resultant anywhere from 1 N to 7 N depending on their relative directions.
Vector addition must account for both magnitude and direction, requiring the triangle law or parallelogram law.
Two vectors are equal only if both magnitude and direction are identical; ordinary algebra has no such directional condition.
The result of multiplying a vector by a negative number reverses its direction, a concept absent in scalar algebra.
Vector subtraction A − B ≠ |A| − |B| in general; it is defined as A + (−B), requiring vectorial treatment.

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Chapter 3: Motion in a Plane | Long Questions

Derive the expression for the magnitude and direction of the resultant of two vectors A and B making an angle θ, using the parallelogram law.

Derive the equations of motion for a projectile launched with speed v0 at angle θ0, and obtain expressions for trajectory, maximum height, time of flight, and horizontal range.

Describe the resolution of a vector into components along two perpendicular axes. How can a vector be completely specified using its components?

Analyse the velocity and acceleration of an object in uniform circular motion and show that the centripetal acceleration has magnitude v²/R directed towards the centre.

Compare the graphical (head-to-tail) method and the analytical (component) method of vector addition, discussing their merits, limitations, and the laws they obey.

Explain how two-dimensional motion with constant acceleration is treated as superposition of two independent one-dimensional motions. Illustrate with component equations.

Derive the expression for maximum height attained by a projectile and explain why the horizontal component of velocity does not change throughout projectile motion.

Prove that for a given initial speed, projectiles launched at angles (45° + α) and (45° − α) have equal horizontal ranges.

Describe position vector, displacement, and velocity of a particle moving in a plane. Explain why instantaneous velocity is always tangent to the path.

Differentiate between scalars and vectors systematically, and explain why the rules of ordinary algebra cannot be applied to vector addition.

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