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

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

Rain is falling vertically at 35 m/s. Wind blows horizontally at 12 m/s from east to west. In which direction should a person hold their umbrella? Justify using vector addition.

The velocity of rain vr = 35 m/s downward and wind velocity vw = 12 m/s horizontal (east to west) must be added vectorially to find the resultant velocity of rain relative to ground.
Magnitude of resultant: R = √(35² + 12²) = √(1225 + 144) = √1369 = 37 m/s.
The angle θ with vertical: tan θ = vw/vr = 12/35 = 0.343, so θ = tan⁻¹(0.343) ≈ 19°.
The umbrella must be held in the vertical plane at about 19° from the vertical, tilted towards the east (into the wind) to intercept the rain.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 2

Applied Concept

A motorboat races north at 25 km/h. The water current flows at 10 km/h in the direction 60° east of south. Find the resultant velocity of the boat.

The angle between vb (north) and vc (60° east of south) is 180° − 60° = 120°.
By the law of cosines: R = √(vb² + vc² + 2vb vc cos 120°) = √(625 + 100 + 2×25×10×(−0.5)) = √(625 + 100 − 250) = √475 ≈ 22 km/h.
Direction using law of sines: sin φ/vc = sin 120°/R, giving sin φ = 10 × sin 120°/22 ≈ 0.397, so φ ≈ 23.4°.
The resultant velocity of the boat is approximately 22 km/h directed at about 23.4° west of north.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

A particle starts from the origin at t = 0 with velocity 5.0 i-hat m/s and moves under constant acceleration (3.0 i-hat + 2.0 j-hat) m/s². Find the y-coordinate when x = 84 m, and the speed at that instant.

Position: r(t) = (5.0t + 1.5t²) i-hat + (1.0t²) j-hat, so x(t) = 5.0t + 1.5t² and y(t) = 1.0t².
Setting x = 84 m: 5.0t + 1.5t² = 84 → t = 6 s.
y-coordinate at t = 6 s: y = 1.0 × 36 = 36 m.
Velocity at t = 6 s: v = (5.0 + 3.0×6) i-hat + (2.0×6) j-hat = 23.0 i-hat + 12.0 j-hat; speed = √(23² + 12²) ≈ 26 m/s.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A hiker stands on a cliff 490 m above the ground and throws a stone horizontally at 15 m/s. Neglecting air resistance, find the time to reach the ground and speed at impact. (g = 9.8 m/s²)

Taking origin at the throw point with y upward, initial conditions: vox = 15 m/s, voy = 0, ay = −9.8 m/s².
Setting y = −490 m: −490 = −(1/2)(9.8)t², giving t² = 100, so t = 10 s.
At impact: vx = 15 m/s (unchanged), vy = 0 − 9.8 × 10 = −98 m/s.
Speed at impact = √(15² + 98²) = √(225 + 9604) = √9829 ≈ 99 m/s.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A cricket ball is thrown at 28 m/s at 30° above the horizontal. Calculate (a) maximum height, (b) time of flight, and (c) horizontal range.

Maximum height: hm = (v0 sin θ0)²/(2g) = (28 × 0.5)²/(2 × 9.8) = (14)²/19.6 = 196/19.6 = 10.0 m.
Time of flight: Tf = 2v0 sin θ0/g = 2 × 28 × sin 30°/9.8 = 28/9.8 ≈ 2.9 s.
Horizontal range: R = v0² sin 2θ0/g = 28² × sin 60°/9.8 = 784 × 0.866/9.8 ≈ 69 m.
These results illustrate the interdependence of range, height, and time through the launch angle and initial speed.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

An insect trapped in a circular groove of radius 12 cm completes 7 revolutions in 100 s. Find the angular speed, linear speed, and magnitude of centripetal acceleration.

Angular speed: ω = 2πν = 2π × (7/100) = 0.44 rad/s.
Linear speed: v = ωR = 0.44 × 12 = 5.3 cm/s.
Centripetal acceleration: ac = ω²R = (0.44)² × 12 = 0.194 × 12 ≈ 2.3 cm/s².
The acceleration is directed towards the centre of the groove at every instant, though its direction continuously changes, so it is not a constant vector.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

The position of a particle is r = 3.0t i-hat + 2.0t² j-hat + 5.0 k-hat m, where t is in seconds. Find the velocity, acceleration, and their magnitudes at t = 1.0 s.

Velocity: v(t) = dr/dt = 3.0 i-hat + 4.0t j-hat m/s; at t = 1.0 s, v = 3.0 i-hat + 4.0 j-hat m/s.
Magnitude of velocity: |v| = √(3² + 4²) = √25 = 5.0 m/s.
Acceleration: a(t) = dv/dt = 4.0 j-hat m/s²; constant at 4.0 m/s² in the y-direction.
The angle of velocity with x-axis: θ = tan⁻¹(4/3) ≈ 53°; the acceleration is perpendicular to neither v nor r, illustrating general 2D motion.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A stone tied to a string 80 cm long is whirled in a horizontal circle and makes 14 revolutions in 25 s. Find the magnitude and direction of acceleration of the stone.

Frequency: ν = 14/25 = 0.56 Hz; angular speed: ω = 2πν = 2π × 0.56 = 3.52 rad/s.
Radius R = 80 cm = 0.80 m; centripetal acceleration: ac = ω²R = (3.52)² × 0.80 ≈ 9.9 m/s².
The direction of this acceleration is always directed horizontally towards the centre of the circular path.
Since the stone moves in a horizontal circle, the centripetal acceleration is horizontal and continuously changes direction as the stone moves.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

An aircraft executes a horizontal loop of radius 1.00 km at 900 km/h. Compare its centripetal acceleration with g (= 9.8 m/s²).

Convert: v = 900 km/h = 250 m/s; R = 1.00 km = 1000 m.
Centripetal acceleration: ac = v²/R = (250)²/1000 = 62500/1000 = 62.5 m/s².
Ratio ac/g = 62.5/9.8 ≈ 6.4; the centripetal acceleration is about 6.4 times the acceleration due to gravity.
This high centripetal acceleration, about 6.4g, explains why pilots in tight high-speed turns experience significant g-forces on their bodies.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

With reference to the independence of perpendicular motions, explain why the horizontal distance covered by a projectile depends only on the initial horizontal speed and time of flight, not on g.

In projectile motion, the horizontal acceleration ax = 0; therefore the horizontal motion is uniform: x = (v0 cos θ0)t.
The vertical motion is independently governed by gravity: y = (v0 sin θ0)t − (1/2)gt².
Since g appears only in the y-equation, it does not directly enter the x-equation; horizontal distance = (v0 cos θ0) × t.
However, g indirectly affects x through the time of flight Tf = 2v0 sin θ0/g; a larger g reduces Tf and thus the range, showing the independence principle has limits.

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Chapter 3: Motion in a Plane | Application Questions

Rain is falling vertically at 35 m/s. Wind blows horizontally at 12 m/s from east to west. In which direction should a person hold their umbrella? Justify using vector addition.

A motorboat races north at 25 km/h. The water current flows at 10 km/h in the direction 60° east of south. Find the resultant velocity of the boat.

A particle starts from the origin at t = 0 with velocity 5.0 i-hat m/s and moves under constant acceleration (3.0 i-hat + 2.0 j-hat) m/s². Find the y-coordinate when x = 84 m, and the speed at that instant.

A hiker stands on a cliff 490 m above the ground and throws a stone horizontally at 15 m/s. Neglecting air resistance, find the time to reach the ground and speed at impact. (g = 9.8 m/s²)

A cricket ball is thrown at 28 m/s at 30° above the horizontal. Calculate (a) maximum height, (b) time of flight, and (c) horizontal range.

An insect trapped in a circular groove of radius 12 cm completes 7 revolutions in 100 s. Find the angular speed, linear speed, and magnitude of centripetal acceleration.

The position of a particle is r = 3.0t i-hat + 2.0t² j-hat + 5.0 k-hat m, where t is in seconds. Find the velocity, acceleration, and their magnitudes at t = 1.0 s.

A stone tied to a string 80 cm long is whirled in a horizontal circle and makes 14 revolutions in 25 s. Find the magnitude and direction of acceleration of the stone.

An aircraft executes a horizontal loop of radius 1.00 km at 900 km/h. Compare its centripetal acceleration with g (= 9.8 m/s²).

With reference to the independence of perpendicular motions, explain why the horizontal distance covered by a projectile depends only on the initial horizontal speed and time of flight, not on g.

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