RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
Once motion leaves a straight line you need vectors, because direction now carries as much information as magnitude. This chapter teaches you to resolve, add and multiply vectors, then applies that machinery to projectiles and to uniform circular motion. The single most useful idea here is that horizontal and vertical motions are completely independent of each other.
Two forces of magnitudes 8 N and 6 N act at a point at right angles to each other. What is the magnitude of their resultant?
Answer
10 N — for perpendicular vectors θ = 90°, so cos θ = 0 and R = √(8² + 6² + 0) = √(64 + 36) = √100 = 10 N. The answers 2 N and 14 N would correspond to antiparallel and parallel forces respectively.
A projectile is fired at 60° to the horizontal. Which other angle of projection gives it the same horizontal range on level ground, with the same launch speed?
Answer
30° — the range R = u² sin 2θ/g is unchanged when θ is replaced by (90° − θ), because sin 2(90° − θ) = sin (180° − 2θ) = sin 2θ. Since 90° − 60° = 30°, the complementary angle is 30°, though the 60° projectile flies higher and stays in the air longer.
If A·B = 0 and neither A nor B is a null vector, then the two vectors must be:
Answer
Perpendicular — the scalar product A·B = AB cos θ vanishes only when cos θ = 0, i.e. θ = 90°, provided A ≠ 0 and B ≠ 0. Parallel vectors would give the maximum value AB, and antiparallel vectors would give −AB.
A stone tied to a 0.5 m string is whirled in a horizontal circle at a constant 4 m/s. What is its centripetal acceleration?
Answer
32 m/s² — centripetal acceleration a_c = v²/r = (4)²/0.5 = 16/0.5 = 32 m/s², directed at every instant towards the centre of the circle. Note that the speed is constant but the velocity is not, which is why an acceleration exists at all.
Assertion (A): At the highest point of its trajectory, the velocity of an obliquely projected projectile is not zero. Reason (R): At the highest point only the vertical component of velocity becomes zero, while the horizontal component remains unchanged throughout the flight.
Answer
Both A and R are true and R is the correct explanation of A — since there is no horizontal force (air resistance being neglected), the horizontal component u cos θ stays constant for the whole flight, while gravity reduces the vertical component to zero at the top. The projectile therefore still moves horizontally at u cos θ at the highest point, which is exactly the reason its velocity there is non-zero.
Find the angle between the vectors A = 3î + 4ĵ and B = 4î − 3ĵ.
Answer
A·B = (3)(4) + (4)(−3) = 12 − 12 = 0. |A| = √(9 + 16) = 5 and |B| = √(16 + 9) = 5, both non-zero. Since A·B = AB cos θ = 0 with A ≠ 0 and B ≠ 0, cos θ = 0, so θ = 90°. The two vectors are perpendicular.
A ball is thrown with speed 30 m/s at 30° above the horizontal from level ground. Taking g = 10 m/s², find its time of flight, maximum height and horizontal range.
Answer
Components: u cos θ = 30 cos 30° = 30 × 0.866 = 25.98 m/s; u sin θ = 30 sin 30° = 15 m/s. Time of flight T = 2u sin θ/g = 2(15)/10 = 3 s. Maximum height H = u² sin²θ/(2g) = (15)²/(2 × 10) = 225/20 = 11.25 m. Range R = (u cos θ) × T = 25.98 × 3 = 77.9 m, which matches u² sin 2θ/g = 900 × sin 60°/10 = 900 × 0.866/10 = 77.9 m.
Derive expressions for the time of flight, maximum height and horizontal range of a projectile launched from level ground with initial speed u at an angle θ to the horizontal. Hence show that the range is maximum when θ = 45°.
Answer
Set up axes with x horizontal and y vertical upward at the launch point. Resolve the initial velocity: uₓ = u cos θ, u_y = u sin θ. Since air resistance is neglected, aₓ = 0 and a_y = −g. The two motions are independent. Time of flight: vertically, y = (u sin θ)t − ½gt². The projectile lands when y = 0 and t ≠ 0: (u sin θ)t = ½gt², so t = 2u sin θ/g. Hence T = 2u sin θ/g. Maximum height: at the top the vertical velocity is zero. Using v_y² = u_y² − 2gH with v_y = 0: 0 = u² sin²θ − 2gH, so H = u² sin²θ/(2g). Horizontal range: horizontally the velocity is constant, so R = uₓT = (u cos θ)(2u sin θ/g) = 2u² sin θ cos θ/g = u² sin 2θ/g, using sin 2θ = 2 sin θ cos θ. Maximum range: for a fixed u, R depends on θ only through sin 2θ, whose greatest value is 1. This occurs when 2θ = 90°, i.e. θ = 45°. The maximum range is therefore R_max = u²/g. Any deviation from 45° reduces sin 2θ and so shortens the range, and angles θ and (90° − θ) give equal ranges because their values of sin 2θ coincide.
(a) Distinguish between the scalar and vector products of two vectors, giving one physical example of each. (b) For A = 2î + 3ĵ and B = 5î − ĵ, compute A·B and the magnitude of A × B, and hence find the angle between them.
Answer
(a) The scalar (dot) product A·B = AB cos θ produces a scalar, is commutative (A·B = B·A), is maximum for parallel vectors and zero for perpendicular ones. Physical example: work done, W = F·s, where only the component of force along the displacement contributes. The vector (cross) product A × B = AB sin θ n̂ produces a vector perpendicular to the plane of A and B, with direction given by the right-hand rule. It is anticommutative (A × B = −B × A), is zero for parallel vectors and maximum for perpendicular ones. Physical example: torque, τ = r × F, where only the component of force perpendicular to the position vector turns the body. (b) With A = 2î + 3ĵ and B = 5î − ĵ: A·B = (2)(5) + (3)(−1) = 10 − 3 = 7. |A| = √(4 + 9) = √13 ≈ 3.606; |B| = √(25 + 1) = √26 ≈ 5.099. cos θ = 7/(3.606 × 5.099) = 7/18.39 = 0.3807, so θ ≈ 67.6°. For the cross product of two vectors in the xy-plane, A × B = (AₓB_y − A_yBₓ)k̂ = [(2)(−1) − (3)(5)]k̂ = (−2 − 15)k̂ = −17k̂, so |A × B| = 17. Check: AB sin θ = 18.39 × sin 67.6° = 18.39 × 0.9245 ≈ 17.0, which agrees.
Read the following and answer the questions that follow: A rescue swimmer needs to cross a river 120 m wide. She can swim at 3 m/s in still water, and the river flows steadily at 4 m/s parallel to its banks. She first tries swimming with her body always pointed straight at the far bank. (a) How long does she take to reach the far bank, and why does the current not change this time? (b) How far downstream is she carried, and what is her resultant speed relative to the ground? (c) If instead she wants to land exactly opposite her starting point, explain whether that is possible here. (d) State the general condition, in terms of swimming speed and current speed, for a swimmer to be able to reach the point directly opposite.
Answer
(a) Time t = width / component of her velocity perpendicular to the bank = 120/3 = 40 s. The current flows parallel to the banks, so it has no component across the river; the across-river and along-river motions are independent, exactly as with a projectile, so the current cannot change the crossing time. (b) Drift = current speed × time = 4 × 40 = 160 m downstream. Her resultant velocity relative to the ground is the vector sum of 3 m/s across and 4 m/s downstream, of magnitude √(3² + 4²) = 5 m/s, at an angle tan⁻¹(4/3) ≈ 53° from the across-river direction. (c) It is not possible. To land directly opposite she would have to aim upstream at angle θ from the across-river direction such that her upstream component cancels the current: 3 sin θ = 4, giving sin θ = 4/3 > 1, which has no solution. Her swimming speed is simply too small to defeat the current. (d) The condition is that the swimmer's speed in still water must be at least equal to the speed of the current, v_swimmer ≥ v_current. Then sin θ = v_current/v_swimmer ≤ 1 has a valid solution and she can aim upstream at that angle; if v_swimmer < v_current, some downstream drift is unavoidable.
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