JEE Physics Kinematics MCQs Set 04

Answer: (a) \( u\cos\theta\sec\alpha \)

Explanation:
1. The horizontal component of velocity is constant: \( v\cos\alpha = u\cos\theta \).
2. So \( v = u\cos\theta\sec\alpha \).

Teacher's Note:
1. Use the conserved horizontal component.
2. At the top α = 0 and v = u cos θ.

Answer: (c) 60 m

Explanation:
1. \( y = 40t - 5t^2 = 75 \Rightarrow t^2 - 8t + 15 = 0 \), so t = 3 s and 5 s.
2. Horizontal separation \( = 30\times(5-3) = 60 \) m.

Teacher's Note:
1. Use vx = 50 cos 53° = 30 m/s and vy = 50 sin 53° = 40 m/s.
2. Both times give the same height, one on the way up and one on the way down.

Answer: (a) 2 s

Explanation:
1. The angle with the plane is 60° − 30° = 30°. Component perpendicular to the plane: \( 10\sqrt3\sin30^\circ = 5\sqrt3 \) m/s.
2. Perpendicular acceleration: \( g\cos30^\circ = 5\sqrt3 \) m/s².
3. \( T = \frac{2\times5\sqrt3}{5\sqrt3} = 2 \) s.

Teacher's Note:
1. Resolve along and perpendicular to the incline.
2. The ball returns to the plane when its perpendicular displacement is again zero.

Answer: (d) \( \frac{2v^2}{g}\tan\theta\sec\theta \)

Explanation:
1. Perpendicular to the slope: initial speed v, acceleration \( g\cos\theta \). Time of flight \( T = \frac{2v}{g\cos\theta} \).
2. Along the slope: initial speed 0, acceleration \( g\sin\theta \) (down the slope). \( R = \tfrac12g\sin\theta\,T^2 = \frac{2v^2\sin\theta}{g\cos^2\theta} \).
3. This is \( \frac{2v^2}{g}\tan\theta\sec\theta \).

Teacher's Note:
1. Take axes along and perpendicular to the incline.
2. Time of flight comes from the perpendicular motion.

Answer: (b) 45 s

Explanation:
1. Speeds in escalator-lengths per minute: escalator = 1, walking = 1/3.
2. Combined speed = 4/3 per minute, so time = 3/4 min = 45 s.

Teacher's Note:
1. Add the speeds because both move in the same direction.
2. Treat the length of the escalator as one unit.

Answer: (c) \( \frac{2u - gt}{t} \)

Explanation:
1. In the lift frame the effective acceleration is \( g + a \), so \( t = \frac{2u}{g+a} \).
2. Therefore \( a = \frac{2u}{t} - g = \frac{2u - gt}{t} \).

Teacher's Note:
1. The lift's upward acceleration adds to g in its own frame.
2. The time of flight is twice the time to the top.

Answer: (d) 11/10 m/s and 1/2 m/s

Explanation:
1. Towards each other: v₁ + v₂ = 16/10 = 1.6 m/s.
2. Same direction: v₁ − v₂ = 3/5 = 0.6 m/s.
3. Hence v₁ = 1.1 m/s = 11/10 m/s and v₂ = 0.5 m/s = 1/2 m/s.

Teacher's Note:
1. Relative speed adds for approach and subtracts for chasing.
2. Check: 1.1 + 0.5 = 1.6 and 1.1 − 0.5 = 0.6.

Answer: (b) 6 s

Explanation:
1. Q starts 8 m ahead of P along the track and moves faster by 4 m/s, so Q must gain a full lap less the initial 8 m: 32 − 8 = 24 m.
2. Time = 24/4 = 6 s. Check: P has covered 36 m and Q 60 m, and both are 4 m from A on AB.

Teacher's Note:
1. Treat it as a chase around a closed track.
2. Q has to catch P from behind, so the gap to close is the perimeter minus the starting separation.

Answer: (a) \( \frac av \)

Explanation:
1. By symmetry the particles stay at the corners of a shrinking square.
2. The velocity of each particle towards its target is v, and the target's velocity is perpendicular to the line joining them, so it does not change the rate of approach. The separation shrinks at v from a, so t = a/v.

Teacher's Note:
1. Take the component of relative velocity along the line joining two particles.
2. For N particles on a regular polygon the time is a / (v(1 − cos(2π/N))).

Answer: (c) 4 minutes

Explanation:
1. Across the river: \( 5\sin127^\circ = 4 \) km/h. Along the flow: \( 4 + 5\cos127^\circ = 4 - 3 = 1 \) km/h downstream.
2. Crossing time \( = \frac{0.2}{4} = 0.05 \) h = 3 min. Drift \( = 1\times0.05 = 0.05 \) km = 50 m.
3. Walking back 50 m at 3 km/h takes 0.05/3 h = 1 min. Total = 4 min.

Teacher's Note:
1. sin 127° = sin 53° = 0.8 and cos 127° = −0.6.
2. The upstream component of the swimmer's velocity is 3 km/h.

Answer: (a) 3

Explanation:
1. Speed along the shortest path = 1 km / 0.25 h = 4 km/h.
2. This equals √(5² − u²), so u² = 25 − 16 = 9 and u = 3 km/h.

Teacher's Note:
1. The shortest path has zero drift, so the boat is aimed upstream.
2. 3-4-5 is a Pythagorean triple.

Answer: (d) 200 m

Explanation:
1. Drift in 10 minutes = 120 m, so the river speed u = 12 m/min and the width w = 10v.
2. For the straight path, w = 12.5√(v² − u²), so 10v = 12.5√(v² − 144). This gives v = 20 m/min.
3. Width = 10 × 20 = 200 m.

Teacher's Note:
1. Both journeys give two equations in v and u.
2. The crossing time for perpendicular heading is the shortest possible time.

Answer: (b) 9.6 h

Explanation:
1. Let the distance be d. Then \( v + u = \frac d8 \) and \( v - u = \frac d{12} \).
2. Adding: \( v = \frac d2\left(\frac18+\frac1{12}\right) = \frac{5d}{48} \). Time \( = \frac{d}{v} = 9.6 \) h.

Teacher's Note:
1. The still-water speed is the mean of the two speeds.
2. The time is the harmonic mean: 2 × 8 × 12 / 20.

Answer: (c) \( \frac{80}{3} \) km

Explanation:
1. Downstream speed 20 km/h, upstream speed 10 km/h. For a one-way distance x: \( \frac{x}{20} + \frac{x}{10} = 4 \).
2. \( \frac{3x}{20} = 4 \Rightarrow x = \frac{80}{3} \) km.

Teacher's Note:
1. The total time is the sum of the two one-way times.
2. The whole 4 hours is used up for the round trip.

Answer: (d) Both will have the same range

Explanation:
1. Equal horizontal components keep the particles in the same vertical line, and the relative velocity is vertical.
2. The range is horizontal speed × time of flight, and the times differ unless the vertical components are equal too. So the statement about equal range is incorrect.

Teacher's Note:
1. Equal horizontal velocities do not mean equal times of flight.
2. Range depends on both components.

Answer: (b) None of the statements is incorrect

Explanation:
1. Equal vertical components give equal times of flight \( T = \frac{2v\sin\theta}{g} \) and equal maximum heights \( \frac{(v\sin\theta)^2}{2g} \).
2. The relative velocity is then purely horizontal, so the relative path is a horizontal straight line. All three statements are correct.

Teacher's Note:
1. Equal vertical components mean equal vertical motion.
2. The relative motion is along the horizontal.

Answer: (a) 20

Explanation:
1. For the second ship to stay due north of the first, the east components of velocity must be equal.
2. v sin 30° = 10, so v = 20 km/h.

Teacher's Note:
1. The relative velocity must have no east-west part.
2. The north component of the second ship is 20 cos 30° = 10√3 km/h.

Answer: (c) \( \tan^{-1}(3) \)

Explanation:
1. In the cart's frame the rain has velocity 6 m/s downward and 2 m/s backward.
2. The pipe must lie along this relative velocity: tan α = 6/2 = 3, so α = tan⁻¹ 3.

Teacher's Note:
1. Rain relative to cart = rain velocity − cart velocity.
2. The angle is measured from the horizontal.

Answer: (a) \( R = 4\sqrt{h_1h_2} \)

Explanation:
1. The angles are θ and 90° − θ. \( h_1 = \frac{u^2\sin^2\theta}{2g} \) and \( h_2 = \frac{u^2\cos^2\theta}{2g} \).
2. \( \sqrt{h_1h_2} = \frac{u^2\sin\theta\cos\theta}{2g} = \frac R4 \), so \( R = 4\sqrt{h_1h_2} \).

Teacher's Note:
1. Equal ranges need complementary angles.
2. R = 2u² sinθ cosθ / g.

Answer: (d) 3 m/s

Explanation:
1. Let each third be d. Times: \( \frac d2, \frac d3, \frac d6 \), total \( d\left(\frac{3+2+1}{6}\right) = d \).
2. Average speed = 3d / d = 3 m/s.

Teacher's Note:
1. Average speed is total distance divided by total time.
2. If the parts were of equal time the answer would be 11/3 m/s instead.

Answer: (b) 1 and 3 only

Explanation:
1. The particle stops after t = 10/5 = 2 s, having moved 10²/(2 × 5) = 10 m. (1) is correct.
2. In the third second it comes back: 0.5 × 5 × 1² = 2.5 m. Distance in 3 s = 10 + 2.5 = 12.5 m, so (3) is correct. (The net displacement is 7.5 m, which is not the distance.)

Teacher's Note:
1. Distance counts the return path as positive.
2. Displacement after 3 s is 10 − 2.5 = 7.5 m.

Answer: (c) 2, 3 and 4 only

Explanation:
1. (1) is wrong: a ball at the top of its path has zero velocity but non-zero acceleration.
2. (2) is correct, as constant velocity (zero) means zero acceleration. (3) is correct: opposite signs mean speed decreasing. (4) is correct: a positive position with negative velocity means the distance from the origin is decreasing.

Teacher's Note:
1. Test statement (1) with a ball at its highest point.
2. Position and velocity of opposite sign mean a decreasing |x|.

Answer: (d) 1 and 3 only

Explanation:
1. (1) is correct: a ball at its highest point. (3) is correct: uniform motion.
2. (2) is false for the same reason as (1). (4) is false: projectile motion has a downward everywhere while v takes all directions.

Teacher's Note:
1. Velocity and acceleration are independent quantities.
2. Look for counterexamples in vertical throws.

Answer: (b) 1 and 3 only

Explanation:
1. (1) is true, since a change in speed changes the velocity vector.
2. (3) is true: uniform circular motion changes velocity but not speed.
3. (2) is false for circular motion. (4) is false: a change in speed with fixed direction changes the velocity.

Teacher's Note:
1. Velocity changes if either magnitude or direction changes.
2. Uniform circular motion is the standard counterexample for (2).

Answer: (a) 1, 2 and 3 only

Explanation:
1. (1) is true, since distance is at least the magnitude of displacement.
2. (2) is true for uniform circular motion. (3) is true for a full circle.
3. (4) is false: on a straight line, zero average velocity means the particle must reverse, and the velocity then passes through zero.

Teacher's Note:
1. Think of uniform circular motion for (2) and (3).
2. On a line, a continuous velocity must pass through zero when it reverses.

Answer: (c) 1, 3 and 4 only

Explanation:
1. \( \frac{dx}{dt} = \sqrt x \Rightarrow 2\sqrt x = t + 4 \), so at t = 2: \( \sqrt x = 3 \) and x = 9 m. (1) is correct.
2. \( a = v\frac{dv}{dx} = \sqrt x\cdot\frac{1}{2\sqrt x} = \frac12 \) m/s², constant. (3) is correct and (2) is wrong.
3. The velocity is always positive, so the particle never reverses. (4) is correct.

Teacher's Note:
1. Use a = v dv/dx.
2. The constant acceleration can be checked: x = (2 + t/2)², so d²x/dt² = 1/2.

Answer: (a) The maximum potential energy during its motion can be 6 J

Explanation:
1. \( R = \frac{20\sin2\theta}{10} = \sqrt3 \Rightarrow \sin2\theta = \frac{\sqrt3}{2} \), so θ = 30° or 60°.
2. θ = 30°: H = 0.25 m, minimum speed \( \sqrt{20}\cos30^\circ = \sqrt{15} \) m/s, T = √(1/5) s. θ = 60°: H = 0.75 m, T = √(3/5) s. So the first three statements can be true.
3. The maximum potential energy is mgH = 2.5 J or 7.5 J, never 6 J. This statement is incorrect.

Teacher's Note:
1. Two angles, 30° and 60°, give the same range.
2. Maximum potential energy = m g H.

Answer: (d) 1, 2, 3 and 4 are all correct

Explanation:
1. (1) The maximum height is \( u^2/2g = h \) for a vertical throw and the maximum range is \( u^2/g = 2h \).
2. (2) \( \frac RH = \frac{4}{\tan\theta} = n \Rightarrow \theta = \tan^{-1}\frac4n \).
3. (3) \( \frac{T^2}{R} = \frac{4u^2\sin^2\theta/g^2}{2u^2\sin\theta\cos\theta/g} = \frac{2\tan\theta}{g} \).
4. (4) The same time of flight means the same vertical component of velocity, so the heights are equal.

Teacher's Note:
1. Check each statement with the standard projectile formulas.
2. The same time of flight needs the same vertical launch speed.