JEE Physics Kinematics MCQs Set 02

Answer: (d) 10 m/s

Explanation:
1. \( v_x = 6 \) and \( v_y = 8 - 10t \), so at t = 0 the components are 6 and 8 m/s.
2. Speed \( = \sqrt{36+64} = 10 \) m/s.

Teacher's Note:
1. Differentiate each coordinate and set t = 0.
2. This is a 3-4-5 type triangle scaled by 2.

Answer: (b) 1 s

Explanation:
1. \( T = \frac{2u\sin\theta}{g} = \frac{2(9.8)(0.5)}{9.8} = 1 \) s.

Teacher's Note:
1. The time of flight depends only on the vertical component of velocity.
2. The ground is at the same level as the point of projection.

Answer: (c) is perpendicular to the acceleration for one instant only

Explanation:
1. The acceleration is g downward throughout.
2. The velocity is horizontal only at the highest point, which is the one instant when it is perpendicular to g.

Teacher's Note:
1. Perpendicularity needs the vertical velocity component to be zero.
2. This happens once, at the top of the path.

Answer: (a) RA < RB

Explanation:
1. \( R \propto \sin2\theta \). For A, \( \sin30^\circ = 0.5 \); for B, \( \sin90^\circ = 1 \).
2. So RA = RB/2 and RA < RB.

Teacher's Note:
1. Equal ranges need complementary angles; 15° and 45° are not complementary.
2. The maximum range occurs at 45°.

Answer: (b) 75°

Explanation:
1. \( R \propto \sin2\theta \) is the same for θ and 90° − θ.
2. So the other angle is 90° − 15° = 75°.

Teacher's Note:
1. The two angles are complementary.
2. The time of flight and the height differ for the two angles.

Answer: (d) E/2

Explanation:
1. At the top only the horizontal component \( u\cos45^\circ \) remains.
2. \( K = \frac12m(u\cos45^\circ)^2 = \frac E2 \).

Teacher's Note:
1. Kinetic energy is proportional to the square of the speed.
2. cos²45° = 1/2.

Answer: (c) 12.8 m

Explanation:
1. Set y = 0: \( x\left(16 - \frac{5x}{4}\right) = 0 \).
2. Range \( x = \frac{16\times4}{5} = 12.8 \) m.

Teacher's Note:
1. The range is the non-zero root of the trajectory equation.
2. The projection angle satisfies tan θ = 16.

Answer: (c) B and D

Explanation:
1. Equal ranges occur for complementary angles that add to 90°.
2. 30° (B) and 60° (D) add up to 90°.

Teacher's Note:
1. Check each pair for a sum of 90°.
2. A at 15° would pair with 75°, which is not present.

Answer: (d) both A and B reach the ground at the same instant

Explanation:
1. The vertical motion of each marble starts from zero vertical velocity and has acceleration g.
2. The horizontal velocity of B does not affect the time of fall.

Teacher's Note:
1. Horizontal and vertical motions are independent.
2. B lands farther from the base, but at the same time.

Answer: (b) 20 s

Explanation:
1. The vertical fall starts from rest: \( 1960 = \tfrac12(9.8)t^2 \).
2. \( t^2 = 400 \), so t = 20 s. (AB = 100 × 20 = 2000 m.)

Teacher's Note:
1. The horizontal speed does not matter for the time of fall.
2. Convert 360 km/h to 100 m/s only if the range is needed.

Answer: (c) \( \frac{ut}{\sqrt3} \)

Explanation:
1. Along the plane the initial speed is \( u\cos30^\circ = \frac{\sqrt3u}{2} \) and the deceleration is \( g\sin30^\circ = \frac g2 \).
2. Perpendicular to the plane the return condition gives \( gt = \frac{2u}{\sqrt3} \).
3. \( AB = \frac{\sqrt3}{2}ut - \frac14gt^2 = \frac{\sqrt3}{2}ut - \frac{ut}{2\sqrt3} = \frac{ut}{\sqrt3} \).

Teacher's Note:
1. Resolve motion along and perpendicular to the incline.
2. The time of flight comes from the perpendicular motion.

Answer: (a) 15°, 30°, 45°, 60°, 75°

Explanation:
1. \( T = \frac{2u\sin\theta}{g} \), which increases with θ for θ up to 90°.
2. So the order of increasing time is the order of increasing angle.

Teacher's Note:
1. Time of flight depends on the vertical component only.
2. Complementary angles have the same range but different times of flight.

Answer: (b) \( \frac17 \)

Explanation:
1. Take east as x and north as y: \( \vec v_A = \frac{80}{\sqrt2}(1,1) \) and \( \vec v_B = \frac{60}{\sqrt2}(1,-1) \).
2. \( \vec v_{AB} = \frac{1}{\sqrt2}(20, 140) \).
3. The angle from north has \( \tan\alpha = \frac{20}{140} = \frac17 \).

Teacher's Note:
1. Relative velocity is vA − vB.
2. tan of the angle with north is the ratio of the east component to the north component.

Answer: (d) 20 s

Explanation:
1. Relative speed of A with respect to B = 10 − 5 = 5 m/s.
2. Time = 100/5 = 20 s.

Teacher's Note:
1. Subtract the speeds for motion in the same direction.
2. In 20 s A moves 200 m and B moves 100 m, so they meet.

Answer: (c) 2 s

Explanation:
1. Relative to the block, the bullet moves with constant speed 25 m/s because both have the same acceleration g.
2. Time = 50/25 = 2 s.

Teacher's Note:
1. Subtracting the two motions cancels the gravitational acceleration.
2. Check: block falls 20 m, bullet rises 50 − 20 = 30 m, and 50 − 20 = 30.

Answer: (c) 40/3 m/s

Explanation:
1. Time of flight \( T = \frac{80}{30} = \frac83 \) s.
2. For vertical projection \( T = \frac{2u}{g} \), so \( u = \frac{gT}{2} = \frac{10}{2}\times\frac83 = \frac{40}{3} \) m/s.

Teacher's Note:
1. The particle keeps the cart's horizontal velocity, so it returns to the cart.
2. Time of flight comes from the vertical motion only.

Answer: (d) 5 m/s

Explanation:
1. The two velocities are perpendicular.
2. Relative speed \( = \sqrt{3^2+4^2} = 5 \) m/s.

Teacher's Note:
1. Relative velocity is a vector difference.
2. For perpendicular velocities use Pythagoras.

Answer: (b) 17 m/s

Explanation:
1. \( \vec v_T = \vec v_{T/C} + \vec v_C = 15\hat j + 8\hat i \).
2. Speed \( = \sqrt{15^2+8^2} = 17 \) m/s.

Teacher's Note:
1. Add the car's velocity to the apparent velocity of the train.
2. 8-15-17 is a Pythagorean triple.

Answer: (c) 0.75 minute

Explanation:
1. The river flow does not affect the crossing time: \( t = \frac{0.5}{40} \) h \( = 0.0125 \) h.
2. = 0.75 minute.

Teacher's Note:
1. Only the velocity component normal to the bank determines the crossing time.
2. The river will carry the boat 0.375 km downstream.

Answer: (a) at an angle \( \sin^{-1}(2/5) \) with PQ, upstream

Explanation:
1. To cancel the drift, the upstream component of the boat's velocity must equal the river speed.
2. \( 10\sin\theta = 4 \Rightarrow \theta = \sin^{-1}\frac25 \), measured from PQ and pointing upstream.

Teacher's Note:
1. Aim upstream to cancel the drift.
2. The crossing then takes longer than the minimum time.

Answer: (b) 12 km/h and 4 km/h

Explanation:
1. Let the swimmer's speed be v and the river's speed u: v + u = 16 and v − u = 8.
2. Adding gives v = 12 km/h; subtracting gives u = 4 km/h.

Teacher's Note:
1. The speeds add downstream and subtract upstream.
2. Check: 12 + 4 = 16 and 12 − 4 = 8.

Answer: (d) 1/3 m/s

Explanation:
1. Minimum time heading: the drift is \( u\times10 = 120 \) m, so u = 12 m/min and the width \( w = 10v \).
2. Shortest route: \( w = \sqrt{v^2-u^2}\times12.5 \), so \( 10v = 12.5\sqrt{v^2-144} \).
3. Squaring, 100v² = 156.25v² − 22500, so v² = 400 and v = 20 m/min = 1/3 m/s.

Teacher's Note:
1. Draw both velocity triangles: heading perpendicular, and heading upstream with zero drift.
2. Convert 20 m/min to m/s by dividing by 60.

Answer: (c) \( 5\sqrt5 \) m/s

Explanation:
1. Velocity component across the river = 60/6 = 10 m/s.
2. The path is at 45° to the flow, so the downstream ground velocity is also 10 m/s. The swimmer's own downstream component is 10 − 5 = 5 m/s.
3. Speed relative to water \( = \sqrt{5^2+10^2} = 5\sqrt5 \) m/s.

Teacher's Note:
1. Ground velocity = swimmer velocity + river velocity.
2. Resolve into components along and across the river.

Answer: (c) \( \frac{2}{\sqrt5} \)

Explanation:
1. Width \( w = 10v \) and the drift in the first crossing is \( 10u \).
2. On the way back he points along the line joining his position to the start, whose length is \( 10\sqrt{u^2+v^2} \). The component of his velocity across the river is \( \frac{v\cdot10v}{10\sqrt{u^2+v^2}} \).
3. Time \( 15 = \frac{10v\sqrt{u^2+v^2}}{v^2} \Rightarrow \sqrt{u^2+v^2} = 1.5v \Rightarrow u^2 = 1.25v^2 \), so \( \frac vu = \frac2{\sqrt5} \).

Teacher's Note:
1. The across-river component of velocity fixes the crossing time.
2. Note that the river flow does not change the time for a given heading.

Answer: (d) \( \sqrt{109} \) km/h

Explanation:
1. Rain relative to man \( = \vec v_r - \vec v_m \), with components 10 down and 3 backwards.
2. Speed \( = \sqrt{10^2+3^2} = \sqrt{109} \) km/h.

Teacher's Note:
1. The man sees the rain slanting towards him.
2. The two components are perpendicular.

Answer: (b) it will always be vertically below the aeroplane

Explanation:
1. At release the bomb has the same velocity as the plane, and the plane's acceleration is zero.
2. The relative acceleration of the bomb with respect to the plane is g, vertically downward, with zero initial relative velocity. So the bomb always remains directly below, whatever the direction of flight.

Teacher's Note:
1. Choose the frame of the aeroplane.
2. The relative motion is a straight vertical fall.

Answer: (c) π

Explanation:
1. \( |\vec v_1 - \vec v_2|^2 = v_1^2 + v_2^2 - 2v_1v_2\cos\theta \).
2. This is greatest when cos θ = −1, that is, θ = π, with value v₁ + v₂.

Teacher's Note:
1. Opposite velocities give the largest relative speed.
2. Equal velocities (θ = 0) give the smallest.

Answer: (a) 35 m/s

Explanation:
1. Relative to the man the stone starts with 5 m/s downward.
2. The stone accelerates 10 m/s² downward and the man 5 m/s² upward, so the relative acceleration is 15 m/s² downward.
3. Relative velocity after 2 s = 5 + 15 × 2 = 35 m/s.

Teacher's Note:
1. Relative acceleration is the difference of the two accelerations.
2. The stone must be thrown after the balloon starts moving up for this to apply.