JEE Physics Kinematics MCQs Set 03

Answer: (b) north-west

Explanation:
1. The wind displacement is the same for both planes, equal to the vector from C to D.
2. D is north of the centre and C is east of it, so CD points to the north-west.

Teacher's Note:
1. Actual displacement = intended displacement + wind displacement.
2. Each plane's intended displacement ends at C, so the wind displacement is CD.

Answer: (d) T/2

Explanation:
1. In the lift frame the effective acceleration is \( g + g = 2g \).
2. Time of flight \( = \frac{2v_0}{2g} = \frac{T}{2} \).

Teacher's Note:
1. Relative to an accelerating lift the stone feels a larger effective gravity.
2. Time of flight is inversely proportional to the effective g.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The relative acceleration is zero, so the relative velocity is constant and non-zero (different velocities). The separation starts at zero and grows steadily, so they never meet again.
2. Both statements are true and R explains A.

Teacher's Note:
1. Different initial velocities mean a non-zero relative velocity.
2. Collision would require the separation to return to zero.

Answer: (a) the information is insufficient to decide the relation of xA with xB

Explanation:
1. The motion reverses at \( t = u/a \), and this time is not given.
2. If the particle turns before 10 s, xB can be smaller or larger than xA depending on a. So no relation can be fixed.

Teacher's Note:
1. Displacement is not the same as distance when the direction reverses.
2. Try u = 10, a = 1 and u = 10, a = 5 to see that the answer changes.

Answer: (c) \( \sqrt{\frac54} \)

Explanation:
1. BD = 10 − 2 = 8 m. In the circle, \( BC^2 = AB\times BD = 16 \), so BC = 4 m.
2. \( AC = \sqrt{AB^2+BC^2} = \sqrt{4+16} = 2\sqrt5 \) m.
3. Equal times give \( \frac{u_1}{u_2} = \frac{AC}{BC} = \frac{2\sqrt5}{4} = \sqrt{\frac54} \).

Teacher's Note:
1. The perpendicular from a point on a diameter to the circle satisfies BC² = AB × BD.
2. Equal times mean speeds are in the ratio of the distances.

Answer: (a) 1/2 unit

Explanation:
1. Distance in the n-th second \( = u + a\left(n - \tfrac12\right) = n \) for every n.
2. Comparing the coefficients of n: a = 1; the constant terms: \( u - \tfrac12a = 0 \), so u = 1/2 unit.

Teacher's Note:
1. The relation must hold for every n, so compare coefficients.
2. The speed in the first second then averages 1 unit.

Answer: (b) all four

Explanation:
1. A particle speeds up when its velocity and acceleration have the same sign.
2. 1: v = −8.1t², a = −16.2t (both negative). 2: v = 8.1t², a = 16.2t (both positive). 3: v = 5.4t, a = 5.4. 4: v = −3.4 − 5.4t, a = −5.4. In every case the signs agree for t > 0.

Teacher's Note:
1. Check the sign of v·a, not the sign of a alone.
2. A negative acceleration with negative velocity still increases speed.

Answer: (b) \( T = \sqrt{\frac{2h}{g}} + \frac hv \)

Explanation:
1. Time for the stone to fall: \( t_1 = \sqrt{2h/g} \).
2. Time for the sound to travel up: \( t_2 = h/v \).
3. The total is T = t₁ + t₂.

Teacher's Note:
1. Two separate motions: free fall, then sound at constant speed.
2. Add the times.

Answer: (a) \( \frac12gt_1t_2 \)

Explanation:
1. Let the time to reach the top be \( t_a \). Time from B to the top is \( t_a - t_1 \) and then \( t_a \) to fall to the ground, so \( t_2 = 2t_a - t_1 \).
2. Height of B \( = \tfrac12gt_a^2 - \tfrac12g(t_a-t_1)^2 = \tfrac12gt_1(2t_a - t_1) = \tfrac12gt_1t_2 \).

Teacher's Note:
1. Use the symmetry of the up and down journeys.
2. The height is the difference between the top height and the extra height above B.

Answer: (d) 60

Explanation:
1. Time to reach the maximum height: \( t = \sqrt{2h/g} = \sqrt{1} = 1 \) s.
2. A ball is thrown every 1 s, so 60 balls per minute.

Teacher's Note:
1. The interval between throws equals the time of ascent.
2. The time of ascent equals the time of descent.

Answer: (d) 60 m

Explanation:
1. The relative speed is constant, 20 m/s, because both have acceleration g.
2. Height = relative speed × time = 20 × 3 = 60 m.

Teacher's Note:
1. The gravitational terms cancel in the relative motion.
2. Check: the first stone falls 45 m, the second rises 60 − 45 = 15 m (20 × 3 − 45 = 15).

Answer: (a) 1.2 m/s

Explanation:
1. \( \tfrac12(v^2 - u^2) = \int a\,dx = 0.4(0.4) + \tfrac12(0.4+0.2)(0.4) + 0.2(0.6) = 0.16+0.12+0.12 = 0.40 \).
2. \( v^2 = 0.64 + 0.80 = 1.44 \), so v = 1.2 m/s.

Teacher's Note:
1. Area under an a-x graph gives half the change in v².
2. Split the area into a rectangle, a trapezium and another rectangle.

Answer: (c) a straight line falling from a positive value to zero at t = T, then a straight line rising to the same value at t = 2T

Explanation:
1. The velocity falls linearly from +v₀ at t = 0 through zero at T to −v₀ at 2T.
2. Speed is the magnitude, so it falls from v₀ to zero at T and rises back to v₀ at 2T, a V shape.

Teacher's Note:
1. Take the modulus of the v-t graph to get the speed-time graph.
2. The speed is zero at the highest point of x.

Answer: (a) t = 5 s

Explanation:
1. Between 4 s and 6 s the velocity falls from 10 to −20 m/s, a rate of 15 m/s². It becomes zero after 10/15 = 0.67 s, that is at t ≈ 4.67 s, not at 5 s.
2. The velocity is also zero at t = 0 and t = 8 s.

Teacher's Note:
1. Find the zero crossing of the straight segment between 4 s and 6 s.
2. Interpolate: 10 − 15(t − 4) = 0.

Answer: (b) −5 m

Explanation:
1. Displacement in 0 to 2 s = area \( = \tfrac12(2)(10) = 10 \) m.
2. Position = −15 + 10 = −5 m.

Teacher's Note:
1. Add the displacement to the initial position.
2. The area under the v-t graph is the displacement.

Answer: (b) 33.3 m

Explanation:
1. Displacement is largest when v becomes zero at t ≈ 4.67 s.
2. Areas: 0 to 2 s: 10 m; 2 to 4 s: 20 m; 4 to 4.67 s: ½ × 0.67 × 10 = 3.33 m. Total = 33.3 m.

Teacher's Note:
1. The particle moves away until its velocity reverses.
2. After 4.67 s it returns towards the origin.

Answer: (a) The mean speed of the car is π/4 m/s

Explanation:
1. Distance travelled = area of the semicircle \( = \frac{\pi(1)^2}{2} = \frac\pi2 \) m.
2. Mean speed \( = \frac{\pi/2}{2} = \frac\pi4 \) m/s.

Teacher's Note:
1. The shape of a v-t graph says nothing about the shape of the path.
2. At the peak of the semicircle the slope is zero, so the acceleration is zero there.

Answer: (d) continuously increases

Explanation:
1. \( v_1 = a_1t \) and \( v_2 = a_2t \), so \( v_{12} = (a_1-a_2)t \).
2. It is zero at t = 0 and grows linearly with time.

Teacher's Note:
1. The relative acceleration is constant and non-zero.
2. Straight lines of different slopes on a v-t graph diverge.

Answer: (d) x = 1 m

Explanation:
1. \( a = v\frac{dv}{dx} = \ln x\cdot\frac1x \).
2. a = 0 when \( \ln x = 0 \), that is, x = 1 m.

Teacher's Note:
1. Zero net force means zero acceleration.
2. Use a = v dv/dx when velocity depends on position.

Answer: (a) \( \left(\frac{2}{3\pi}, \frac{2}{3\pi^2}\right) \)

Explanation:
1. \( v_x = \frac{2}{3\pi}(1-\cos3\pi t) \) and \( x = \frac{2}{3\pi}\left(t - \frac{\sin3\pi t}{3\pi}\right) \). At t = 1: \( x = \frac{2}{3\pi} \).
2. \( v_y = \frac{\sin3\pi t}{\pi} \) and \( y = \frac{1-\cos3\pi t}{3\pi^2} \). At t = 1: \( y = \frac{2}{3\pi^2} \).

Teacher's Note:
1. Integrate a = F/m twice with zero initial conditions.
2. Do each component separately.

Answer: (c) \( \frac12 < t < 1 \)

Explanation:
1. \( a = 2t - 1 \) and \( v = t(t-1) \).
2. The particle retards when v and a have opposite signs. For \( \frac12 < t < 1 \), v < 0 and a > 0. For t < 1/2 both are negative and for t > 1 both are positive, so it speeds up there.

Teacher's Note:
1. Retardation means speed decreasing, not acceleration negative.
2. Compare the signs of v and a.

Answer: (a) \( \frac1a\ln(1+aut) \)

Explanation:
1. \( v\frac{dv}{dx} = -av^2 \Rightarrow \frac{dv}{v} = -a\,dx \Rightarrow v = ue^{-ax} \).
2. \( \frac{dx}{dt} = ue^{-ax} \Rightarrow \int e^{ax}dx = \int u\,dt \Rightarrow \frac{e^{ax}-1}{a} = ut \).
3. So \( x = \frac1a\ln(1+aut) \).

Teacher's Note:
1. The speed decays but never reaches zero in finite time.
2. For small a, x ≈ ut, as expected.

Answer: (d) A is false, but R is true.

Explanation:
1. Statement (R) is true for constant acceleration.
2. Statement (A) is false: if a particle is thrown up, u and v are equal and opposite over a suitable interval, and the average velocity (u + v)/2 is zero.

Teacher's Note:
1. A particle that returns to its starting point has zero displacement.
2. Zero average velocity is possible when the velocity reverses.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. At t = 2 s, v = 4 − 12 + 8 = 0, the least possible speed. So (A) is true.
2. (R) is a true general statement. But at t = 2 s, dv/dt = 2t − 6 = −2, not zero, so (R) does not explain (A).

Teacher's Note:
1. A speed of zero is a minimum regardless of the slope.
2. Zero acceleration is where v is extreme, which for this motion is at t = 3 s (v = −1 m/s).

Answer: (b) \( -\frac{2}{\pi^2}\sin\pi t + \frac{2t}{\pi} \)

Explanation:
1. \( v = \frac2\pi(1-\cos\pi t) \) is never negative, so distance equals \( \int_0^tv\,dt \).
2. \( = \frac{2t}{\pi} - \frac{2}{\pi^2}\sin\pi t \).

Teacher's Note:
1. The term sin πt averages out over each period.
2. For large t the distance grows at the mean speed 2/π m/s.

Answer: (c) 0° < θ < 180°

Explanation:
1. Initially the velocity points upward at an angle with g, so θ is greater than 90° (but less than 180°).
2. At the top θ = 90°; afterwards θ is less than 90° but greater than 0°. So the angle covers values on both sides of 90° and lies in (0°, 180°).

Teacher's Note:
1. θ is exactly 90° at only one instant.
2. θ never reaches 0° or 180° because there is always a horizontal velocity.

Answer: (d) 180y = 240x − x²

Explanation:
1. \( y = x\tan53^\circ - \frac{gx^2}{2u^2\cos^253^\circ} = \frac43x - \frac{10x^2}{2(2500)(0.36)} \).
2. \( = \frac43x - \frac{x^2}{180} \), so 180y = 240x − x².

Teacher's Note:
1. The coefficient of x² is negative for a path that is concave downward.
2. cos 53° = 3/5, tan 53° = 4/3.

Answer: (b) \( \frac{2ba}{c} \)

Explanation:
1. The vertical velocity is zero at \( t = b/c \), so the time of flight is \( \frac{2b}{c} \).
2. Range \( = a\times\frac{2b}{c} = \frac{2ab}{c} \).

Teacher's Note:
1. The horizontal component a is constant.
2. The time of flight is twice the time to the top.