JEE Physics Kinematics MCQs Set 06

Answer: (d) \( \frac{5v_1v_2}{3v_1+2v_2} \)

Explanation:
1. Time = \( (2d/5)/v1 + (3d/5)/v2 = d(2v2 + 3v1)/(5v1v2) \).
2. Average speed = d/time = \( 5v1v2/(3v1 + 2v2) \).

Teacher's Note:
1. Average speed is total distance divided by total time.
2. Do not average the speeds directly.

Answer: (a) 4.0 m/s

Explanation:
1. Let each half be \( d \). Time for the first half = \( d/3 \).
2. Second half in equal time intervals: average speed = \( (4.5 + 7.5)/2 = 6 m/s \), so time = \( d/6 \).
3. Average speed = \( 2d/(d/3 + d/6) = 4 m/s \).

Teacher's Note:
1. For equal time intervals, the average speed is the arithmetic mean.
2. Total distance over total time.

Answer: (c) in the ratio of the differences of the square roots of integers, i.e. \( (\sqrt1-\sqrt0):(\sqrt2-\sqrt1):(\sqrt3-\sqrt2):(\sqrt4-\sqrt3) \)

Explanation:
1. From \( s \) = (1/2)gt², the time to fall \( n \) metres is \( t_n = \sqrt{2n/g} \).
2. The time for the n-th metre = \( t_n - t_{n-1} ∝ \sqrt{n} - \sqrt{n - 1} \).

Teacher's Note:
1. Time to fall \( n \) metres varies as \( \sqrt{n} \).
2. Subtract consecutive totals.

Answer: (b) 1 : 16

Explanation:
1. Stopping distance = \( u^{2}/(2a) \), proportional to the square of the speed.
2. Ratio = \( u^{2} \) : \( 16u^{2} = 1 \) : 16.

Teacher's Note:
1. Use \( v^{2} = u^{2} \) + 2as with \( v = 0 \).
2. The distance varies as the square of the speed.

Answer: (c) both strike the ground with the same velocity

Explanation:
1. A returns to the top with speed \( v \) downward, so it then behaves exactly like \( B \).
2. Both therefore reach the ground with the same speed.

Teacher's Note:
1. Use \( v^{2} = u^{2} \) + 2gh for each ball.
2. The initial speed is the same and the height is the same.

Answer: (c) \( (\text{velocity})^{3/2} \)

Explanation:
1. \( v = dx/dt = -(t + 5)^{-2} \) and \( a \) = dv/dt = \( 2(t + 5)^{-3} \).
2. \( v^{3/2} ∝ (t + 5)^{-3} \), so a \( ∝ v^{3/2} \).

Teacher's Note:
1. Differentiate twice.
2. Compare the powers of \( (t + 5) \).

Answer: (a) 1 m

Explanation:
1. From \( A \) to \( B \): \( (v/2)^{2} = v^{2} - 2g(3) \), so \( v^{2} = 80 \) and the speed at \( B \) is \( v/2 = \sqrt{20} \).
2. Height above \( B = (v/2)^{2}/(2g) = 20/20 = 1 m \).

Teacher's Note:
1. Use \( v^{2} = u^{2} \) - 2gh between \( A \) and \( B \).
2. Then find the further rise from \( B \).

Answer: (d) 26.1 m/s

Explanation:
1. Time for the first ball to fall = \( \sqrt{2 \times 122.5/9.8} = 5 s \).
2. The second ball has \( 3 s \): \( 122.5 = 3u + (1/2)(9.8)(9) = 3u + 44.1 \).
3. \( u = 26.1 m/s \).

Teacher's Note:
1. The second ball must fall the same height in \( 5 - 2 = 3 s \).
2. Use \( s \) = ut + (1/2)gt².

Answer: (d) 5 : 9

Explanation:
1. Distance of \( A \) in its 5th second = \( a1(2 \times 5 - 1)/2 = 4.5a1 \).
2. The 5th second of \( A \) is the 3rd second of \( B \): distance = \( a2(2 \times 3 - 1)/2 = 2.5a2 \).
3. \( 4.5a1 = 2.5a2 \), so \( a1 \) : \( a2 = 5 \) : 9.

Teacher's Note:
1. Use \( s_n = (a/2)(2n - 1) \).
2. \( B \) has been moving for only \( 3 s \) at the end of \( A's \) 5th second.

Answer: (d) \( \frac{2v}{a} \)

Explanation:
1. (1/2)at² = vt, so \( t = 2v/a \) (excluding \( t = 0 \)).

Teacher's Note:
1. Equate the two displacements.
2. Reject the trivial solution \( t = 0 \).

Answer: (d) \( \frac{1}{x^3} \)

Explanation:
1. Differentiating \( x^{2} = 1 + t^{2} \): xv = \( t \), so \( v = t/x \).
2. \( a \) = dv/dt = \( (1/x) - t v/x^{2} = (1/x) - t^{2}/x^{3} = (x^{2} - t^{2})/x^{3} = 1/x^{3} \).

Teacher's Note:
1. Differentiate implicitly twice.
2. Use \( x^{2} - t^{2} = 1 \).

Answer: (b) 80 m

Explanation:
1. First ball: \( h1 = (1/2)(10)(5^{2}) = 125 m \).
2. Second ball falls for \( 3 s \): \( h2 = (1/2)(10)(3^{2}) = 45 m \).
3. Difference = \( 80 m \).

Teacher's Note:
1. Time of fall of the second ball is \( 5 - 2 = 3 s \).
2. Use \( h \) = (1/2)gt².

Answer: (b) 2 s

Explanation:
1. \( a = g \sin 30^{\circ} = 4.9 m/s^{2} \).
2. \( 9.8 = (1/2)(4.9)t^{2} \), so \( t = 2 s \).

Teacher's Note:
1. Acceleration along a smooth incline is \( g \sin \theta \).
2. Use \( s \) = (1/2)at².

Answer: (a) \( \frac12gt^2 \)

Explanation:
1. Reverse the motion: the last \( t \) seconds of ascent are like a fall from rest for \( t \) seconds.
2. Distance = (1/2)gt².

Teacher's Note:
1. Time-reverse the final part of the motion.
2. The ball is momentarily at rest at the top.

Answer: (c) \( \frac T2+\frac L{gT} \)

Explanation:
1. Separation = \( (1/2)g[t^{2} - (t - T)^{2}] \) = gTt - (1/2)gT².
2. Setting this equal to \( L \) gives \( t \) = L/(gT) \( + T/2 \).

Teacher's Note:
1. Subtract the two distances fallen.
2. Solve the linear equation in \( t \).

Answer: (a) 15 m/s

Explanation:
1. \( v \) = ∫2(t - 1) dt = t² - 2t.
2. At \( t = 5 \): \( v = 25 - 10 = 15 m/s \).

Teacher's Note:
1. Integrate from rest.
2. Substitute \( t = 5 \).

Answer: (c) \( \frac{v_0}{\sqrt{2v_0^2kt+1}} \)

Explanation:
1. ∫dv/v³ = -∫k \( dt \) gives \( 1/v^{2} = 1/v0^{2} \) + 2kt.
2. v = v0/√(1 + 2kv0²t).

Teacher's Note:
1. Separate the variables and integrate.
2. Check that \( v = v0 \) at \( t = 0 \).

Answer: (d) \( \frac{m}{3P}(v_2^3-v_1^3) \)

Explanation:
1. \( v \) dv/dx = P/(mv), so \( v^{2} \) dv = \( (P/m) dx \).
2. \( x \) = (m/P)∫v² dv = \( (m/3P)(v2^{3} - v1^{3}) \).

Teacher's Note:
1. Use \( a = v \) dv/dx.
2. Integrate \( v^{2} \) between \( v1 \) and \( v2 \).

Answer: (d) 1 : 3

Explanation:
1. Velocity = slope = \( \tan \theta \).
2. vA : vB = \( \tan 30^{\circ} \) : \( \tan 60^{\circ} = (1/\sqrt{3}) \) : \( \sqrt{3} = 1 \) : 3.

Teacher's Note:
1. The slope of an \( x-t \) graph is the velocity.
2. Use equal scales on both axes.

Answer: (a) 4 cm/s²

Explanation:
1. Accelerations: 1 cm/s² \( (0-20 s), 0 (20-30 s), 40/10 = 4 \) cm/s² \( (30-40 s) \), and \( -60/30 = -2 \) cm/s² \( (40-70 s) \).
2. The maximum is 4 cm/s².

Teacher's Note:
1. Acceleration is the slope of the \( v-t \) graph.
2. Compare slopes of all the segments.

Answer: (d) -2 m/s²

Explanation:
1. \( v \) at \( A = \tan 45^{\circ} = +1 m/s \) and \( v \) at \( B = \tan 135^{\circ} = -1 m/s \).
2. Average acceleration = \( (-1 - 1)/(2 - 1) = -2 m/s^{2} \).

Teacher's Note:
1. Slope of the \( x-t \) graph is the velocity.
2. Average acceleration = change in velocity/time.

Answer: (c) \( \frac{v^2}{2m} \)

Explanation:
1. Acceleration = \( m \), so \( v^{2} \) = 2ms.
2. Distance \( s = v^{2}/(2m) \).

Teacher's Note:
1. The slope of the \( v-t \) line is the acceleration.
2. Use \( v^{2} \) = 2as.