JEE Physics Motion In A Plane MCQs Set 09

Answer: (c) \(100\sqrt{3}\ \text{m}\)

Explanation:
1. The swimmer is slower than the river, so the drift cannot be zero. Minimum drift occurs when the swimmer's velocity relative to water is perpendicular to the resultant velocity.
2. That gives \(\sin\theta = \frac{5}{10}\) where \(\theta\) is the angle between the resultant and the river flow, so the resultant makes \(30^\circ\) with the bank.
3. Drift \(= \text{width}\times\cot 30^\circ = 100\sqrt{3}\ \text{m}\) (equivalently \(d\,\frac{\sqrt{u^2 - v^2}}{v} = 100\cdot\frac{\sqrt{75}}{5}\)).

Teacher's Note:
1. When the river speed is greater than the swimmer's speed, drift can only be minimised, not eliminated.
2. The minimum drift formula is \(d\,\frac{\sqrt{u^2 - v^2}}{v}\) for \(u > v\).

Answer: (a) \(y = 2x - 5x^2\)

Explanation:
1. Here \(u_x = 1\) and \(u_y = 2\), so \(x = t\) and \(y = 2t - 5t^2\).
2. Eliminating \(t\): \(y = 2x - 5x^2\).

Teacher's Note:
1. Use \(y = x\tan\theta - \frac{gx^2}{2u_x^2}\).
2. Here \(\tan\theta = 2\) and \(\frac{g}{2u_x^2} = 5\).

Answer: (a) \(3\)

Explanation:
1. The shortest path is straight across, so the resultant velocity is perpendicular to the bank: \(v_{res} = \frac{1\ \text{km}}{0.25\ \text{h}} = 4\ \text{km h}^{-1}\).
2. The boat must point upstream so that its component along the flow cancels the river speed: \(v_{res}^2 + u^2 = 5^2\).
3. \(u = \sqrt{25 - 16} = 3\ \text{km h}^{-1}\).

Teacher's Note:
1. Convert minutes to hours before dividing.
2. The boat speed is the hypotenuse of the velocity triangle.

Answer: (d) \((R + H),\ H\)

Explanation:
1. The horizontal acceleration does not affect the vertical motion, so the time of flight \(T\) and the maximum height \(H\) stay the same.
2. The extra horizontal displacement is \(\frac{1}{2}\cdot\frac{g}{4}T^2 = \frac{gT^2}{8}\).
3. Since \(H = \frac{1}{2}g\left(\frac{T}{2}\right)^2 = \frac{gT^2}{8}\), the extra range is \(H\), so the range becomes \(R + H\).

Teacher's Note:
1. Horizontal and vertical motions are independent.
2. Express the extra range in terms of \(H\) using the time of flight.

Answer: (a) \(1.15\ \text{s}\)

Explanation:
1. \(v_x = 20\cos 60^\circ = 10\ \text{m s}^{-1}\) is constant. The velocity makes \(30^\circ\) with the horizontal when \(|v_y| = 10\tan 30^\circ = \frac{10}{\sqrt{3}}\ \text{m s}^{-1}\).
2. One instant is on the way up (\(v_y = +\frac{10}{\sqrt{3}}\)) and the other on the way down (\(v_y = -\frac{10}{\sqrt{3}}\)).
3. The time gap is \(\frac{2\times 10/\sqrt{3}}{10} = \frac{2}{\sqrt{3}} = 1.15\ \text{s}\).

Teacher's Note:
1. The same angle occurs twice, once rising and once falling.
2. The time gap equals the change in \(v_y\) divided by \(g\).

Answer: (b) \(\frac{\sin\theta_2}{\cos\theta_1}\)

Explanation:
1. Both balls have the same downward acceleration \(g\), so in the frame of one ball the other moves in a straight line at constant velocity.
2. For a collision their horizontal positions must be equal at all times: \(v_0\sin\theta_2\cdot t = kv_0\cos\theta_1\cdot t\).
3. So \(k = \frac{\sin\theta_2}{\cos\theta_1}\), because \(A\)'s horizontal component uses the angle from the vertical.

Teacher's Note:
1. Equal accelerations cancel in relative motion, so only the horizontal condition decides \(k\).
2. Check which angle is measured from the vertical before choosing sine or cosine.

Answer: (c) \(\frac{a^2}{4b},\ \tan^{-1}(a)\)

Explanation:
1. The slope at the origin is \(\tan\theta = a\), so \(\theta = \tan^{-1}(a)\).
2. The height is maximum when \(\frac{dy}{dx} = a - 2bx = 0\), so \(x = \frac{a}{2b}\).
3. \(H = a\cdot\frac{a}{2b} - b\cdot\frac{a^2}{4b^2} = \frac{a^2}{4b}\).

Teacher's Note:
1. Compare with \(y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}\).
2. The maximum of a downward parabola lies at \(x = \frac{a}{2b}\).

Answer: (c) \(16\sqrt{2}\ \text{m}\)

Explanation:
1. Each segment is \(10\times 0.8 = 8\ \text{m}\). Turns of \(90^\circ\) make consecutive segments perpendicular.
2. The largest displacement comes from turning alternately right and left: east, north, east, north.
3. Displacement \(= \sqrt{(16)^2 + (16)^2} = 16\sqrt{2}\ \text{m}\). A straight \(32\ \text{m}\) is not possible because the person cannot go straight on or reverse.

Teacher's Note:
1. Two segments along the same line add up to \(16\ \text{m}\).
2. Perpendicular components combine as the hypotenuse.

Answer: (c) \(\frac{v}{2}\sqrt{1 + 3\cos^2\theta}\)

Explanation:
1. The horizontal component of average velocity is \(v\cos\theta\) (constant). The vertical component goes from \(v\sin\theta\) to \(0\), so its average is \(\frac{v\sin\theta}{2}\).
2. \(v_{avg} = \sqrt{v^2\cos^2\theta + \frac{v^2\sin^2\theta}{4}}\).
3. \(= \frac{v}{2}\sqrt{4\cos^2\theta + \sin^2\theta} = \frac{v}{2}\sqrt{1 + 3\cos^2\theta}\).

Teacher's Note:
1. Average velocity is displacement divided by time, not the average of speeds.
2. Vertical velocity changes uniformly, so its mean is half the initial value.

Answer: (b) \(10\)

Explanation:
1. The maximum range of one jump is at \(45^\circ\): \(R_{max} = \frac{v^2}{g} = \frac{10}{10} = 1\ \text{m}\).
2. Distance \(= 10\ \text{m}\), so the number of jumps \(= \frac{10}{1} = 10\).

Teacher's Note:
1. Maximum range is \(\frac{u^2}{g}\).
2. Landing and taking off again repeats the same motion each time.

Answer: (b) \(51\) units

Explanation:
1. \(v_x = \frac{dx}{dt} = 4t + 6 = 46\) and \(v_y = \frac{dy}{dt} = 2t + 2 = 22\) at \(t = 10\).
2. Speed \(= \sqrt{46^2 + 22^2} = \sqrt{2116 + 484} = \sqrt{2600}\).
3. \(\sqrt{2600} \approx 51\) units.

Teacher's Note:
1. Differentiate each coordinate separately and then combine the components.
2. Speed is the magnitude of the velocity vector.

Answer: (c) Both will hit simultaneously

Explanation:
1. The vertical motion of both bullets is free fall from the same height with zero initial vertical velocity.
2. The time of fall is \(t = \sqrt{\frac{2h}{g}}\), independent of horizontal speed and mass.
3. So both hit the ground at the same time.

Teacher's Note:
1. Horizontal velocity does not influence the time of fall.
2. Mass does not matter in free fall.

Answer: (a) \(15^\circ\)

Explanation:
1. Maximum range is \(R = \frac{u^2}{g}\). The range at angle \(\theta\) is \(\frac{u^2\sin 2\theta}{g}\).
2. Setting \(\frac{u^2\sin 2\theta}{g} = \frac{R}{2}\) gives \(\sin 2\theta = \frac{1}{2}\).
3. \(2\theta = 30^\circ\), so \(\theta = 15^\circ\) (or \(75^\circ\), but only \(15^\circ\) is given).

Teacher's Note:
1. Two complementary angles give the same range.
2. The option set has only one of them.

Answer: (b) \(8\ \text{m s}^{-2}\)

Explanation:
1. \(a_x = \frac{d^2x}{dt^2} = 8\ \text{m s}^{-2}\) and \(a_y = \frac{d^2y}{dt^2} = 0\).
2. The acceleration is constant, so it is \(8\ \text{m s}^{-2}\) at \(t = 5\ \text{s}\) as well.

Teacher's Note:
1. Differentiate twice.
2. A term linear in time gives zero acceleration.

Answer: (d) \(20\sqrt{5}\ \text{m s}^{-1}\)

Explanation:
1. At \(t = 4\ \text{s}\) the particle is at the top, so \(u_y = g\times 4 = 40\ \text{m s}^{-1}\).
2. At \(t = 2\ \text{s}\), \(v_y = 40 - 20 = 20\ \text{m s}^{-1}\), and the angle is \(45^\circ\), so \(v_x = 20\ \text{m s}^{-1}\).
3. \(u = \sqrt{20^2 + 40^2} = 20\sqrt{5}\ \text{m s}^{-1}\).

Teacher's Note:
1. Horizontal motion means the vertical velocity is zero.
2. The horizontal component stays constant throughout.

Answer: (d) \(10\sqrt{2}\ \text{s}\)

Explanation:
1. The bomb's initial vertical velocity is zero, so \(980 = \frac{1}{2}(9.8)t^2\).
2. \(t^2 = 200\), so \(t = 10\sqrt{2}\ \text{s}\).
3. The horizontal speed does not affect the time of fall.

Teacher's Note:
1. Only the vertical motion decides the time.
2. Substitute directly into \(h = \frac{1}{2}gt^2\).

Answer: (d) \(8.66\ \text{m}\)

Explanation:
1. The ball returns to the level of the throw after a time \(T = \frac{2u\sin\theta}{g} = \frac{2\times 10\times 0.5}{10} = 1\ \text{s}\).
2. Horizontal distance \(= u\cos\theta\times T = 10\times 0.866\times 1\).
3. \(= 8.66\ \text{m}\), which is the same as \(\frac{u^2\sin 2\theta}{g}\).

Teacher's Note:
1. The height of the building does not matter at the level of the throw.
2. Use the range formula for the same level.

Answer: (a) \(20\ \text{m}\)

Explanation:
1. The ball also has the truck's horizontal velocity \(10\ \text{m s}^{-1}\) at the instant of throw, and after leaving the hand there is no horizontal acceleration.
2. Time of flight \(= \frac{2\times 10}{10} = 2\ \text{s}\).
3. Horizontal distance from the pole \(= 10\times 2 = 20\ \text{m}\).

Teacher's Note:
1. A projected object keeps the horizontal velocity of its launcher.
2. The truck's acceleration does not act on the ball.

Answer: (b) \(1\ \text{m s}^{-1}\)

Explanation:
1. Rain moves forward with a horizontal component \(v\sin 30^\circ\) (as it seems to fall on his back). For it to appear vertical to the moving man, this horizontal component must equal his speed.
2. \(v\sin 30^\circ = 0.5\), so \(v = 1\ \text{m s}^{-1}\).
3. The vertical component is \(\frac{\sqrt{3}}{2}\ \text{m s}^{-1}\).

Teacher's Note:
1. Rain looks vertical when the man's velocity equals the rain's horizontal velocity.
2. Use the angle with the vertical, so the horizontal part has the sine.

Answer: (b) \(7.5\)

Explanation:
1. Let the distance be \(D\), boat speed \(v\) and stream speed \(u\). \(v + u = \frac{D}{6}\) and \(v - u = \frac{D}{10}\).
2. Adding: \(2v = D\left(\frac{1}{6} + \frac{1}{10}\right) = \frac{4D}{15}\), so \(v = \frac{2D}{15}\).
3. Time in still water \(= \frac{D}{v} = 7.5\ \text{h}\).

Teacher's Note:
1. Add the two speed equations to eliminate the stream speed.
2. The answer is the harmonic mean of the two times.

Answer: (d) \(y^2 = x^2 + \text{constant}\)

Explanation:
1. \(\frac{dx}{dt} = ky\) and \(\frac{dy}{dt} = kx\), so \(\frac{dy}{dx} = \frac{x}{y}\).
2. \(y\,dy = x\,dx\).
3. Integrating: \(y^2 = x^2 + \text{constant}\).

Teacher's Note:
1. Divide the two components to remove time.
2. Separate the variables and integrate.

Answer: (c) \(120^\circ\)

Explanation:
1. To go straight across, the swimmer's component along the flow must cancel the stream: \(4\cos\theta = -2\), where \(\theta\) is measured from the direction of flow.
2. \(\cos\theta = -\frac{1}{2}\).
3. \(\theta = 120^\circ\).

Teacher's Note:
1. The angle with respect to the flow is more than \(90^\circ\) because the swimmer must head upstream.
2. The component along the bank has to be zero.

Answer: (d) \(\tan^{-1}\left(\frac{v\sin\theta - gt}{v\cos\theta}\right)\)

Explanation:
1. At time \(t\), \(v_x = v\cos\theta\) and \(v_y = v\sin\theta - gt\).
2. The angle with the horizontal satisfies \(\tan\alpha = \frac{v_y}{v_x}\).
3. \(\alpha = \tan^{-1}\left(\frac{v\sin\theta - gt}{v\cos\theta}\right)\).

Teacher's Note:
1. The tangent of the angle is vertical component over horizontal component.
2. The angle decreases from \(\theta\) to zero at the top and becomes negative afterwards.