JEE Physics Motion In A Plane MCQs Set 14

Answer: (b) \(9\hat i+12\hat j\)

Explanation:
1. \(\vec v_{bw}=\vec v_b-\vec v_w=(3\hat i+4\hat j)-(-6\hat i-8\hat j)\)
2. \(=9\hat i+12\hat j\)

Teacher's Note:
1. Velocity relative to water equals velocity relative to ground minus water velocity.
2. Subtracting a negative gives an addition.

Answer: (b) \(13\sqrt2\) m/s

Explanation:
1. \(\vec v_1=3\hat i+8t\hat j\) and \(\vec v_2=8t\hat i+3\hat j\)
2. At t = 2 s: \(\vec v_1=(3,16)\), \(\vec v_2=(16,3)\)
3. \(\vec v_1-\vec v_2=(-13,13)\), relative speed \(=13\sqrt2\) m/s

Teacher's Note:
1. Differentiate the position vectors to get velocities.
2. Then take the magnitude of the difference.

Answer: (c) 4.2 m/s

Explanation:
1. Relative velocity has components 3 m/s (horizontal) and 3 m/s (vertical)
2. \(v=\sqrt{3^2+3^2}=3\sqrt2\approx4.2\) m/s

Teacher's Note:
1. The two components are perpendicular.
2. The rain appears to slant at 45 degrees.

Answer: (d) \(5\sqrt5\) km/h

Explanation:
1. Components: 5 km/h horizontal and 10 km/h vertical
2. \(v=\sqrt{5^2+10^2}=\sqrt{125}=5\sqrt5\) km/h

Teacher's Note:
1. Horizontal and vertical velocities are perpendicular.
2. Do not add the speeds directly.

Answer: (b) \(\tan^{-1}\left(\dfrac32\right)\)

Explanation:
1. The rain relative to the man has horizontal component 30 m/s and vertical component 20 m/s
2. The angle with the vertical satisfies \(\tan\theta=\dfrac{30}{20}=\dfrac32\)
3. \(\theta=\tan^{-1}\left(\dfrac32\right)\)

Teacher's Note:
1. The umbrella must be held along the relative velocity of the rain.
2. The angle with the vertical uses horizontal over vertical.

Answer: (c) \(\tan^{-1}\left(\dfrac uv\right)\)

Explanation:
1. The velocity of rain relative to the boy has horizontal component u and vertical component v
2. The tube must be along this direction: \(\tan\theta=\dfrac uv\)
3. \(\theta=\tan^{-1}\left(\dfrac uv\right)\)

Teacher's Note:
1. The tube must be parallel to the relative velocity of the drops.
2. The angle is measured from the vertical.

Answer: (c) \(\sqrt{164}\) km/h

Explanation:
1. Swimming across is perpendicular to the flow, so the velocities are perpendicular
2. \(v=\sqrt{10^2+8^2}=\sqrt{164}\) km/h

Teacher's Note:
1. Perpendicular velocities add by the Pythagoras theorem.
2. The swimmer is also carried downstream.

Answer: (c) 120° with downstream

Explanation:
1. To go straight south he must cancel the eastward drift, so he swims partly upstream (towards the west)
2. \(\sin\alpha=\dfrac{5}{10}\Rightarrow\alpha=30^\circ\) with the south direction, towards the west
3. Measured from the downstream (east) direction this is \(90^\circ+30^\circ=120^\circ\)

Teacher's Note:
1. For the shortest path the component of swimming velocity against the river must equal the river speed.
2. Check the angle reference carefully.

Answer: (c) 135°

Explanation:
1. To reach the opposite point the upstream component must equal the river speed: \(\sqrt2\sin\alpha=1\Rightarrow\alpha=45^\circ\) from the perpendicular, towards upstream
2. With respect to the flow direction the angle is \(90^\circ+45^\circ=135^\circ\)

Teacher's Note:
1. Upstream component of swimming velocity cancels the river flow.
2. Angle with the flow is measured from the downstream direction.

Answer: (d) \(\dfrac12\) h

Explanation:
1. He must head upstream so that the river drift is cancelled: resultant speed across \(=\sqrt{10^2-6^2}=8\) km/h
2. \(t=\dfrac{4}{8}=\dfrac12\) h

Teacher's Note:
1. Shortest path means no drift, so the resultant is perpendicular to the banks.
2. The resultant speed is found from a right triangle.

Answer: (c) \(\dfrac x{2u}\)

Explanation:
1. Horizontal component of particle 1: \(\sqrt3u\cos30^\circ=\dfrac{3u}2\); of particle 2: \(u\cos60^\circ=\dfrac u2\)
2. They move towards each other horizontally, so the closing speed is \(\dfrac{3u}2+\dfrac u2=2u\)
3. \(t=\dfrac x{2u}\)

Teacher's Note:
1. Horizontal motions are independent of the vertical motions.
2. Relative horizontal speed is the sum when they approach each other.

Answer: (a) 3.5 m/s²

Explanation:
1. An identical trajectory means the same range and height at the same angle: \(\dfrac{u^2}{g}\) must be equal
2. \(\dfrac{25}{9.8}=\dfrac9{g'}\)
3. \(g'=\dfrac{9\times9.8}{25}=3.5\) m/s²

Teacher's Note:
1. Range and height both depend on \(u^2/g\) for a fixed angle.
2. The slower projectile needs weaker gravity.

Answer: (d) \(\dfrac{11}5(\hat i+\hat j)\)

Explanation:
1. Displacement from t = 0 to 5 s \(=(13-2)\hat i+(14-3)\hat j=11\hat i+11\hat j\)
2. \(\vec v_{av}=\dfrac{11}{5}(\hat i+\hat j)\)

Teacher's Note:
1. Average velocity depends only on the initial and final positions.
2. The position at t = 2 s is not needed.

Answer: (b) \(\dfrac{\vec r_1-\vec r_2}{|\vec r_1-\vec r_2|}=\dfrac{\vec v_2-\vec v_1}{|\vec v_2-\vec v_1|}\)

Explanation:
1. For collision, the relative velocity of A with respect to B must point along the line joining A to B
2. The direction of \(\vec r_2-\vec r_1\) must equal the direction of \(\vec v_1-\vec v_2\), which is the direction of \(-(\vec r_1-\vec r_2)\)
3. Hence \(\dfrac{\vec r_1-\vec r_2}{|\vec r_1-\vec r_2|}=\dfrac{\vec v_2-\vec v_1}{|\vec v_2-\vec v_1|}\)

Teacher's Note:
1. Collision needs the relative velocity to be directed along the line joining the particles.
2. Unit vectors on both sides compare directions only.

Answer: (a) 30° west

Explanation:
1. For the shortest path (straight north) the westward component of the swimming velocity must cancel the river's eastward 10 m/s
2. \(20\sin\theta=10\Rightarrow\sin\theta=\dfrac12\Rightarrow\theta=30^\circ\)
3. He must aim 30° west of north

Teacher's Note:
1. The shortest path is straight across, so no drift is allowed.
2. The angle is measured from north towards west.

Answer: (c) \(1:\sqrt3\)

Explanation:
1. Retardation along the plane \(=g\sin\alpha\), so \(x=\dfrac{u^2}{2g\sin\alpha}\)
2. \(\dfrac{x_1}{x_2}=\dfrac{\sin30^\circ}{\sin60^\circ}=\dfrac{1/2}{\sqrt3/2}\)
3. \(x_1:x_2=1:\sqrt3\)

Teacher's Note:
1. Along a smooth incline the acceleration is \(g\sin\alpha\).
2. A steeper plane stops the object sooner.

Answer: (c) \(\tan^{-1}\dfrac12\)

Explanation:
1. At the highest point: horizontal distance \(=\dfrac R2\) and height \(H=\dfrac R4\) at 45°
2. \(\tan\alpha=\dfrac{H}{R/2}=\dfrac{R/4}{R/2}=\dfrac12\)
3. \(\alpha=\tan^{-1}\dfrac12\)

Teacher's Note:
1. At 45° the height is a quarter of the range.
2. The highest point lies at half the range.

Answer: (d) \(\tan^{-1}(4)\)

Explanation:
1. \(\dfrac RH=4\cot\theta=1\)
2. \(\tan\theta=4\)
3. \(\theta=\tan^{-1}4\)

Teacher's Note:
1. Use \(R/H=4\cot\theta\).
2. A range equal to the height means a steep launch.

Answer: (b) \(\tan^{-1}\left(\dfrac12\right)\)

Explanation:
1. \(u_x=10\) m/s and \(u_y=5\) m/s (from \(-10=u_y-10\times1.5\))
2. \(\tan\theta=\dfrac{u_y}{u_x}=\dfrac5{10}=\dfrac12\)

Teacher's Note:
1. The launch angle comes from the initial components.
2. Do not use the final velocity components for the angle of projection.

Answer: (b) Particle B

Explanation:
1. Vertical launch component of A \(=2V\sin30^\circ=V\); of B \(=V\sin60^\circ=0.866V\)
2. The time of flight (from the same height) increases with the upward launch component
3. B has the smaller upward component, so it hits the ground earlier

Teacher's Note:
1. The time to land depends on the vertical component of the launch velocity.
2. The horizontal components do not matter for the landing time.

Answer: (b) 30°

Explanation:
1. The maximum height must not exceed 40 m: \(\dfrac{u^2\sin^2\theta}{2g}\le40\)
2. \(\sin^2\theta\le\dfrac{40\times19.6}{3136}=0.25\Rightarrow\theta\le30^\circ\)
3. The range increases with \(\theta\) up to 45°, so the best allowed angle is 30°

Teacher's Note:
1. The ceiling limits the angle that can be used.
2. Range grows with angle below 45°, so choose the largest allowed angle.

Answer: (b) 19.6 m

Explanation:
1. Equal heights at 1 s and 3 s are symmetric about the highest point, so the highest point is reached at \(t=\dfrac{1+3}2=2\) s
2. \(H=\dfrac{u_y^2}{2g}\) with \(u_y=g\times2\)
3. \(H=\tfrac12g(2)^2=19.6\) m

Teacher's Note:
1. Points at equal heights are symmetric about the top of the path.
2. The time of ascent gives the vertical launch speed directly.

Answer: (b) 0, 10 m/s

Explanation:
1. After a full lap the displacement is zero, so the average velocity is 0
2. Distance per lap \(=2\pi(100)=628\) m in 62.8 s
3. Average speed \(=\dfrac{628}{62.8}=10\) m/s

Teacher's Note:
1. Average velocity needs the displacement; average speed needs the path length.
2. They differ for every complete circle.