JEE Physics Motion In A Plane MCQs Set 13

Answer: (a) 4 : 3

Explanation:
1. Ball 1: \(H_1=\dfrac{u^2}{2g}\). Ball 2 makes 60° with the horizontal: \(H_2=\dfrac{u^2\sin^2 60^\circ}{2g}=\dfrac34\cdot\dfrac{u^2}{2g}\)
2. Potential energy \(\propto H\): \(\dfrac{PE_1}{PE_2}=\dfrac{1}{3/4}=4:3\)

Teacher's Note:
1. Potential energy at the top is mgH.
2. Convert the angle with the vertical to the angle with the horizontal.

Answer: (c) \(u\cot\theta\)

Explanation:
1. The horizontal component of v is \(u\cos\theta\) always
2. Let the velocity vector make angle \(\phi\) with the horizontal; being perpendicular to u means \(\phi=\theta-90^\circ\)
3. \(v\cos(90^\circ-\theta)=u\cos\theta\Rightarrow v\sin\theta=u\cos\theta\Rightarrow v=u\cot\theta\)

Teacher's Note:
1. The horizontal component of velocity never changes.
2. Perpendicular directions differ by 90 degrees.

Answer: (b) \(54\hat i+18\hat j\)

Explanation:
1. \(\vec a=\dfrac{\vec F}m=6t^2\hat i+4t\hat j\)
2. \(\vec v=\int\vec a\,dt=2t^3\hat i+2t^2\hat j\)
3. At t = 3 s: \(\vec v=54\hat i+18\hat j\)

Teacher's Note:
1. Integrate acceleration to get velocity, with zero initial velocity.
2. Here m = 1 kg so a equals F numerically.

Answer: (b) 40 m

Explanation:
1. Time of fall \(t=\sqrt{\dfrac{2(80)}{10}}=4\) s
2. \(x=10\times4=40\) m

Teacher's Note:
1. A horizontal throw has zero initial vertical velocity.
2. The time depends only on the height.

Answer: (b) 250 m

Explanation:
1. \(90\) km/h \(=25\) m/s
2. \(t=\sqrt{\dfrac{2(490)}{9.8}}=10\) s
3. \(x=25\times10=250\) m

Teacher's Note:
1. Convert the speed into m/s first.
2. The packet keeps the plane's horizontal velocity.

Answer: (d) 6.2 m/s

Explanation:
1. Fall time \(t=\sqrt{\dfrac{2(4.9)}{9.8}}=1\) s
2. Minimum speed \(=\dfrac{6.2}{1}=6.2\) m/s

Teacher's Note:
1. He must cover the horizontal gap in the time it takes to fall.
2. A smaller speed would make him land short of the roof.

Answer: (b) 300 m

Explanation:
1. \(t=\sqrt{\dfrac{2(19.6)}{9.8}}=2\) s
2. Horizontal distance \(=150\times2=300\) m

Teacher's Note:
1. The bomb keeps the horizontal speed of the aeroplane.
2. Release it before reaching the target.

Answer: (c) 4 m/s

Explanation:
1. The vertical drop is \(25-5=20\) m, so \(t=\sqrt{\dfrac{2(20)}{10}}=2\) s
2. Minimum speed \(=\dfrac82=4\) m/s

Teacher's Note:
1. Only the difference of the roof heights matters.
2. The time of fall fixes the minimum horizontal speed.

Answer: (a) 10 m/s

Explanation:
1. Drop \(=39.2-19.6=19.6\) m, so \(t=\sqrt{\dfrac{2(19.6)}{9.8}}=2\) s
2. Minimum speed \(=\dfrac{20}2=10\) m/s

Teacher's Note:
1. Subtract the heights to get the vertical drop.
2. Use the fall time with the horizontal gap.

Answer: (b) 6 m/s

Explanation:
1. Vertical drop \(=150-27.5=122.5\) m
2. \(t=\sqrt{\dfrac{2(122.5)}{9.8}}=5\) s
3. Speed \(=\dfrac{30}5=6\) m/s

Teacher's Note:
1. Take the height difference between the two windows.
2. Horizontal distance divided by time gives the speed.

Answer: (b) 5 : 2

Explanation:
1. Both fall through the same height, so the times of fall are equal
2. Horizontal range \(=ut\propto u\)
3. Ratio \(=100:40=5:2\)

Teacher's Note:
1. Equal heights give equal times of fall.
2. Range then depends only on the horizontal speed.

Answer: (d) 3 s

Explanation:
1. Fall of 45 m from 90 m: \(45=\tfrac12(10)t^2\)
2. \(t^2=9\Rightarrow t=3\) s

Teacher's Note:
1. Horizontal speed has no effect on the vertical motion.
2. Use the vertical drop only.

Answer: (b) \(\sqrt{u^2t^2+\dfrac14g^2t^4}\)

Explanation:
1. Horizontal displacement \(x=ut\) and vertical displacement \(y=\tfrac12gt^2\)
2. Displacement \(=\sqrt{x^2+y^2}=\sqrt{u^2t^2+\dfrac14g^2t^4}\)

Teacher's Note:
1. The displacement is the vector sum of the horizontal and vertical displacements.
2. Option (c) is dimensionally wrong.

Answer: (d) 30°

Explanation:
1. Vertical velocity at the ground \(v_y=gt=9.8\times10=98\) m/s
2. \(\tan\phi=\dfrac{v_y}{v_x}=\dfrac{98}{98\sqrt3}=\dfrac1{\sqrt3}\)
3. \(\phi=30^\circ\)

Teacher's Note:
1. The horizontal velocity stays constant at \(98\sqrt3\) m/s.
2. The angle with the horizontal comes from the ratio of vertical to horizontal speed.

Answer: (d) \(\dfrac23\) s

Explanation:
1. \(\tan30^\circ=\dfrac{gt}{u}\)
2. \(\dfrac1{\sqrt3}=\dfrac{10t}{20/\sqrt3}\Rightarrow10t=\dfrac{20}3\)
3. \(t=\dfrac23\) s

Teacher's Note:
1. The angle with the initial (horizontal) velocity satisfies \(\tan\phi=gt/u\).
2. The horizontal velocity is unchanged.

Answer: (b) \(\dfrac{40}{3\sqrt3}\) m

Explanation:
1. The angle becomes 30° at \(t=\dfrac23\) s (since \(\tan30^\circ=gt/u\))
2. \(x=ut=\dfrac{20}{\sqrt3}\times\dfrac23=\dfrac{40}{3\sqrt3}\) m

Teacher's Note:
1. First find the time at which the angle is 30°.
2. Horizontal displacement is horizontal speed times time.

Answer: (c) 7 s

Explanation:
1. Vertical motion: \(u_y=50\sin30^\circ=25\) m/s and displacement \(-70\) m
2. \(-70=25t-5t^2\Rightarrow t^2-5t-14=0\)
3. \((t-7)(t+2)=0\Rightarrow t=7\) s

Teacher's Note:
1. Take upward as positive and the final displacement as negative.
2. Reject the negative root.

Answer: (c) 5 m/s

Explanation:
1. The velocities are perpendicular: the monkey moves vertically and the dog horizontally
2. Relative velocity components are 3 m/s and 4 m/s at right angles
3. Relative speed \(=\sqrt{3^2+4^2}=5\) m/s

Teacher's Note:
1. Perpendicular velocities combine like the sides of a right triangle.
2. Relative speed uses the difference of the vectors, not the sum of the speeds.

Answer: (b) west-north direction

Explanation:
1. \(\vec v_{CT}=\vec v_C-\vec v_T=(0,v)-(v,0)=(-v,\,v)\)
2. The result points west and north equally, so the car appears to move towards the west-north direction

Teacher's Note:
1. Relative velocity is the vector difference.
2. The reversed train velocity points towards west.

Answer: (b) \(5\sqrt2\) km/min towards north-west

Explanation:
1. \(\vec v_B=(-5,0)\) and \(\vec v_A=(0,-5)\) (east, north positive)
2. \(\vec v_{BA}=\vec v_B-\vec v_A=(-5,+5)\)
3. Magnitude \(5\sqrt2\) km/min towards north-west

Teacher's Note:
1. Subtract vectors component by component.
2. Westward and northward components make a north-west vector.

Answer: (a) \(\dfrac du\)

Explanation:
1. Consider the line PQ. P moves along PQ with speed u, while Q moves towards R, which is perpendicular to PQ, so the motion of Q does not change the distance PQ at that instant
2. By symmetry the figure remains a square, so the gap PQ closes at the constant rate u
3. Time \(=\dfrac du\)

Teacher's Note:
1. In a square, the pursued person moves at right angles to the line joining them.
2. That means the distance between them decreases at rate u only.

Answer: (c) \(40\sqrt2\) km/h, S-W

Explanation:
1. \(\vec v_{BT}=\vec v_B-\vec v_T=(0,-40)-(40,0)=(-40,-40)\)
2. Magnitude \(40\sqrt2\) km/h, directed towards the south-west

Teacher's Note:
1. The train's velocity is subtracted from that of the bird.
2. Both components are negative, which points south-west.

Answer: (a) 50 km/h in the N-E direction

Explanation:
1. \(\vec v_{BT}=(0,40)-(-30,0)=(30,40)\) (east, north)
2. Magnitude \(\sqrt{30^2+40^2}=50\) km/h
3. Both components are positive: north-east

Teacher's Note:
1. Subtracting the train's westward velocity adds an eastward component.
2. The speed is a 3-4-5 triangle scaled by 10.