JEE Physics Motion In A Plane MCQs Set 05

Answer: (b) \( \frac1{\sqrt2} \) s

Explanation:
1. Relative to the lift the bolt's acceleration is \( g+a=38 \) ft/s\( ^2 \).
2. \( 9.5=\frac12(38)t^2\Rightarrow t^2=\frac12\Rightarrow t=\frac1{\sqrt2} \) s.

Teacher's Note:
1. Use the acceleration relative to the elevator.

Answer: (d) 39.2 m

Explanation:
1. Time of flight \( =\frac{2\times19.6}{9.8}=4 \) s.
2. Range \( =9.8\times4=39.2 \) m.

Teacher's Note:
1. Range is horizontal velocity times the time of flight.

Answer: (b) \( \frac{t_1t_2}{t_2+t_1} \)

Explanation:
1. Speeds add: \( \frac1{t}=\frac1{t_1}+\frac1{t_2} \).
2. \( t=\frac{t_1t_2}{t_1+t_2} \).

Teacher's Note:
1. Add the rates (reciprocals of times).

Answer: (b) \( 5\sqrt5 \) m/s

Explanation:
1. Horizontal velocity \( =10 \) m/s throughout.
2. Final vertical velocity \( =-10 \): \( -10=u_y-10(1.5)\Rightarrow u_y=5 \) m/s (upward).
3. \( u=\sqrt{10^2+5^2}=5\sqrt5 \) m/s.

Teacher's Note:
1. The horizontal velocity does not change.

Answer: (a) 8 s

Explanation:
1. Bus position: \( 96+t^2 \). Cyclist position: \( 20t \).
2. \( 96+t^2=20t\Rightarrow t^2-20t+96=0\Rightarrow t=8 \) or 12.
3. He first overtakes at 8 s (after which the bus catches up again at 12 s).

Teacher's Note:
1. Equate the positions and take the first root.

Answer: (a) \( 5\sqrt2 \) km/min towards South-East

Explanation:
1. \( \vec v_{BA}=\vec v_B-\vec v_A=(5,0)-(0,5)=(5,-5) \).
2. Magnitude \( 5\sqrt2 \) km/min towards south-east.

Teacher's Note:
1. Subtract A's velocity from B's.

Answer: (c) 4H

Explanation:
1. For the maximum range \( \theta=45^\circ \): \( R=\frac{u^2}g \) and \( H=\frac{u^2}{4g} \).
2. So \( R=4H \).

Teacher's Note:
1. Maximum range is at 45 degrees.

Answer: (c) \( \sqrt{109} \) km/h

Explanation:
1. The rain's velocity relative to the man has a vertical part 10 km/h and a horizontal part 3 km/h (opposite to his motion).
2. \( v=\sqrt{10^2+3^2}=\sqrt{109} \) km/h.

Teacher's Note:
1. Combine the perpendicular components.

Answer: (b) 7 s

Explanation:
1. Relative speed \( =25+5=30 \) m/s.
2. Time \( =\frac{210}{30}=7 \) s.

Teacher's Note:
1. Opposite directions: add the speeds.

Answer: (b) \( \frac{40}{\sqrt3} \) m

Explanation:
1. The time is 2 s (from \( \tan60^\circ=\frac{gt}u \)).
2. \( x=ut=\frac{20}{\sqrt3}\times2=\frac{40}{\sqrt3} \) m.

Teacher's Note:
1. Horizontal velocity is constant.

Answer: (c) 30 m

Explanation:
1. \( H\propto\sin^2\theta \).
2. \( H_{30}=90\times\frac{\sin^230^\circ}{\sin^260^\circ}=90\times\frac{1/4}{3/4}=30 \) m.

Teacher's Note:
1. Compare the two heights by their sine squared.

Answer: (d) \( \frac{u^2\cos^2\theta}{2g} \)

Explanation:
1. The vertical component is \( u\cos\theta \) (angle measured from the vertical).
2. \( H=\frac{u^2\cos^2\theta}{2g} \).

Teacher's Note:
1. Take the component along the vertical.

Answer: (c) \( \frac{2v_1v_2}g \)

Explanation:
1. The time of flight is \( T=\frac{2v_1}g \).
2. The shell also moves horizontally with the ship's speed \( v_2 \), so the range is \( v_2T=\frac{2v_1v_2}g \).

Teacher's Note:
1. The horizontal velocity of the shell equals that of the ship.

Answer: (a) 45 km/h

Explanation:
1. Let the speed be v. The relative speed is \( v+30 \) km/h.
2. The second car is reached after covering the 5 km gap: \( (v+30)\times\frac4{60}=5\Rightarrow v+30=75 \).
3. \( v=45 \) km/h.

Teacher's Note:
1. Use the relative speed to cover the gap in 4 minutes.

Answer: (b) 400 m/s

Explanation:
1. \( R_{max}=\frac{u^2}g\Rightarrow u^2=16000\times10=1.6\times10^5 \).
2. \( u=400 \) m/s.

Teacher's Note:
1. Maximum range is u squared over g.

Answer: (b) 17 m/s towards East

Explanation:
1. The person's velocity relative to the train is 3 m/s towards west.
2. Relative to the ground: \( 20-3=17 \) m/s towards east.

Teacher's Note:
1. Add the velocities with signs.

Answer: (d) \( \frac12\sin^{-1}\left(\frac{Rg}{v^2}\right) \)

Explanation:
1. \( R=\frac{v^2\sin2\theta}g\Rightarrow\sin2\theta=\frac{Rg}{v^2} \).
2. \( \theta=\frac12\sin^{-1}\left(\frac{Rg}{v^2}\right) \).

Teacher's Note:
1. Invert the range formula.

Answer: (c) \( 6\hat i+8\hat j \)

Explanation:
1. \( \vec v_{BW}=\vec v_{BG}-\vec v_{WG}=(3\hat i+4\hat j)-(-3\hat i-4\hat j)=6\hat i+8\hat j \).

Teacher's Note:
1. Relative velocity is the difference of velocities.

Answer: (a) Their horizontal ranges will be the same

Explanation:
1. The angles are complementary, so \( \sin2\theta \) is the same and the ranges are equal.
2. The heights (\( \propto\sin^2\theta \)) and times of flight (\( \propto\sin\theta \)) differ. The landing velocities have the same speed but different directions.

Teacher's Note:
1. Complementary angles give equal ranges.

Answer: (d) 36 s

Explanation:
1. Rates add: \( \frac1{90}+\frac1{60}=\frac{2+3}{180}=\frac1{36} \).
2. Time \( =36 \) s.

Teacher's Note:
1. Add the speeds (reciprocals of times).

Answer: (a) \( \frac{\sqrt3u^2}{2g} \)

Explanation:
1. Speed at the top \( =u\cos\theta=\frac{\sqrt3}2u\Rightarrow\theta=30^\circ \).
2. \( R=\frac{u^2\sin60^\circ}g=\frac{\sqrt3u^2}{2g} \).

Teacher's Note:
1. At the top only the horizontal component remains.

Answer: (d) 1 s

Explanation:
1. \( T=\frac{2u\sin\theta}g=\frac{2\times9.8\times0.5}{9.8}=1 \) s.

Teacher's Note:
1. Use the time of flight formula.

Answer: (b) 9.6 metre

Explanation:
1. Time of flight \( =\frac{2\times8}{10}=1.6 \) s.
2. Range \( =6\times1.6=9.6 \) m.

Teacher's Note:
1. Horizontal velocity times time of flight.

Answer: (a) 1

Explanation:
1. The range is maximum for \( \theta=45^\circ \), so \( \tan\theta=1 \).

Teacher's Note:
1. Maximum range at 45 degrees.

Answer: (a) \( b=2a \)

Explanation:
1. \( R=\frac{2ab}g \) and \( H=\frac{b^2}{2g} \).
2. \( R=2H\Rightarrow\frac{2ab}g=\frac{b^2}g\Rightarrow b=2a \).

Teacher's Note:
1. Write R and H using the components.

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

Explanation:
1. If the perpendicular components differ, the separation perpendicular to the line keeps changing and they cannot meet; the assertion is true.
2. Relative velocity is the vector difference of the velocities and it is not just a component along the line, so the reason is false.

Teacher's Note:
1. Relative velocity is a vector difference.