JEE Physics Motion In A Plane MCQs Set 06

Answer: (a) 15 km/h

Explanation:
1. \( v_{rel}=80-65=15 \) km/h.

Teacher's Note:
1. Subtract the velocities for the same direction.

Answer: (b) 1.0 km

Explanation:
1. \( R_{75}=\frac{u^2\sin150^\circ}g=\frac{u^2}{2g}=0.5 \) km, so \( \frac{u^2}g=1 \) km.
2. \( R_{45}=\frac{u^2}g=1.0 \) km.

Teacher's Note:
1. Use R = u squared sin 2 theta over g.

Answer: (c) \( 40\sqrt2 \) km/h towards south-west

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

Teacher's Note:
1. Subtract the train's velocity from the bird's.

Answer: (b) \( \tan^{-1}\frac12 \)

Explanation:
1. \( \tan\theta=\frac{u_y}{u_x}=\frac5{10}=\frac12 \).

Teacher's Note:
1. Use the components of the initial velocity.

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

Explanation:
1. \( R=H\Rightarrow\frac{2u^2\sin\theta\cos\theta}g=\frac{u^2\sin^2\theta}{2g} \).
2. \( 4\cos\theta=\sin\theta\Rightarrow\tan\theta=4 \).

Teacher's Note:
1. Equate the two formulas.

Answer: (a) \( \frac xu \)

Explanation:
1. Horizontal component of particle 1: \( \frac u{\sqrt3}\cos30^\circ=\frac u2 \). Horizontal component of particle 2: \( u\cos60^\circ=\frac u2 \).
2. The horizontal separation closes at \( u \), so the time is \( \frac xu \).

Teacher's Note:
1. Horizontal velocities are constant for both particles.

Answer: (c) 600 m

Explanation:
1. Time of fall \( =\sqrt{\frac{2\times78.4}{9.8}}=4 \) s.
2. Horizontal distance \( =150\times4=600 \) m.

Teacher's Note:
1. The bomb keeps the horizontal speed of the plane.

Answer: (a) \( \sqrt{117} \)

Explanation:
1. The two velocities are perpendicular: \( v=\sqrt{9^2+6^2}=\sqrt{117} \) km/h.

Teacher's Note:
1. Add the perpendicular velocities.

Answer: (d) 80 m

Explanation:
1. \( u^2=2gH=80g \).
2. Maximum horizontal range \( =\frac{u^2}g=80 \) m.

Teacher's Note:
1. The same speed gives maximum range when thrown at 45 degrees.

Answer: (a) 1 : 1

Explanation:
1. The two angles add up to \( 90^\circ \), i.e. they are complementary.
2. Complementary angles give equal ranges, so the ratio is 1 : 1.

Teacher's Note:
1. Check whether the angles add up to 90 degrees.

Answer: (d) both (a) and (b)

Explanation:
1. For a man with velocity \( v\hat i \): the rain relative to him is \( (2-v)\hat i-4\hat j \), of speed 5 m/s.
2. \( (2-v)^2+16=25\Rightarrow2-v=\pm3\Rightarrow v=-1 \) or \( 5 \).
3. So both \( -\hat i \) and \( 5\hat i \) work.

Teacher's Note:
1. Set the magnitude of the relative velocity to 5.

Answer: (b) \( -\vec r_0 \)

Explanation:
1. \( \vec v=\frac{d\vec r}{dt}=\vec r_0(1-2at) \).
2. At \( t=\frac1a \): \( \vec v=\vec r_0(1-2)=-\vec r_0 \).

Teacher's Note:
1. Differentiate r and substitute the return time.

Answer: (d) 45 km/h

Explanation:
1. Relative speed \( =\frac{100}{7.2}=13.9 \) m/s = 50 km/h.
2. Train speed \( =50-5=45 \) km/h.

Teacher's Note:
1. Relative speed is the sum of the two speeds.

Answer: (c) 31.25 m

Explanation:
1. Time to the top \( =2.5 \) s. \( H=\frac12g(2.5)^2=\frac12\times10\times6.25=31.25 \) m.

Teacher's Note:
1. The time of ascent is half of the flight time.

Answer: (d) \( \pi\left(\frac{u^2}g\right)^2 \)

Explanation:
1. The farthest a bullet can land is the maximum range \( \frac{u^2}g \) in every direction.
2. The area is a circle: \( \pi R_{max}^2=\pi\left(\frac{u^2}g\right)^2 \).

Teacher's Note:
1. The covered region is a circle of radius equal to the maximum range.

Answer: (a) \( \frac du \)

Explanation:
1. Along the line joining a person to the next, the pursued person's velocity is perpendicular (zero component), so the approach speed is u.
2. Time \( =\frac du \).

Teacher's Note:
1. Use the component of velocity along the line joining them.

Answer: (b) \( 45^\circ \)

Explanation:
1. \( R=\frac{2u^2\sin\theta\cos\theta}g \) and \( T^2=\frac{4u^2\sin^2\theta}{g^2} \).
2. \( R=5T^2\Rightarrow2\sin\theta\cos\theta=\frac{20\sin^2\theta}g=2\sin^2\theta\Rightarrow\tan\theta=1 \), so \( \theta=45^\circ \).

Teacher's Note:
1. Write both quantities in terms of u and theta.

Answer: (b) -4 m/s\( ^2 \)

Explanation:
1. \( a_x=\frac{d^2x}{dt^2}=-4 \) and \( a_y=0 \).
2. The acceleration is constant at \( -4 \) m/s\( ^2 \).

Teacher's Note:
1. Differentiate each coordinate twice.

Answer: (b) \( \frac2{5\sqrt3} \) h

Explanation:
1. He heads upstream at angle \( \alpha \) to the perpendicular with \( \sin\alpha=\frac5{10}=\frac12 \), so \( \alpha=30^\circ \).
2. The crossing speed is \( 10\cos30^\circ=5\sqrt3 \) km/h.
3. Time \( =\frac2{5\sqrt3} \) h.

Teacher's Note:
1. The downstream drift must be cancelled by his heading.

Answer: (b) \( 40\sqrt2 \) km/h in the north-east direction

Explanation:
1. Take east as x and north as y: \( \vec v_B=(0,40) \), \( \vec v_T=(-40,0) \).
2. \( \vec v_{BT}=(40,40) \): magnitude \( 40\sqrt2 \) towards north-east.

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

Answer: (b) \( 45^\circ \)

Explanation:
1. \( v_x=98 \) m/s and \( v_y=gt=98 \) m/s.
2. \( \tan\theta=1\Rightarrow\theta=45^\circ \).

Teacher's Note:
1. Compare the final vertical and horizontal speeds.

Answer: (d) 6.2

Explanation:
1. Time of fall \( =\sqrt{\frac{2\times4.9}{9.8}}=1 \) s.
2. Minimum speed \( =\frac{6.2}1=6.2 \) m/s.

Teacher's Note:
1. He must cover 6.2 m horizontally during the fall.

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

Explanation:
1. \( R=\frac{u^2\sin2\theta}g \) and \( \sin2\theta=\sin(180^\circ-2\theta) \), so the assertion is true.
2. The range depends on the angle through \( \sin2\theta \), so the reason is false.

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

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In the frame of one particle the other must move straight towards it to reach it, so the relative velocity must lie along the line joining them.
2. This follows from the meaning of relative velocity given in the reason.

Teacher's Note:
1. Think in the frame of one particle.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The acceleration g is constant and makes an oblique angle with the velocity (except at the extremes), so both the magnitude and direction of the velocity change.
2. The reason gives this explanation.

Teacher's Note:
1. A force that is oblique to the velocity changes both its magnitude and its direction.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both have zero initial vertical velocity and the same vertical acceleration, so they take the same time.
2. This is because the horizontal velocity does not influence the vertical motion.

Teacher's Note:
1. The two motions are independent.