JEE Physics Motion In A Plane MCQs Set 04

Answer: (a) 3.75 m

Explanation:
1. \( v_y^2=u_y^2+2gH \) (taking H as the drop): \( 100=25+20H\Rightarrow H=3.75 \) m.

Teacher's Note:
1. Use the vertical velocity equation between launch and landing.

Answer: (d) All of the above

Explanation:
1. Equal maximum heights mean equal vertical components, so the same time of flight \( T=\frac{2u_y}g \).
2. With the same T, the range \( u_xT \) is largest for C, so its horizontal component is the largest.
3. The launch speed \( \sqrt{u_x^2+u_y^2} \) is also the largest for C. So all statements are true.

Teacher's Note:
1. Height fixes the vertical component and the time of flight.

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

Explanation:
1. Initial velocity \( (u\cos\theta,u\sin\theta) \); velocity at the top \( (u\cos\theta,0) \).
2. Change \( =(0,-u\sin\theta) \), of magnitude \( u\sin\theta \).

Teacher's Note:
1. Subtract the two velocity vectors.

Answer: (d) the ground

Explanation:
1. The velocity of the stone relative to the earth is its ordinary velocity.
2. It is largest in magnitude at the ground (launch and landing) and zero at the highest point.

Teacher's Note:
1. Speed is greatest where the potential energy is least.

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

Explanation:
1. For a collision, the relative velocity of B with respect to A must point along the line from B to A.
2. So the unit vectors of \( \vec r_1-\vec r_2 \) and \( \vec v_2-\vec v_1 \) must be equal.

Teacher's Note:
1. The relative velocity must point along the line of separation.

Answer: (c) 1000 km/h

Explanation:
1. The exhaust moves backward relative to the plane: \( v=500-1500=-1000 \) km/h.
2. The speed is 1000 km/h (backwards).

Teacher's Note:
1. Take the plane's direction as positive.

Answer: (b) 4 s

Explanation:
1. Relative speed \( =25 \) m/s and the total length to be covered is 100 m.
2. Time \( =\frac{100}{25}=4 \) s.

Teacher's Note:
1. Add the lengths and the speeds.

Answer: (a) 25 m/s

Explanation:
1. \( u_y=g\frac T2=10\times2.5=25 \) m/s.

Teacher's Note:
1. The vertical velocity vanishes after half the flight time.

Answer: (b) 15 s

Explanation:
1. In the belt's frame the distance is 60 m at 4 m/s, so the time is 15 s.

Teacher's Note:
1. The belt's motion does not matter in its own frame.

Answer: (b) 40 m/s

Explanation:
1. \( u_x=\frac RT=\frac{200}5=40 \) m/s.

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

Answer: (d) 10 s

Explanation:
1. Relative speed \( =10+5=15 \) m/s.
2. Time \( =\frac{150}{15}=10 \) s.

Teacher's Note:
1. They move in opposite directions.

Answer: (d) \( 60^\circ \)

Explanation:
1. \( u\cos\theta=\frac u2\Rightarrow\cos\theta=\frac12\Rightarrow\theta=60^\circ \).

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

Answer: (a) 5 h

Explanation:
1. Take A at the origin moving west and B initially at \( (0,-100) \). Relative velocity of B with respect to A is \( (10,10) \).
2. Relative position at time t: \( (10t,-100+10t) \). The distance is least when position is perpendicular to velocity: \( 10t\cdot10+(-100+10t)\cdot10=0\Rightarrow t=5 \) h.

Teacher's Note:
1. The separation is least when the relative position is perpendicular to the relative velocity.

Answer: (a) 250 metre

Explanation:
1. \( R_{max}=\frac{u^2}g=1000 \) m at \( 45^\circ \).
2. \( H=\frac{u^2\sin^245^\circ}{2g}=\frac{u^2}{4g}=250 \) m.

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

Answer: (a) \( mv\sqrt2 \)

Explanation:
1. The horizontal component is unchanged; the vertical component reverses from \( +v\sin45^\circ \) to \( -v\sin45^\circ \).
2. \( |\Delta p|=2mv\sin45^\circ=\sqrt2mv \).

Teacher's Note:
1. Only the vertical momentum changes.

Answer: (a) 3.5

Explanation:
1. Identical trajectories mean the same range: \( \frac{u^2\sin2\theta}g \) must be equal.
2. \( \frac{25}{9.8}=\frac9{g'}\Rightarrow g'=\frac{9\times9.8}{25}=3.5 \) m/s\( ^2 \).

Teacher's Note:
1. Equate the ranges for the same angle.

Answer: (b) \( \frac{2u_0v_0}g \)

Explanation:
1. Time of flight \( =\frac{2v_0}g \).
2. Maximum x displacement (the range) \( =u_0\cdot\frac{2v_0}g=\frac{2u_0v_0}g \).

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

Answer: (a) 15 m above the ground

Explanation:
1. Relative to each other the balls approach with speed 40 m/s (gravity acts equally on both), over a gap of 40 m.
2. Time \( =1 \) s. A is at \( 20(1)-5(1)^2=15 \) m above the ground.

Teacher's Note:
1. Gravity cancels in the relative motion.

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

Explanation:
1. At the top: height \( H=\frac{u^2\sin^2\theta}{2g} \) and horizontal distance \( \frac R2=\frac{u^2\sin\theta\cos\theta}g \).
2. \( \tan\phi=\frac{H}{R/2}=\frac{\tan\theta}2=\frac12 \).

Teacher's Note:
1. Compare the height with half the range.

Answer: (b) \( 30^\circ \)

Explanation:
1. The maximum height must not exceed 40 m: \( \frac{u^2\sin^2\theta}{2g}\le40\Rightarrow\sin^2\theta\le\frac{80\times9.8}{3136}=\frac14 \).
2. So \( \theta\le30^\circ \). The range increases with angle up to \( 45^\circ \), so the best allowed angle is \( 30^\circ \).

Teacher's Note:
1. The range is greatest at the largest allowed angle below 45 degrees.

Answer: (c) \( 2h \)

Explanation:
1. Time of fall \( =\sqrt{\frac{2h}g} \).
2. \( x=\sqrt{2gh}\times\sqrt{\frac{2h}g}=2h \).

Teacher's Note:
1. Horizontal distance is velocity times time of fall.

Answer: (d) 10 m/s due north

Explanation:
1. Train: 54 km/h = 15 m/s north. Monkey relative to train: 5 m/s south.
2. \( v=15-5=10 \) m/s due north.

Teacher's Note:
1. Add the velocities as vectors.

Answer: (b) Particle B

Explanation:
1. The vertical components are \( 2V\sin30^\circ=V \) for A and \( V\sin60^\circ=0.87V \) for B.
2. B starts upward with the smaller vertical speed, so it is in the air for a shorter time and hits the ground earlier.

Teacher's Note:
1. The time of flight depends on the vertical component.

Answer: (a) 9.8 m/s

Explanation:
1. A 45 degree line means the horizontal distance equals the height: 19.6 m.
2. Time of fall \( =\sqrt{\frac{2\times19.6}{9.8}}=2 \) s, so \( u=\frac{19.6}2=9.8 \) m/s.

Teacher's Note:
1. Equal horizontal and vertical displacements.

Answer: (a) 1 hr and 0 minute

Explanation:
1. The crossing time is least when the boat heads perpendicular to the banks.
2. \( t=\frac{4.0}{4}=1 \) h, independent of the river's flow.

Teacher's Note:
1. The flow does not affect the time of crossing.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both projectiles have the same acceleration g, so their relative acceleration is zero and the relative motion is along a straight line: true.
2. The reason is true but concerns equal ranges, which does not explain the assertion.

Teacher's Note:
1. Relative acceleration is zero.