JEE Physics Motion In A Plane MCQs Set 03

Answer: (d) four times

Explanation:
1. \( R\propto u^2 \), so doubling u makes the range 4 times.

Teacher's Note:
1. Range is proportional to the square of the speed.

Answer: (a) \( \tan^{-1}\left(\frac vu\right) \)

Explanation:
1. Rain relative to the boy has components u downwards and v backwards.
2. The tube must point along this relative velocity, at angle \( \theta \) to the vertical with \( \tan\theta=\frac vu \).

Teacher's Note:
1. Rain relative to the boy has a backward component equal to v.

Answer: (d) \( 2\hat i-3\hat j \)

Explanation:
1. The horizontal component stays \( 2 \) m/s. At the same level the vertical component has the same magnitude but is reversed.
2. \( \vec v_B=2\hat i-3\hat j \).

Teacher's Note:
1. Vertical speed is the same at the same height.

Answer: (b) \( 120^\circ \)

Explanation:
1. To go straight across, the upstream component of his swimming velocity must cancel the flow: \( 2\cos\theta=-1 \), measuring \( \theta \) from the flow direction.
2. \( \cos\theta=-\frac12\Rightarrow\theta=120^\circ \).

Teacher's Note:
1. His velocity component along the river must be zero.

Answer: (c) \( 5\sqrt2 \) m/s

Explanation:
1. \( \vec v_1=3\hat i+8t\hat j\to(3,8) \) and \( \vec v_2=8t\hat i+3\hat j\to(8,3) \) at t = 1 s.
2. \( \vec v_{12}=(-5,5) \), magnitude \( 5\sqrt2 \) m/s.

Teacher's Note:
1. Differentiate the positions and subtract.

Answer: (d) \( \tan^{-1}\left(\frac{v\sin\theta-gt}{v\cos\theta}\right) \)

Explanation:
1. At time t: \( v_x=v\cos\theta \) and \( v_y=v\sin\theta-gt \).
2. The angle is \( \tan^{-1}\frac{v_y}{v_x} \).

Teacher's Note:
1. Take the ratio of vertical to horizontal velocity.

Answer: (c) 64 m

Explanation:
1. At the end of the flight \( y=0 \): \( 16x=\frac{x^2}4\Rightarrow x=64 \) m.

Teacher's Note:
1. Set y = 0.

Answer: (c) g - a, g

Explanation:
1. The ground observer sees the ball fall freely with g.
2. The lift accelerates down with a, so relative to the lift the ball's acceleration is \( g-a \).

Teacher's Note:
1. Subtract the lift's acceleration for the relative value.

Answer: (d) 2 m/s

Explanation:
1. Drop in height \( =25-5=20 \) m, so the time of fall is \( \sqrt{\frac{2\times20}{10}}=2 \) s.
2. He must cover 4 m in 2 s: \( u=\frac42=2 \) m/s.

Teacher's Note:
1. He must clear the 4 m gap during the fall.

Answer: (d) 5 m/s\( ^2 \)

Explanation:
1. \( \Delta\vec v=40\hat j-30\hat i \), magnitude \( \sqrt{40^2+30^2}=50 \) m/s.
2. \( a=\frac{50}{10}=5 \) m/s\( ^2 \).

Teacher's Note:
1. Take the vector change of velocity.

Answer: (b) \( t_1t_2\propto R \)

Explanation:
1. The angles are \( \theta \) and \( 90^\circ-\theta \): \( t_1t_2=\frac{2u\sin\theta}g\cdot\frac{2u\cos\theta}g=\frac{2}{g}\cdot\frac{u^2\sin2\theta}g=\frac{2R}g \).
2. So \( t_1t_2\propto R \).

Teacher's Note:
1. Multiply the two times of flight.

Answer: (c) \( \frac{r_0}{2a} \)

Explanation:
1. The particle moves out until \( v=0 \): \( 1-2at=0\Rightarrow t=\frac1{2a} \), at distance \( r_0\cdot\frac1{2a}\cdot\frac12=\frac{r_0}{4a} \).
2. It then returns the same way, so the total distance is \( 2\times\frac{r_0}{4a}=\frac{r_0}{2a} \).

Teacher's Note:
1. The path is a straight line out and back.

Answer: (c) \( \frac83 \) m

Explanation:
1. The range is \( R=8+4=12 \) m. For 45 degrees, \( y=x\left(1-\frac xR\right) \).
2. At \( x=8 \): \( y=8\left(1-\frac8{12}\right)=\frac83 \) m.

Teacher's Note:
1. Use the trajectory equation y = x tan theta (1 - x/R).

Answer: (b) (200, 100) m

Explanation:
1. Maximum range means 45 degrees and \( \frac{u^2}g=400 \) m.
2. The speed is least at the top, at \( x=\frac R2=200 \) m and \( y=H=\frac{u^2}{4g}=100 \) m.

Teacher's Note:
1. The velocity is minimum at the highest point.

Answer: (a) 6 m/s

Explanation:
1. Along the belt's motion the speeds add: \( 4+2=6 \) m/s.

Teacher's Note:
1. Add the speed of the belt to the speed relative to the belt.

Answer: (c) 500 m

Explanation:
1. 180 km/h = 50 m/s. Time of fall \( =\sqrt{\frac{2\times490}{9.8}}=10 \) s.
2. Range \( =50\times10=500 \) m.

Teacher's Note:
1. Convert the speed to m/s first.

Answer: (d) 4 : 1

Explanation:
1. First: \( h_1=\frac{u^2}{2g} \). Second: the vertical component is \( u\cos60^\circ=\frac u2 \), so \( h_2=\frac{u^2}{8g} \).
2. The potential energies are in the ratio of the heights: \( 4:1 \).

Teacher's Note:
1. Find the vertical component of the second ball.

Answer: (b) The ranges of A and C are equal and less than that of B

Explanation:
1. Range \( \propto\sin2\theta \): \( \sin60^\circ,\sin90^\circ,\sin120^\circ \).
2. A and C have equal ranges (complementary angles) and B (45 degrees) has the maximum range.

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

Answer: (a) due north

Explanation:
1. The time to cross is \( \frac{\text{width}}{v\cos\alpha} \), where \( \alpha \) is measured from the perpendicular, and it is smallest for \( \alpha=0 \).
2. So he should head due north, perpendicular to the banks.

Teacher's Note:
1. Time depends only on the component perpendicular to the river.

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

Explanation:
1. The projectile lands with the same speed but in a different direction (the vertical component is reversed), so the velocity vector is not the same: the assertion is false.
2. The vertical component on landing is equal in magnitude but opposite in direction to the initial one, so the reason is false as stated.

Teacher's Note:
1. Velocity is a vector.

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

Explanation:
1. The component of velocity across the river is largest when he heads perpendicular to the banks, so the crossing time is least. The assertion is true.
2. The river flow only carries him downstream and does not help in crossing, so the reason is false.

Teacher's Note:
1. The flow affects only the drift.

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

Explanation:
1. In the train frame the coin has only vertical motion, a straight line: true.
2. The ground observer sees a horizontal velocity plus free fall, a parabola: true. But this is not the reason for the first statement.

Teacher's Note:
1. Change the frame of reference.

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

Explanation:
1. If the initial velocity is zero or along the vertical the path is a straight line; otherwise it is a parabola.
2. The shape depends on the initial conditions, as the reason states.

Teacher's Note:
1. The initial velocity decides the shape.

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

Explanation:
1. To go straight across he must swim upstream with \( v\sin\alpha=u \), which is impossible if \( u>v \): the assertion is true.
2. For \( u\lt v \) the crossing speed is \( \sqrt{v^2-u^2} \), so the time is \( \frac d{\sqrt{v^2-u^2}} \): true, but it does not explain why the crossing is impossible for u > v.

Teacher's Note:
1. The upstream component of v must equal u.

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

Explanation:
1. \( x=2t \) and \( y=-4t^2 \), so \( y=-x^2 \), which is a parabola.
2. The reason gives exactly this relation.

Teacher's Note:
1. Eliminate t between x and y.

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

Explanation:
1. A projectile is a freely falling body: only gravity acts on it, so the assertion is true.
2. Its velocity and hence its momentum change continuously because of gravity, so the reason is false.

Teacher's Note:
1. The force of gravity changes the momentum.