JEE Physics Motion In A Plane MCQs Set 20

Answer: (c) 40 km/h, 60° N of E

Explanation:
1. \( \vec v_T=\vec v_{T/C}+\vec v_C=(0,20\sqrt3)+(20,0)=(20,20\sqrt3) \).
2. Magnitude \( =\sqrt{400+1200}=40 \) km/h.
3. \( \tan\theta=\frac{20\sqrt3}{20}=\sqrt3\Rightarrow\theta=60^\circ \) north of east.

Teacher's Note:
1. Actual velocity equals relative velocity plus velocity of the observer. Add vector components rather than magnitudes.

Answer: (a) 5 h

Explanation:
1. Take A as origin. Relative position of B: \( (0,100) \). Relative velocity of B with respect to A: \( (-10,-10) \).
2. Separation squared: \( d^2=(10t)^2+(100-10t)^2=200t^2-2000t+10000 \).
3. It is minimum when \( \frac{d}{dt}=0\Rightarrow400t=2000\Rightarrow t=5 \) h.

Teacher's Note:
1. Work in the frame of one ship, then minimise the separation. The line of relative motion has its closest approach where the displacement is perpendicular to the relative velocity.

Answer: (d) \( g \)

Explanation:
1. A projectile moves only under gravity (air resistance neglected).
2. So the acceleration is \( g \) downward at every point of the path.
3. At the top only the vertical velocity is zero, not the acceleration.

Teacher's Note:
1. Velocity can be zero while acceleration is not. At the highest point the acceleration is still g.

Answer: (a) \( u\cos\theta \)

Explanation:
1. The ball's horizontal velocity is constant at \( u\cos\theta \).
2. To catch it at the same horizontal position at every instant, the boy must match this horizontal velocity.
3. So he should run at \( u\cos\theta \).

Teacher's Note:
1. If the boy's horizontal velocity equals the ball's, he stays directly below the ball throughout its flight.

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

Explanation:
1. Vertical motion: \( y=u\sin\theta\,t-\frac12gt^2 \Rightarrow 5=25\sin\theta(2)-5(4) \).
2. \( 50\sin\theta=25\Rightarrow\sin\theta=\frac12 \).
3. \( \theta=30^\circ \).

Teacher's Note:
1. Only the vertical motion decides the height at a given time. Substitute the time into the vertical displacement equation.

Answer: (c) \( \frac{40}{3} \) m/s

Explanation:
1. Time for the cart to move 80 m: \( t=\frac{80}{30}=\frac83 \) s.
2. The projectile must return to the cart, so relative to the cart it has no horizontal motion and is fired vertically upward with speed \( u \): \( t=\frac{2u}{g} \).
3. \( u=\frac{gt}{2}=\frac{10\times8/3}{2}=\frac{40}{3} \) m/s.

Teacher's Note:
1. In the cart's frame the horizontal velocity of the projectile is zero. It rises and falls vertically, and the time of flight is 2u/g.

Answer: (a) \( R=4\sqrt{H_1H_2} \)

Explanation:
1. \( H_1=\frac{u^2\sin^2\theta}{2g} \) and \( H_2=\frac{u^2\cos^2\theta}{2g} \).
2. \( \sqrt{H_1H_2}=\frac{u^2\sin\theta\cos\theta}{2g} \).
3. \( R=\frac{2u^2\sin\theta\cos\theta}{g}=4\sqrt{H_1H_2} \).

Teacher's Note:
1. Complementary angles give the same range. The product of their heights links directly to \( \sin\theta\cos\theta \).

Answer: (c) 80 m

Explanation:
1. Net acceleration \( =\sqrt{6^2+8^2}=10 \) m/s² along a fixed straight line.
2. Distance \( =\frac12at^2=\frac12(10)(16) \).
3. \( =80 \) m.

Teacher's Note:
1. Starting from rest with constant acceleration components, the body moves along a straight line in the direction of the net acceleration.

Answer: (d) The magnitude of the velocity of the particle is 8 m/s

Explanation:
1. \( |\vec R|=2\sqrt{\sin^2+\cos^2}=2 \) m, so the path is a circle of radius 2 m.
2. \( \vec a=-(2\pi)^2\vec R \) is along \( -\vec R \) with magnitude \( \frac{v^2}{R} \), so statements (a), (b) and (c) are correct.
3. Speed \( v=R\omega=2\times2\pi=4\pi\approx12.6 \) m/s, not 8 m/s, so (d) is wrong.

Teacher's Note:
1. For circular motion speed is radius times angular velocity. Here omega is 2π rad/s, so the speed is 4π m/s.

Answer: (b) \( -4 \) m/s²

Explanation:
1. \( a_x=\frac{d^2x}{dt^2}=-4 \) m/s² and \( a_y=\frac{d^2y}{dt^2}=0 \).
2. The acceleration is constant, independent of time.
3. So at \( t=2 \) s the acceleration is \( -4 \) m/s².

Teacher's Note:
1. Differentiate each coordinate twice. A linear term in a coordinate gives no contribution to the acceleration.

Answer: (a) \( \frac{2hu^2}{gb^2} \)

Explanation:
1. At the \( n \)th step the horizontal and vertical distances are \( nb \) and \( nh \).
2. \( nb=ut \) and \( nh=\frac12gt^2 \).
3. Eliminating \( t \): \( nh=\frac{g}{2}\left(\frac{nb}u\right)^2\Rightarrow n=\frac{2hu^2}{gb^2} \).

Teacher's Note:
1. Treat the stairs as the point where the vertical drop equals n times the step height. Eliminate time between the two equations.

Answer: (a) 3 km/h

Explanation:
1. The umbrella must point along the velocity of rain relative to the man, which makes \( 53^\circ \) with the vertical.
2. Horizontal component of relative rain velocity \( =4 \) km/h (equal to the man's speed), and \( \tan53^\circ=\frac{4}{v_r} \).
3. \( v_r=\frac{4}{\tan53^\circ}=4\times\frac34=3 \) km/h.

Teacher's Note:
1. The raindrops' true speed is vertical. The man's speed gives the horizontal part of the relative velocity, and the tangent of the angle links the two.

Answer: (b) 50 s

Explanation:
1. To go straight across, the boat's upstream component must cancel the current: \( 5\sin\theta=3 \Rightarrow\sin\theta=\frac35 \), \( \theta=37^\circ \).
2. Effective speed across \( =5\cos37^\circ=4 \) m/s.
3. Time \( =\frac{200}4=50 \) s.

Teacher's Note:
1. Minimum crossing time (40 s, heading straight across) is not the same as the time on the shortest path (50 s). Read which is asked.