JEE Physics Relative Motion MCQs Set 03

Answer: (b) \( -10\ \text{m} \)

Explanation:
1. \( x_{AC}=x_A-x_C=0-10=-10\ \text{m} \).
2. The minus sign says that A lies 10 m to the left of C.

Teacher's Note:
1. Position of A with respect to C is x of A minus x of C. Swapping the order flips the sign.

Answer: (a) \( -15\ \text{m/s} \)

Explanation:
1. \( v_{AB}=v_A-v_B=5-20=-15 \) m/s.
2. The negative sign shows A appears to move backwards as seen from B (and \( v_{BA}=-v_{AB}=+15 \) m/s).

Teacher's Note:
1. Relative velocity is velocity of the object minus velocity of the observer, both measured from the ground.

Answer: (d) \( 22\ \text{m/s} \)

Explanation:
1. Take A's direction as positive: \( v_A=+10 \), \( v_B=-12 \).
2. \( v_{AB}=10-(-12)=22 \) m/s.

Teacher's Note:
1. For head-on motion the relative speed is the sum of the speeds.

Answer: (c) \( 15\ \text{m/s} \)

Explanation:
1. \( v_{AB}=20-5=15 \) m/s, and the separation is decreasing.
2. So the velocity of approach is \( |v_{AB}|=15 \) m/s.

Teacher's Note:
1. In one dimension the velocity of approach is the magnitude of the relative velocity when the separation is shrinking.

Answer: (b) 5 s

Explanation:
1. C approaches A and B with relative speed \( 10+10=20 \) m/s.
2. Since C meets B first, then A, which is 100 m behind B: \( t=\frac{100}{20}=5 \) s.

Teacher's Note:
1. The velocity of approach of C with each particle is the same, so the delay is the gap divided by that relative speed.

Answer: (a) 5 m

Explanation:
1. Relative acceleration \( a_{BA}=-g-(-g)=0 \), so the relative velocity stays \( 10-5=5 \) m/s.
2. Separation \( 5\times1=5 \) m. (Check: A is at \( 5-5=0 \) and B at \( 10-5=5 \).)

Teacher's Note:
1. When both bodies accelerate equally the relative motion is uniform, and a single multiplication gives the separation.

Answer: (d) 3 s

Explanation:
1. Relative acceleration is zero, and the relative speed is \( 20+30=50 \) m/s.
2. Time \( =\frac{150}{50}=3 \) s.

Teacher's Note:
1. Because gravity acts equally on both balls, the closing speed is constant and is the sum of the launch speeds.

Answer: (b) \( 10\sqrt2\ \text{m/s} \)

Explanation:
1. \( \vec v_A=10\hat i,\ \vec v_B=10\hat j \) so \( \vec v_{AB}=10\hat i-10\hat j \).
2. Magnitude \( =\sqrt{10^2+10^2}=10\sqrt2 \) m/s.

Teacher's Note:
1. Subtract velocities as vectors.
2. For perpendicular velocities of equal size the relative speed is root 2 times the common speed.

Answer: (a) 50 m

Explanation:
1. The velocity vectors make an angle \( 180^\circ-37^\circ-53^\circ=90^\circ \) with each other.
2. Relative acceleration is zero, so \( v_{AB}=\sqrt{30^2+40^2}=50 \) m/s and the separation after 1 s is \( 50 \) m.

Teacher's Note:
1. Both bodies fall equally, so their separation grows uniformly at the relative launch velocity.

Answer: (d) \( \dfrac{2u}{g+a} \)

Explanation:
1. In the lift frame the ball's acceleration is \( -g-a \), of magnitude \( g+a \).
2. Returning to the hand: \( 0=uT-\tfrac12(g+a)T^2 \), so \( T=\frac{2u}{g+a} \).

Teacher's Note:
1. An accelerating lift acts like a change in gravity: effective gravity is g plus a for an upward accelerating lift.

Answer: (c) \( \dfrac{u^2}{2(g+a)} \)

Explanation:
1. Relative to the lift the acceleration is \( g+a \) downward.
2. So \( 0-u^2=-2(g+a)H \) and \( H=\frac{u^2}{2(g+a)} \).

Teacher's Note:
1. Use the lift as the reference frame with effective gravity g + a.

Answer: (b) \( \dfrac{2dv}{v^2-u^2} \)

Explanation:
1. Downstream speed \( v+u \), upstream speed \( v-u \).
2. Time \( =\frac{d}{v+u}+\frac{d}{v-u}=\frac{2dv}{v^2-u^2} \).

Teacher's Note:
1. The upstream leg takes longer than the downstream leg saves, so the trip always takes longer than in still water.

Answer: (a) 40 s

Explanation:
1. Only the component perpendicular to the flow carries the boat across: \( t=\frac{400}{10}=40 \) s.

Teacher's Note:
1. The river current does not change the crossing time when the boat heads straight across.

Answer: (d) 80 m

Explanation:
1. Crossing time \( 40 \) s and drift speed \( 2 \) m/s.
2. Drift \( 2\times40=80 \) m downstream.

Teacher's Note:
1. Drift is the displacement along the river during the crossing time.

Answer: (c) \( \sin^{-1}\frac35\approx37^\circ \)

Explanation:
1. Zero drift requires the upstream component of his swimming velocity to cancel the river: \( 5\sin\phi=3 \).
2. So \( \sin\phi=\frac35 \) and \( \phi=37^\circ \) from the perpendicular (i.e. \( 127^\circ \) from the river flow direction).

Teacher's Note:
1. Heading upstream by exactly the right angle makes his ground velocity perpendicular to the banks.
2. This works only when his swimming speed exceeds the river speed.

Answer: (b) 15 min

Explanation:
1. Resultant speed across \( =\sqrt{5^2-3^2}=4 \) km/h.
2. Time \( =\frac{1}{4}\ \text{h}=15 \) min.

Teacher's Note:
1. Only the perpendicular component of the ground velocity, root of v squared minus u squared, matters for the crossing time.

Answer: (a) \( \sqrt{u^2+v^2+2uv\cos\theta} \)

Explanation:
1. \( \vec v_M=\vec v_{MR}+\vec v_R=(v\cos\theta+u)\hat i+v\sin\theta\,\hat j \).
2. Magnitude \( =\sqrt{(u+v\cos\theta)^2+v^2\sin^2\theta}=\sqrt{u^2+v^2+2uv\cos\theta} \).

Teacher's Note:
1. Add the swimmer's velocity relative to water to the water's velocity, component by component.

Answer: (b) \( \tan^{-1}\dfrac35 \)

Explanation:
1. Velocity of rain relative to the man: \( \vec v_{rm}=-10\hat j-6\hat i \).
2. Angle with the vertical: \( \tan\phi=\frac{6}{10}=\frac35 \) (the umbrella points along the apparent rain, tilted forward).

Teacher's Note:
1. The umbrella must be held opposite to the rain's velocity relative to the man.

Answer: (a) \( 5\sqrt5\ \text{m/s},\ \tan^{-1}\frac12 \)

Explanation:
1. \( \vec v_r=\vec v_{rm}+\vec v_m \): vertical component \( 10 \) m/s and horizontal component \( 5 \) m/s (equal to the man's velocity).
2. Speed \( =\sqrt{100+25}=5\sqrt5 \) m/s and \( \tan\phi=\frac5{10}=\frac12 \).

Teacher's Note:
1. If the man sees vertical rain, then the rain's horizontal velocity equals his horizontal velocity.

Answer: (d) \( 10\ \text{m/s} \)

Explanation:
1. Rain velocity components: horizontal \( 20\sin30^\circ=10 \) m/s, vertical \( 20\cos30^\circ=10\sqrt3 \) m/s.
2. For vertical relative rain, the man must match the horizontal component: \( 10 \) m/s in the same direction.

Teacher's Note:
1. Relative to the man the horizontal part of the rain velocity must vanish.

Answer: (b) \( v\cos\theta \)

Explanation:
1. \( \vec v_{BA}=\vec v \) since A is at rest.
2. The velocity of separation is the component along the line joining them: \( v\cos\theta \).

Teacher's Note:
1. Only the component along the line joining the particles changes their separation.

Answer: (a) \( v_1\sin\theta_1=v_2\sin\theta_2 \)

Explanation:
1. They collide only if their relative velocity points along the line joining them.
2. So the components perpendicular to the line must cancel: \( v_1\sin\theta_1=v_2\sin\theta_2 \).

Teacher's Note:
1. No sideways relative velocity means the line of sight between the particles keeps its direction until they meet.

Answer: (d) \( \dfrac{d}{v_1\cos\theta_1+v_2\cos\theta_2} \)

Explanation:
1. The velocity of approach is the sum of the components along the line: \( v_1\cos\theta_1+v_2\cos\theta_2 \).
2. Time \( =\frac{d}{v_{app}} \).

Teacher's Note:
1. Both particles close the gap along the joining line, so their line components add.