JEE Physics Relative Motion MCQs Set 02

Answer: (a) 53 m

Explanation:
1. Ball's velocity w.r.t. ground = 3 + 5 = 8 m/s upward.
2. Ball: \( y = 50 + 8t - 5t^2 \); lift: \( y = 50 + 3t \).
3. Equal when \( 5t = 5t^2 \), so \( t = 1 \) s (after the top at 0.8 s, so on the way down).
4. Lift height = 50 + 3(1) = 53 m.

Teacher's Note:
1. Equate the ground positions of ball and lift, or use the relative motion which has acceleration g.
2. Check that the meeting time is after the top of the throw (0.8 s).

Answer: (d) 5 km/h

Explanation:
1. Time between meeting runners: \( 20/(15+v) \); between meeting cyclists: \( 30/(25+v) \) (in consistent units).
2. Equal intervals: \( 20(25+v) = 30(15+v) \), so \( 500+20v = 450+30v \).
3. \( v = 5 \) km/h.

Teacher's Note:
1. Meeting both together every time means the two meeting rates are equal.
2. The observer's speed adds to each group's speed because he moves oppositely.

Answer: (c) 15 s

Explanation:
1. Let speeds be \( u_1,u_2 \) and train length L. Passing means covering 2L relative: \( 2L/(u_1+u_2)=3 \).
2. With one speed raised 50% (the slower, \( u_1 \)): \( 2L/(1.5u_1+u_2)=2.5 \), so \( 1.5u_1+u_2 = 1.2(u_1+u_2) \), giving \( u_2 = 1.5u_1 \).
3. Then \( 2L = 3(2.5u_1) = 7.5u_1 \).
4. Same direction: relative speed \( = 0.5u_1 \), time \( = 7.5u_1/0.5u_1 = 15 \) s.

Teacher's Note:
1. Passing time equals total length (2L) divided by relative speed.
2. Opposite directions add speeds, same direction subtracts them.

Answer: (b) 15 m/s

Explanation:
1. A-C separation grows from 2d to 3d at the rate \( 10-5 = 5 \) m/s: \( 5t = d \), so \( t = d/5 \).
2. In this time B gains d on C: \( (u-10)t = d \), so \( u-10 = 5 \).
3. \( u = 15 \) m/s.

Teacher's Note:
1. Use the same time interval for both gaps: it is fixed by the A-C separation change.
2. The gap B-C closes at the speed difference.

Answer: (a) 0 m

Explanation:
1. In the truck frame, the pseudo-acceleration is 10 m/s\(^2\) backwards, together with g downwards.
2. Initial velocity components: 5 m/s forward and 5 m/s upward.
3. After 1 s: \( x = 5(1) - \tfrac12(10)(1)^2 = 0 \); \( y = 5(1) - \tfrac12(10)(1)^2 = 0 \).
4. The ball is back at the man's hand: displacement 0.

Teacher's Note:
1. An accelerating truck frame has an effective gravity that is tilted backwards.
2. Here the throw is exactly opposite to the effective gravity, so the ball returns.

Answer: (d) 5

Explanation:
1. Across: \( v_y = 60/6 = 10 \) m/s. The path is at 45\(^\circ\), so \( v_x = 10 \) m/s along the flow.
2. Velocity relative to water = \( (10-5, 10) = (5, 10) \), magnitude \( 5\sqrt5 \) m/s.
3. Hence \( n = 5 \).

Teacher's Note:
1. Ground velocity = swimmer relative to water + river velocity, so subtract the river velocity to get the swimmer's own velocity.
2. The 45 degrees applies to the ground path.

Answer: (c) 15

Explanation:
1. Component of the wind perpendicular to AB: \( 20\sin30^\circ = 10 \) m/s.
2. The plane must cancel it with its own sideways component: \( 150\sin\phi = 10 \).
3. \( \sin\phi = 1/15 \), so \( n = 15 \).

Teacher's Note:
1. Only the wind component across the track needs to be cancelled.
2. The angle between wind (north) and line AB is 30 degrees.

Answer: (b) 5

Explanation:
1. The apparent horizontal component of rain is \( r_x - v \). The same angle on opposite sides of vertical means \( r_x - 7 = -(r_x-25) \), so \( r_x = 16 \) m/s.
2. Then \( \tan37^\circ = 9/r_y = 3/4 \), so \( r_y = 12 \) m/s.
3. Rain speed \( = \sqrt{16^2+12^2} = 20 = 4n \), so \( n = 5 \).

Teacher's Note:
1. The rain's apparent direction flips side because the car overtakes the rain's horizontal speed.
2. Equal angles on opposite sides give equal and opposite horizontal components.

Answer: (a) 26

Explanation:
1. At time t the boy's speed is \( 2t \); \( \tan\theta = 2t/2 = t \).
2. Differentiate: \( \sec^2\theta\,\dfrac{d\theta}{dt} = 1 \), so \( \dfrac{d\theta}{dt} = \dfrac{1}{1+t^2} \).
3. At t = 5 s: \( 1/26 \) rad/s, so \( n = 26 \).

Teacher's Note:
1. Relative to the boy the rain has a growing horizontal component, so its direction rotates.
2. Use calculus on tan theta = t rather than computing angles at two instants.

Answer: (d) 3 s

Explanation:
1. Q is 8 m ahead of P along the track direction, so to meet Q must gain a full lap minus that: 32 - 8 = 24 m.
2. Gain rate = 10 - 2 = 8 m/s.
3. \( t = 24/8 = 3 \) s.

Teacher's Note:
1. On a closed track, meeting for the first time means gaining the track length minus the initial lead.
2. Check: in 3 s Q covers 30 m and P covers 6 m; 30 = 6 + 24.

Answer: (c) 45

Explanation:
1. Distance from O at time t (hours after noon): first \( 5t \), second \( 5(t-1.5) \).
2. \( d^2 = 25[t^2 + (t-1.5)^2] \) is least when \( 2t + 2(t-1.5) = 0 \), i.e. \( t = 0.75 \) h.
3. That is 12:45 PM, so T = 45.

Teacher's Note:
1. With equal speeds on perpendicular tracks, the closest approach is midway between the two times of reaching O.
2. Midway between 12:00 and 1:30 is 12:45.

Answer: (b) 20 s

Explanation:
1. Component of velocity along PO (towards O) is \( 10\cos60^\circ = 5 \) m/s, which is constant.
2. \( t = 100/5 = 20 \) s.

Teacher's Note:
1. Only the radial component of the velocity changes the distance from O.
2. The perpendicular component moves P around O but does not affect the time to reach it.

Answer: (a) All four: (1), (2), (3) and (4)

Explanation:
1. Both stones have acceleration g, so their relative velocity is constant: (1) true.
2. The first stone returns to terrace level after 4 s with 20 m/s downward, exactly the state of the second stone at its launch, only 2 s later in time. So both reach the ground with the same speed (2) true, and 2 s apart (3) true.
3. Equal impact speeds and elastic rebounds give equal rise heights (4) true.

Teacher's Note:
1. The first stone passing the terrace on its way down is identical in state to the second stone at its throw.
2. That one observation proves three of the four statements.

Answer: (d) (1), (3) and (4) only

Explanation:
1. The ball is a free projectile: acceleration g in the ground frame (1) true.
2. Relative acceleration is \( g+a \) in the first case and \( g-a \) in the second: \( t_1 = 2v/(g+a) \), \( t_2 = 2v/(g-a) \).
3. Adding reciprocals: \( 1/t_1 + 1/t_2 = g/v \), so \( v = g\,t_1t_2/(t_1+t_2) \): (4) true, (2) false.
4. Subtracting: \( 1/t_1 - 1/t_2 = a/v \), so \( a = g(t_2-t_1)/(t_1+t_2) \): (3) true.

Teacher's Note:
1. Relative to the lift, the effective gravity is g+a (lift going up) or g-a (lift going down).
2. Time of flight equals twice the relative speed divided by effective gravity.

Answer: (c) (1), (2) and (3) only

Explanation:
1. \( \vec v_{BA} = (4-3)\hat i + (-3-4)\hat j = \hat i - 7\hat j \): (1) true. Both velocities are constant, so (3) true.
2. Line AB direction: \( (\hat i+\hat j)/\sqrt2 \); component of \( \vec v_{BA} \) along it: \( (1-7)/\sqrt2 = -3\sqrt2 \), so approach speed \( 3\sqrt2 \): (2) true.
3. The direction of the line joining A and B changes with time, so its component of relative velocity (approach velocity) changes: (4) false.

Teacher's Note:
1. Approach velocity is the component of relative velocity along the line joining the particles.
2. That line rotates as they move, so the approach velocity varies even if the relative velocity is constant.

Answer: (b) (1), (2) and (3)

Explanation:
1. Flight time: \( 60/15 = 4 \) s.
2. In the truck frame the ball goes straight up and down: \( u = g t/2 = 20 \) m/s upward: (1) and (2) true.
3. Ground speed \( = \sqrt{15^2+20^2} = 25 \) m/s: (3) true.

Teacher's Note:
1. The ball shares the truck's horizontal velocity, so in the truck frame it simply goes up and down.
2. Find the flight time from the distance travelled by the truck.

Answer: (a) All four are correct

Explanation:
1. Ground velocity of the ball = 40 m/s: maximum height \( = 10 + 40^2/20 = 90 \) m (1) true.
2. Relative to the lift (constant velocity, so an inertial frame): \( 30^2/20 = 45 \) m (2) true; return after \( 2\times30/10 = 6 \) s (3) true.
3. Ground velocity at t = 6 s: \( 40 - 60 = -20 \) m/s, speed 20 m/s (4) true.

Teacher's Note:
1. A lift moving at constant velocity is an inertial frame, so projectile formulas work directly in it.
2. Add the lift's velocity only when you need ground quantities.

Answer: (d) 20 s

Explanation:
1. In the truck's frame the car starts at rest and accelerates at 0.5 m/s\(^2\).
2. Relative distance to cover: 40 + 17 + 3 + 40 = 100 m.
3. \( 100 = \tfrac12(0.5)t^2 \), so \( t = 20 \) s.

Teacher's Note:
1. Work in the truck's frame: the truck is at rest and the car starts from rest.
2. Add the gaps and both vehicle lengths to get the relative distance.

Answer: (c) 500 m

Explanation:
1. Distance covered by the truck in 20 s = 400 m.
2. Car's distance = 400 + 100 = 500 m (or \( 20\times20 + \tfrac12(0.5)(400) = 500 \) m).

Teacher's Note:
1. Ground distance = truck's distance plus the relative distance.
2. Verify with the direct kinematic formula.

Answer: (b) 30 m/s

Explanation:
1. \( v = u + at = 20 + 0.5\times20 = 30 \) m/s.

Teacher's Note:
1. The car accelerates at its maximum for the full 20 s.
2. Its speed relative to the truck at the end is 10 m/s.

Answer: (a) a straight line parallel to the x-axis

Explanation:
1. Both have acceleration g, so the relative acceleration is zero.
2. Relative velocity of A w.r.t. B: \( (2u_0\cos\theta, 0) \), horizontal and constant.
3. So A moves in a straight horizontal line in B's frame.

Teacher's Note:
1. Equal accelerations cancel in the relative frame, leaving uniform relative motion.
2. The relative velocity has no vertical part because both have the same vertical velocity.

Answer: (d) h

Explanation:
1. In B's frame A moves along a horizontal line that is always h above B.
2. The closest approach is the perpendicular distance, h.

Teacher's Note:
1. Closest approach in a straight-line relative motion is the perpendicular distance to that line.
2. The line here is horizontal, offset by h.

Answer: (c) \( \dfrac{\ell}{2u_0\cos\theta} \)

Explanation:
1. The horizontal gap \( \ell \) closes at \( 2u_0\cos\theta \).
2. The separation is least when the horizontal gap vanishes: \( t = \ell/(2u_0\cos\theta) \).

Teacher's Note:
1. The minimum occurs when the particles are vertically aligned.
2. Use the horizontal closing speed.

Answer: (b) 10 m/s

Explanation:
1. Rain velocity components: \( 10\sqrt2\sin45^\circ = 10 \) m/s horizontal and 10 m/s vertical.
2. Rain appears vertical, so the man's velocity equals the horizontal component: 10 m/s.

Teacher's Note:
1. Apparent vertical fall means the horizontal rain component matches the man's speed.
2. Resolve the rain velocity along horizontal and vertical.

Answer: (a) 20 m/s

Explanation:
1. Earlier apparent (vertical) speed = 10 m/s; the new one = \( 10\sqrt3 \) m/s.
2. The horizontal component is still 10 m/s (man's speed). Speed \( = \sqrt{10^2 + (10\sqrt3)^2} = 20 \) m/s.

Teacher's Note:
1. The new rain again has horizontal component equal to the man's speed, since it again appears vertical.
2. Only the vertical component changes.

Answer: (d) 15

Explanation:
1. Initial angle with the vertical: 45\(^\circ\). Final: \( \tan\phi = 10/(10\sqrt3) \), so \( \phi = 30^\circ \).
2. Angle between them = 45 - 30 = 15\(^\circ\).

Teacher's Note:
1. Both velocities are on the same side of the vertical.
2. Subtract the two angles from the vertical.

Answer: (c) 5

Explanation:
1. The relative velocity of B is perpendicular to A's velocity, so the components of both velocities along A's direction are equal: \( v_B\cos30^\circ = 100\sqrt3 \), giving \( v_B = 200 \) m/s.
2. The relative speed is the perpendicular component: \( v_B\sin30^\circ = 100 \) m/s.
3. B must cover the 500 m separation at this relative speed: \( t_0 = 500/100 = 5 \) s.

Teacher's Note:
1. The observer sees B moving across, not towards or away, so the along-track velocities match.
2. Then the relative motion is a straight line across A's path covering the stated distance.