JEE Physics Relative Motion MCQs Set 01

Answer: (b) 10 m/s\(^2\)

Explanation:
1. Height of plane: \( y_p = 1000 + 500t + \tfrac12 a t^2 \); height of shot: \( y_s = 700t - 5t^2 \).
2. They meet when \( (\tfrac a2 + 5)t^2 - 200t + 1000 = 0 \).
3. The plane escapes if this has no real root, and the limiting case is zero discriminant: \( 200^2 = 4(\tfrac a2+5)(1000) \).
4. So \( \tfrac a2 + 5 = 10 \), giving \( a = 10\ \text{m/s}^2 \).

Teacher's Note:
1. Escape means the two position curves never intersect, so test the discriminant of the meeting equation.
2. The limiting case (grazing contact) gives the minimum acceleration.

Answer: (a) 100 m/s

Explanation:
1. Time to reach the top for the first stone: \( t = 50/10 = 5 \) s, where its velocity is 0.
2. Second stone after 5 s: \( v = 50 + 10\times5 = 100 \) m/s downwards.
3. Relative velocity = 100 - 0 = 100 m/s.

Teacher's Note:
1. Both stones have the same acceleration g, so their relative velocity is constant at 100 m/s from the start.
2. Check: initially 50 down minus 50 up gives 100 m/s, the same value.

Answer: (d) It is a non-zero constant

Explanation:
1. A straight x-t line means constant velocity for each child.
2. Relative velocity = difference of two constant velocities, hence constant.
3. The slopes are different, so the difference is not zero.

Teacher's Note:
1. The slope of an x-t graph is velocity.
2. Two straight lines of unequal slope give a constant non-zero relative velocity.

Answer: (c) It continuously increases

Explanation:
1. Velocities: \( v_1 = a_1 t,\ v_2 = a_2 t \) with \( a_1 > a_2 \).
2. Relative velocity \( v_1 - v_2 = (a_1 - a_2)t \), which grows linearly with time.

Teacher's Note:
1. Straight v-t lines through the origin mean constant accelerations.
2. Different accelerations make the relative velocity change steadily, here increasing.

Answer: (b) 60 km

Explanation:
1. \( x_A = 50t + 10t^2 \), \( x_B = 50t - 10t^2 \).
2. They cross when \( x_A + x_B = 100 \): \( 100t = 100 \), so \( t = 1 \) h.
3. \( x_A = 50 + 10 = 60 \) km.

Teacher's Note:
1. The accelerations cancel in the sum, so the closing distance grows at a steady 100 km/h.
2. Then use the time in A's own displacement equation.

Answer: (a) 1000 km/h from west to east

Explanation:
1. Take east as positive: jet velocity = -500 km/h.
2. Gases leave backwards, i.e. towards east, at +1500 km/h relative to the jet.
3. Ground velocity of gases = -500 + 1500 = +1000 km/h, i.e. 1000 km/h towards east.

Teacher's Note:
1. Add the jet's velocity to the gas velocity relative to the jet, with signs.
2. Gases ejected backwards still move forward relative to the ground only if their relative speed exceeds the jet speed; here it does, hence eastward.

Answer: (d) south-west

Explanation:
1. Take east = +x, north = +y: \( \vec v_H = -50\hat j \), \( \vec v_T = 50\hat i \).
2. \( \vec v_{HT} = \vec v_H - \vec v_T = -50\hat i - 50\hat j \).
3. Both components are negative: south-west.

Teacher's Note:
1. Subtract the observer's velocity, i.e. add the reverse of the train's velocity.
2. The reversed train velocity points west, which combines with south to give south-west.

Answer: (c) \( \pi \)

Explanation:
1. \( |\vec v_1-\vec v_2|^2 = v_1^2+v_2^2-2v_1v_2\cos\theta \).
2. This is largest when \( \cos\theta=-1 \), i.e. \( \theta=\pi \).

Teacher's Note:
1. Relative velocity is a vector subtraction, so opposite directions add magnitudes.
2. The maximum is \( v_1+v_2 \), the minimum \( |v_1-v_2| \).

Answer: (b) 20 km/h

Explanation:
1. For the second ship to stay due north of the first, their eastward velocity components must be equal.
2. \( v\sin30^\circ = 10 \), so \( v = 20 \) km/h.

Teacher's Note:
1. Staying on the same north-south line means zero relative velocity along east-west.
2. Equate the east components and solve.

Answer: (a) 3

Explanation:
1. Shortest path: net velocity is perpendicular to the banks, \( v = 1/0.25 = 4 \) km/h.
2. \( v_r = \sqrt{5^2-4^2} = 3 \) km/h.

Teacher's Note:
1. Shortest path means the boat's upstream component cancels the river speed.
2. Then \( v_b^2 = v_r^2 + v_{\text{net}}^2 \).

Answer: (d) 15 km/h

Explanation:
1. The two velocities are perpendicular: \( v = \sqrt{9^2+12^2} = 15 \) km/h.

Teacher's Note:
1. Perpendicular components add by Pythagoras.
2. 9-12-15 is a scaled 3-4-5 triangle.

Answer: (c) 160 s

Explanation:
1. Shortest path: net speed across \( = \sqrt{5^2-4^2} = 3 \) m/s.
2. \( t = 480/3 = 160 \) s.

Teacher's Note:
1. For the shortest path the boat heads upstream at an angle so that the drift is cancelled.
2. Only the cross-river component of velocity counts for the crossing time.

Answer: (b) 100 km/h, 37\(^\circ\) west of south

Explanation:
1. Ground velocity = (150/0.5) W + (40/0.5) S = 300 km/h W and 80 km/h S.
2. Air velocity of the plane = 240 km/h W.
3. Wind = ground - air = 60 km/h W and 80 km/h S, magnitude \( \sqrt{60^2+80^2}=100 \) km/h.
4. Angle from south toward west: \( \tan\theta = 60/80 \), \( \theta = 37^\circ \).

Teacher's Note:
1. Ground velocity = air velocity + wind velocity, so wind = ground - air.
2. Name the angle carefully: here it is measured from south towards west.

Answer: (a) 5 km/h

Explanation:
1. The vertical rain velocity and horizontal walking velocity are perpendicular.
2. \( v_{rel} = \sqrt{3^2+4^2} = 5 \) km/h.

Teacher's Note:
1. Rain relative to man = rain velocity minus man's velocity.
2. The man's horizontal velocity makes the rain appear to slant towards him.

Answer: (d) \( 10\sqrt3 \) m/s

Explanation:
1. \( v_{rel}^2 = v_{rain}^2 + v_c^2 \).
2. \( v_c = \sqrt{20^2-10^2} = 10\sqrt3 \) m/s.

Teacher's Note:
1. Relative speed is the hypotenuse because the rain velocity and cyclist velocity are perpendicular.
2. The raindrops then make 30 degrees with the vertical.

Answer: (c) \( \dfrac{2\ell}{\sqrt{V^2-\upsilon^2}} \)

Explanation:
1. The plane must head slightly into the wind to cancel the sideways drift, so its ground speed along AB is \( \sqrt{V^2-\upsilon^2} \).
2. Same on the way back.
3. Total time \( = 2\ell/\sqrt{V^2-\upsilon^2} \).

Teacher's Note:
1. A cross wind reduces the useful ground speed along the track on both legs.
2. Do not use \( V\pm\upsilon \) here; that is for wind along the track.

Answer: (b) \( -1 \)

Explanation:
1. Relative position \( \vec r_{BA} = 4\hat i+4\hat j \); relative velocity \( \vec v_{BA} = (x-3)\hat i - 4\hat j \).
2. For collision, \( \vec v_{BA} \) must be directed opposite to \( \vec r_{BA} \): \( (x-3, -4) = -k(4,4) \).
3. From the y-component, k = 1, so \( x-3 = -4 \) and \( x = -1 \).

Teacher's Note:
1. Collision requires the line joining the particles to keep its direction while the distance shrinks to zero.
2. So the relative velocity must be antiparallel to the relative position.

Answer: (a) \( \dfrac{d\,v_1}{\sqrt{v_1^2+v_2^2}} \)

Explanation:
1. Relative position of B from A: \( (-v_1t,\ -d+v_2t) \), i.e. B starts d below A with relative velocity \( (-v_1, v_2) \).
2. Least distance = perpendicular distance from A's frame origin to the line of relative velocity: \( d_{min} = \dfrac{|\vec r\times\vec v|}{|\vec v|} = \dfrac{d\,v_1}{\sqrt{v_1^2+v_2^2}} \).

Teacher's Note:
1. In the frame of A, B moves in a straight line, and the closest approach is the perpendicular drop to that line.
2. Use \( d\sin\theta \) where \( \tan\theta = v_1/v_2 \).

Answer: (d) 45 km/h

Explanation:
1. Relative speed of the third car with respect to the cars = \( 30+v \).
2. It covers the 5 km gap in 4 min = 1/15 h: \( 30+v = 5\times15 = 75 \).
3. \( v = 45 \) km/h.

Teacher's Note:
1. The time interval between meeting two cars equals gap divided by closing speed.
2. Convert minutes to hours before using km/h.

Answer: (c) 20 m/s

Explanation:
1. Relative speed needed = 1000/100 = 10 m/s.
2. Scooter speed = 10 + 10 = 20 m/s.

Teacher's Note:
1. Overtaking means closing the gap, so the relative speed equals gap over time.
2. Add the bus speed to get the scooter speed.

Answer: (b) 4 s

Explanation:
1. Acceleration of coin relative to lift = 10 - 11 = -1 m/s\(^2\), i.e. 1 m/s\(^2\) upwards relative to the lift.
2. The coin therefore rises relative to the lift and strikes the ceiling after covering 10 - 2 = 8 m.
3. \( 8 = \tfrac12(1)t^2 \), so \( t = 4 \) s.

Teacher's Note:
1. When the lift accelerates down faster than g, objects fall upward relative to it.
2. So the coin hits the ceiling, not the floor.

Answer: (a) 105 m/s

Explanation:
1. Van speed = 30/3.6 = 8.33 m/s; thief speed = 192/3.6 = 53.33 m/s.
2. Bullet speed relative to ground = 150 + 8.33 = 158.33 m/s.
3. Relative to the thief: 158.33 - 53.33 = 105 m/s.

Teacher's Note:
1. The muzzle speed is relative to the gun, so first add the van's speed.
2. Convert km/h to m/s by dividing by 3.6.

Answer: (d) The bus may be moving in any direction between south and east

Explanation:
1. The flag points along the wind velocity relative to the bus: \( \vec v_w - \vec v_b = \) north.
2. So \( \vec v_b = \vec v_w - (\text{north}) \): an eastward component plus a southward component.
3. The bus therefore moves somewhere between south and east.

Teacher's Note:
1. A flag streams in the direction of the wind relative to the bus, not relative to the ground.
2. Write the equation as vectors and solve for the bus velocity.

Answer: (c) 20 m/s towards west

Explanation:
1. \( \vec v_{DA} = \vec v_{DC} + \vec v_{CB} + \vec v_{BA} \).
2. \( = 20\hat j + (-20\hat i) + (-20\hat j) = -20\hat i \), i.e. 20 m/s towards west.

Teacher's Note:
1. Chain the relative velocities so that the middle letters cancel: DC + CB + BA = DA.
2. Remember \( \vec v_{CB} = -\vec v_{BC} \).

Answer: (b) \( \dfrac{15}{2} \) m/s

Explanation:
1. Relative to the man, a drop moves down 10 m/s and back at his speed u.
2. To clear the face, the drop must travel 12 cm horizontally while falling at most 16 cm: \( u/10 \le 12/16 \).
3. \( u_{max} = 7.5 = 15/2 \) m/s.

Teacher's Note:
1. Use the slope of the relative path: horizontal over vertical equals u divided by 10.
2. The brim edge to the face bottom forms a 12-16 right triangle.