JEE Physics Motion In A Plane MCQs Set 15

Answer: (b) \( \hat B=-\hat A \)

Explanation:
1. Multiplying a vector by a negative number reverses its direction, so B points opposite to A and the unit vectors satisfy \( \hat B=-\hat A \).
2. The magnitude is multiplied by 1.5, so |B| = 1.5|A|, which rules out options (c) and (d).
3. B and A are different vectors, so option (a) is wrong.

Teacher's Note:
1. A negative scalar reverses direction and its modulus scales the magnitude.
2. Unit vectors carry only direction information.

Answer: (d) Figure (d)

Explanation:
1. By the triangle law, the vectors A and B are placed head to tail and R = A + B joins the tail of A to the head of B.
2. In figure (d), A goes from O to P, B goes from P to Q, and R goes from O to Q.
3. In figure (a) R points from Q to O, in figure (b) B points from Q to P, and in figure (c) A points from P to O, so none of these shows the head-to-tail arrangement.

Teacher's Note:
1. Check each arrowhead: the head of A must meet the tail of B.
2. The resultant starts at the tail of the first vector and ends at the head of the last one.

Answer: (a) Figure (a)

Explanation:
1. The component along X is A cos θ î and along Y is A sin θ ĵ, with θ measured from the X-axis.
2. Figure (a) shows exactly this. In figure (b) the angle θ is marked from the Y-axis, so the labelling would have to be swapped, and in figure (c) the labels cos θ and sin θ are interchanged.

Teacher's Note:
1. The component adjacent to the angle uses cosine and the opposite component uses sine.
2. Always note which axis the angle is measured from.

Answer: (a) \( \cos^{-1}(3/5) \)

Explanation:
1. Since 15² = 12² + 9², B and C are perpendicular, and A is the hypotenuse of the right triangle.
2. The angle between A and C: \( \cos\theta=\frac{C}{A}=\frac{9}{15}=\frac35 \).

Teacher's Note:
1. Check whether the magnitudes form a Pythagorean triple (9, 12, 15).
2. In the right triangle, the adjacent side to the angle between A and C is C.

Answer: (b) P and Q are parallel but R is not parallel

Explanation:
1. Q = 1.2 P (3.6/3 = 4.8/4 = 1.2), so P and Q are parallel.
2. For R: the ratio of components is (3/4)/3 = 1/4 for î but (4/3)/4 = 1/3 for ĵ, which are unequal, so R is not parallel to P.

Teacher's Note:
1. Two vectors are parallel if their corresponding components are in the same ratio.
2. Equal vectors need equal magnitude as well as direction.

Answer: (c) \( \sin\theta=\frac{12}{13} \)

Explanation:
1. To reach the opposite point, the upstream component of his swimming velocity must cancel the current: \( 13\cos\theta=5 \).
2. So cos θ = 5/13 and \( \sin\theta=\sqrt{1-\frac{25}{169}}=\frac{12}{13} \).

Teacher's Note:
1. For the shortest (perpendicular) path, the net velocity along the river must be zero.
2. The angle with the perpendicular to the bank would instead satisfy sin = 5/13.

Answer: (b) 6 m/s

Explanation:
1. Let the rocket's velocity relative to the ground be v (vertical). The car's velocity is 8 m/s horizontal.
2. Velocity of rocket relative to the car has components v (vertical) and 8 (horizontal): \( v^2+8^2=10^2 \).
3. \( v=\sqrt{100-64}=6\ \text{m/s} \).

Teacher's Note:
1. Relative velocity is a vector difference, and perpendicular components combine as in Pythagoras.
2. The car's motion adds a horizontal component the stationary observer does not see.

Answer: (d) 40 m

Explanation:
1. Maximum range occurs at 45°: \( R_{max}=\frac{u^2}{g}=80\ \text{m} \).
2. Maximum height is for vertical throw: \( H_{max}=\frac{u^2}{2g}=40\ \text{m} \).

Teacher's Note:
1. The greatest height with a given speed is obtained by throwing vertically upwards.
2. H_max is always half of R_max.

Answer: (c) Ranges of A and C are equal and less than that of B

Explanation:
1. \( R=\frac{u^2\sin2\theta}{g} \).
2. sin 60° = sin 120°, so the 30° and 60° angles (complementary) give equal ranges.
3. The range is greatest at 45° (sin 90° = 1), so A and C have equal ranges less than that of B.

Teacher's Note:
1. Angles that add up to 90° give the same range for the same speed.
2. 45° gives the maximum range.

Answer: (d) None of these

Explanation:
1. For maximum range the angle of projection is 45°.
2. Check the ceiling: the maximum height reached at 45° is \( \frac{u^2\sin^2 45^\circ}{2g}=\frac{3600\times0.5}{20}=90\ \text{m} \), which is below 100 m, so the ceiling does not interfere.
3. So θ = 45°, and sin θ = 1/√2, which is none of the given values.

Teacher's Note:
1. The ceiling only matters if the maximum height at 45° exceeds 100 m.
2. Maximum range always needs θ = 45° on level ground.

Answer: (c) \( \tan^{-1}\frac{4H}{R} \)

Explanation:
1. \( \frac{H}{R}=\frac{u^2\sin^2\theta/2g}{2u^2\sin\theta\cos\theta/g}=\frac{\tan\theta}{4} \).
2. So \( \tan\theta=\frac{4H}{R} \).

Teacher's Note:
1. Divide the formula for H by the formula for R.
2. This relation is used to find the angle from the observed trajectory.

Answer: (b) 0, 10 m/s

Explanation:
1. After one full lap the displacement is zero, so the average velocity is 0.
2. Average speed = distance/time = \( \frac{2\pi\times100}{62.8}=10\ \text{m/s} \).

Teacher's Note:
1. Average velocity uses displacement, average speed uses path length.
2. For a full circle, displacement is always zero.

Answer: (b) \( u\sin\theta \)

Explanation:
1. Initial velocity: \( u\cos\theta\,\hat i+u\sin\theta\,\hat j \). At the highest point: \( u\cos\theta\,\hat i \).
2. Change in velocity: \( -u\sin\theta\,\hat j \), whose magnitude is u sin θ.

Teacher's Note:
1. The horizontal component does not change; only the vertical component drops from u sin θ to zero.
2. A magnitude cannot be negative, so option (c) is not a magnitude.

Answer: (c) 500 m

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

Teacher's Note:
1. The packet keeps the horizontal speed of the plane.
2. Convert km/h to m/s by multiplying by 5/18.

Answer: (b) 20 km/h

Explanation:
1. While standing, the rain falls at 30° to the vertical, so its horizontal component is \( v\sin30^\circ \).
2. When he runs at 10 km/h, the rain appears vertical, so the horizontal component equals his speed: \( v\sin30^\circ=10 \).
3. So v = 20 km/h.

Teacher's Note:
1. The rain appears vertical when the man's horizontal velocity equals the rain's horizontal component.
2. The umbrella is held along the direction opposite to the apparent rain velocity.

Answer: (d) 17 m/s

Explanation:
1. Velocity of train relative to car (north) = 15 m/s: \( \vec v_T-\vec v_C=15\,\hat j \).
2. \( \vec v_T=8\,\hat i+15\,\hat j \) so \( |v_T|=\sqrt{64+225}=17\ \text{m/s} \).

Teacher's Note:
1. Add the velocity of the observer to the relative velocity to get the actual velocity.
2. 8-15-17 is a Pythagorean triple.

Answer: (d) 12 min, 600 m

Explanation:
1. Time to cross: \( t=\frac{1.0}{5.0}=0.2\ \text{h}=12\ \text{min} \).
2. Drift down the river: \( 3.0\times0.2=0.6\ \text{km}=600\ \text{m} \).

Teacher's Note:
1. With strokes normal to the current, the crossing time depends only on the swimming speed and the width.
2. The drift equals the river speed multiplied by the crossing time.

Answer: (b) 3 km/h

Explanation:
1. Shortest path: the boat's resultant velocity is perpendicular to the banks: \( v=\frac{1\ \text{km}}{0.25\ \text{h}}=4\ \text{km/h} \).
2. \( v_r=\sqrt{5^2-4^2}=3\ \text{km/h} \).

Teacher's Note:
1. For the shortest path, the boat's upstream component cancels the river speed.
2. 3-4-5 is a Pythagorean triple.

Answer: (c) 7 s

Explanation:
1. Vertical component: uᵧ = 50 sin 30° = 25 m/s upwards.
2. Taking upward as positive, displacement = −70 m: \( -70=25t-5t^2 \).
3. \( t^2-5t-14=0\Rightarrow(t-7)(t+2)=0 \), so t = 7 s.

Teacher's Note:
1. Only the vertical motion decides the time of flight.
2. Reject the negative root.

Answer: (c) 5.49 s

Explanation:
1. Horizontal component: \( 147\cos60^\circ=73.5\ \text{m/s} \). Initial vertical: \( 147\sin60^\circ=127.3\ \text{m/s} \).
2. At 45°, vertical velocity equals the horizontal one: 73.5 m/s.
3. \( t=\frac{127.3-73.5}{9.8}=5.49\ \text{s} \).

Teacher's Note:
1. The velocity makes 45° when |vᵧ| = vₓ.
2. Equivalent: t = (u/g)(sin60° − cos60°).

Answer: (b) 20 m

Explanation:
1. Equal ranges with equal speeds mean complementary angles, so the other angle is 60°.
2. \( H_1+H_2=\frac{u^2}{2g}(\sin^230^\circ+\sin^260^\circ)=\frac{400}{20}(1)=20\ \text{m} \).

Teacher's Note:
1. sin² 30° + sin² 60° = 1/4 + 3/4 = 1.
2. Equal range at the same speed means θ and 90° − θ.

Answer: (c) \( \sqrt2\,a\omega \)

Explanation:
1. \( v_x=-a\omega\sin\omega t,\ v_y=a\omega\cos\omega t,\ v_z=a\omega \).
2. \( v=\sqrt{a^2\omega^2+a^2\omega^2}=\sqrt2\,a\omega \).

Teacher's Note:
1. Differentiate each coordinate with respect to time and combine the components.
2. sin² + cos² = 1 simplifies the first two terms.

Answer: (d) 1 : 16

Explanation:
1. The bullets can reach any point within a circle of radius equal to the maximum range: \( R_{max}=\frac{u^2}{g} \).
2. Area \( =\pi R_{max}^2\propto u^4 \).
3. Ratio = 1⁴ : 2⁴ = 1 : 16.

Teacher's Note:
1. Area is proportional to the square of the radius, and the radius to the square of the speed.
2. So doubling the speed multiplies the area by 16.

Answer: (c) 2.8 m

Explanation:
1. At t = 1 s: vₓ = 5 m/s, \( v_y=10\sin60^\circ-10=-1.34\ \text{m/s} \). Then \( v^2=25+1.8=26.8 \).
2. The normal component of g is \( g\cos\alpha=g\frac{v_x}{v} \) where α is the angle of the velocity with the horizontal.
3. \( R=\frac{v^2}{g\cos\alpha}=\frac{v^3}{g\,v_x}=\frac{26.8^{3/2}}{10\times5}\approx2.8\ \text{m} \).

Teacher's Note:
1. The radius of curvature is v² divided by the component of acceleration perpendicular to the velocity.
2. The perpendicular component of g is g cos α.

Answer: (b) 10 m/s

Explanation:
1. \( |\Delta\vec v|=2v\sin\frac\theta2=2(10)\sin30^\circ=10\ \text{m/s} \).

Teacher's Note:
1. The speed is the same, but the direction rotates by 60°, so the velocity vector changes.
2. The two velocity vectors and their difference form an equilateral triangle for 60°.

Answer: (c) \( 31\ \text{km/h at }\theta=\tan^{-1}\frac{1}{2\sqrt3} \) east of north

Explanation:
1. Boat relative to water: 25 km/h towards north. Current: \( 10\sin60^\circ=5\sqrt3 \) east and \( 10\cos60^\circ=5 \) north.
2. Resultant: east component 8.66 km/h, north component 25 + 5 = 30 km/h.
3. \( v=\sqrt{8.66^2+30^2}=\sqrt{975}\approx31\ \text{km/h} \), \( \tan\theta=\frac{8.66}{30}=\frac{1}{2\sqrt3} \) east of north.

Teacher's Note:
1. Add the boat's velocity relative to water and the velocity of the current as vectors.
2. Resolve the current into east and north components first.

Answer: (b) 5 m/s²

Explanation:
1. Time of flight: \( T=\frac{2\times10\sin60^\circ}{10}=\sqrt3\ \text{s} \).
2. Range on the ground: \( 10\cos60^\circ\times\sqrt3=5\sqrt3=8.66\ \text{m} \).
3. Range inside the train = ground range − distance moved by the train relative to the ball's horizontal motion: \( 8.66-\tfrac12a(\sqrt3)^2=1.15 \), so 1.5a = 7.51 and a ≈ 5 m/s².

Teacher's Note:
1. In the train's frame, the ball also feels a backward pseudo-acceleration a, which reduces its range.
2. The time of flight does not change.

Answer: (d) \( \sqrt{16+t^4}\ \text{m s}^{-2} \)

Explanation:
1. Speed: \( v=\frac{dx}{dt}=4t \). Tangential acceleration: \( a_t=4\ \text{m/s}^2 \).
2. Centripetal acceleration: \( a_c=\frac{v^2}{r}=\frac{16t^2}{16}=t^2 \).
3. \( a=\sqrt{a_t^2+a_c^2}=\sqrt{16+t^4} \).

Teacher's Note:
1. In non-uniform circular motion the acceleration has a tangential and a radial part at right angles.
2. Add them as vectors, not directly.

Answer: (d) All of these

Explanation:
1. Initial velocity: \( (u\cos\theta,\ u\sin\theta) \). After t: \( (u\cos\theta,\ u\sin\theta-gt) \).
2. Perpendicular: \( u^2\cos^2\theta+u\sin\theta(u\sin\theta-gt)=0\Rightarrow gt=\frac{u}{\sin\theta} \).
3. Then \( v^2=(gt)^2-u^2 \) gives \( v^2+u^2=g^2t^2 \); and \( v=u\cot\theta \) gives \( u=v\tan\theta \) and \( t=\frac{u}{g\sin\theta}=\frac{v\tan\theta}{g\sin\theta}=\frac{v}{g\cos\theta} \).
4. All three relations hold.

Teacher's Note:
1. Use the dot product of the two velocities, which must be zero.
2. The change in velocity gt is the hypotenuse of the right triangle formed by u and v.