JEE Physics Motion In A Plane MCQs Set 08

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

Explanation:
1. The horizontal component of velocity stays constant at \( u\cos\theta \).
2. At the given instant the horizontal component is also \( v\cos\alpha \).
3. So \( v\cos\alpha = u\cos\theta \) and \( v = u\cos\theta\sec\alpha \).

Teacher's Note:
1. In projectile motion only the vertical component changes.
2. Equate the horizontal components at the two instants.

Answer: (d) fourth

Explanation:
1. After time t the ball has moved x = 1.8t horizontally and fallen y = 5t².
2. It reaches the edge of the nth step when x = 0.2n, so t = 0.2n/1.8, and has then fallen \( y = 5\left(\frac{0.2n}{1.8}\right)^2 \approx 0.0617n^2 \) m.
3. The ball is below that step edge when \( 0.0617n^2 \ge 0.2n \), that is n ≥ 3.24, so the first step it can reach is the fourth.

Teacher's Note:
1. Compare the fall of the ball with the depth of the nth step at horizontal distance 0.2n.
2. The first whole n that satisfies the condition is the answer.

Answer: (a) t = 3 s, x = 45 m

Explanation:
1. Vertical motion: \( 45 = \frac{1}{2} \times 10 \times t^2 \), so t = 3 s.
2. Horizontal motion starts from rest with a = 10 m/s²: \( x = \frac{1}{2} \times 10 \times 3^2 = 45 \) m.

Teacher's Note:
1. The horizontal and vertical motions are independent.
2. Time of fall depends only on the vertical motion.

Answer: (b) \( \frac{37}{7} \) m

Explanation:
1. The trajectory is \( y = x\tan\theta\left(1 - \frac{x}{R}\right) \).
2. Using (3, 4): \( 4 = 3\tan\theta\left(1 - \frac{3}{R}\right) \) and using (4, 3): \( 3 = 4\tan\theta\left(1 - \frac{4}{R}\right) \).
3. Solving the two equations gives \( \tan\theta = \frac{37}{12} \) and \( R = \frac{37}{7} \) m.

Teacher's Note:
1. Use the trajectory equation in terms of range.
2. Two points give two equations for two unknowns.

Answer: (b) 60 m

Explanation:
1. Along the incline the component of g is \( g\sin30^\circ = 5 \) m/s² and perpendicular to it \( g\cos30^\circ \).
2. Time of flight: \( T = \frac{2u\sin30^\circ}{g\cos30^\circ} = \frac{2 \times 30 \times 0.5}{10 \times 0.866} = \frac{30}{8.66} \) s.
3. Range along the incline: \( R = u\cos30^\circ T - \frac{1}{2}(5)T^2 = \frac{2u^2\sin30^\circ\cos60^\circ}{g\cos^2 30^\circ} = 60 \) m.

Teacher's Note:
1. Resolve motion along and perpendicular to the incline.
2. The time of flight is found from the perpendicular motion.

Answer: (d) 39.0 m

Explanation:
1. Range \( R = \frac{u^2\sin120^\circ}{g} \) is proportional to \( u^2 \).
2. Nominal R = 400 × 0.866 / 10 = 34.64 m.
3. A ±5% change in u gives about ±10% change in R, so R lies between about 31.2 m and 38.1 m, and 39.0 m is outside this range.

Teacher's Note:
1. Range depends on the square of the speed, so percentage error doubles.
2. Check each option against the allowed band.

Answer: (d) a straight line

Explanation:
1. Both particles have the same acceleration g, so the relative acceleration is zero.
2. Their relative velocity is therefore constant, and relative to the first particle the other moves along a straight line.

Teacher's Note:
1. Equal accelerations cancel in relative motion.
2. Relative motion with constant velocity is a straight line.

Answer: (d) \( 5\sqrt{3} \) km/h

Explanation:
1. Rain appears vertical to the running man, so its horizontal velocity equals the man's speed, 5 km/h.
2. For the man at rest, the rain makes 60° with the horizontal: \( \tan60^\circ = \frac{v_y}{5} \), so \( v_y = 5\sqrt{3} \) km/h.
3. Relative to the running man the rain is vertical, so its speed is \( 5\sqrt{3} \) km/h.

Teacher's Note:
1. The horizontal component of rain velocity equals the man's speed.
2. The relative velocity of vertical rain is just its vertical component.

Answer: (c) R = 4H

Explanation:
1. \( H = \frac{u^2\sin^2\theta}{2g} = \frac{u^2}{4g} \) and \( R = \frac{u^2\sin2\theta}{g} = \frac{u^2}{g} \) for θ = 45°.
2. Therefore R = 4H.

Teacher's Note:
1. General relation: \( \tan\theta = \frac{4H}{R} \).
2. At 45° the tangent is 1, giving R = 4H.

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

Explanation:
1. The horizontal component stays constant: 30 cos60° = 15 m/s.
2. When the velocity is at 45° to the horizontal, \( v\cos45^\circ = 15 \).
3. So \( v = 15\sqrt{2} \) m/s.

Teacher's Note:
1. The horizontal component never changes.
2. Use it to find the speed at any angle.

Answer: (c) 15 m

Explanation:
1. A: vertical component 10 sin60° = 5√3 m/s, so time of flight \( T = \frac{2 \times 5\sqrt{3}}{10} = \sqrt{3} \) s.
2. Its range = 10 cos60° × √3 = 5√3 m.
3. B moves at 5 m/s horizontally and must also take √3 s to land at the same point, so \( h = \frac{1}{2} \times 10 \times 3 = 15 \) m.

Teacher's Note:
1. Both particles must have the same time of flight to meet on the ground.
2. The same time then fixes the height of B.

Answer: (d) never possible

Explanation:
1. The angle with the horizontal is 30°, so \( u_x = 15\sqrt{3} \) m/s and \( u_y = 15 \) m/s.
2. The condition \( \vec{u}\cdot\vec{v} = 0 \) gives \( 675 + 15(15 - 10t) = 0 \), so t = 6 s.
3. The time of flight is only \( \frac{2 \times 15}{10} = 3 \) s, so the projectile lands before t = 6 s and the condition is never met.

Teacher's Note:
1. Always compare the answer from the algebra with the time of flight.
2. A time longer than the flight time is not physically possible.

Answer: (b) 6.0 s

Explanation:
1. Let the height be h and the common speed u, with downward taken as positive.
2. Stone thrown up: \( h = -ut_1 + \frac{1}{2}gt_1^2 \). Stone thrown down: \( h = ut_2 + \frac{1}{2}gt_2^2 \).
3. Multiply the first by \( t_2 \), the second by \( t_1 \) and add: \( h(t_1+t_2) = \frac{1}{2}g\,t_1t_2(t_1+t_2) \), so \( h = \frac{1}{2}g\,t_1t_2 \). For a dropped stone \( h = \frac{1}{2}gt_3^2 \), so \( t_3 = \sqrt{t_1t_2} = 6 \) s.

Teacher's Note:
1. Eliminate u between the two equations of motion.
2. The result \( t_3 = \sqrt{t_1t_2} \) is worth remembering.

Answer: (b) \( \frac{4}{\sqrt{3}} \)

Explanation:
1. For a projectile that grazes the vertex and lands at the other end, \( \tan\theta = \tan\alpha + \tan\beta \) where α and β are the base angles.
2. This follows from \( y = x\tan\theta(1 - x/R) \) at the vertex.
3. \( \tan\theta = \tan30^\circ + \tan60^\circ = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{4}{\sqrt{3}} \).

Teacher's Note:
1. The vertex at height h divides the base into two parts whose tangents add to tan θ.
2. Use the trajectory equation.

Answer: (c) \( \frac{v}{a\sqrt{2}} \)

Explanation:
1. Wind velocity: \( \frac{v}{\sqrt{2}} \) towards the east and \( \frac{v}{\sqrt{2}} \) towards the north.
2. The man's northward velocity is at.
3. Relative to the man the northward component is \( \frac{v}{\sqrt{2}} - at \), which is zero when \( t = \frac{v}{a\sqrt{2}} \).

Teacher's Note:
1. Wind appears to blow towards east when the relative north component vanishes.
2. Use relative velocity components.

Answer: (d) \( (\sqrt{3}-1) \) s

Explanation:
1. Horizontal component = 20 cos60° = 10 m/s, so at 45° the vertical component must be 10 m/s upward.
2. Initial vertical component = 20 sin60° = 10√3 m/s.
3. \( 10 = 10\sqrt{3} - 10t \), so \( t = (\sqrt{3} - 1) \) s.

Teacher's Note:
1. At 45° the vertical and horizontal components are equal in size.
2. The projectile passes 45° twice, once rising and once falling; the first is asked here.

Answer: (d) \( \frac{10}{\sqrt{3}} \) m/s

Explanation:
1. Horizontal component = 10 cos60° = 5 m/s, which stays constant.
2. At 30° above the horizontal, \( v\cos30^\circ = 5 \).
3. \( v = \frac{5}{\cos30^\circ} = \frac{10}{\sqrt{3}} \) m/s.

Teacher's Note:
1. Use the constant horizontal component.
2. Divide it by the cosine of the new angle.

Answer: (c) its velocity makes zero angle with the horizontal at its maximum height

Explanation:
1. At the highest point only the horizontal component remains, so the velocity is horizontal and its angle with the horizontal is zero.
2. The speed there is \( u\cos\theta \), not zero.
3. The velocity is perpendicular to the acceleration only at the highest point, and it is never along g unless θ = 90°.
4. So (c) is the correct statement.

Teacher's Note:
1. At the top the vertical component is zero but the horizontal component is not.
2. The angle between velocity and g changes throughout the flight.

Answer: (d) \( \frac{80}{3\sqrt{3}} \) m

Explanation:
1. Initial velocity (10, 10√3). A velocity (10, \( v_y \)) is perpendicular to it when \( 100 + 10\sqrt{3}v_y = 0 \), so \( v_y = -\frac{10}{\sqrt{3}} \).
2. Speed \( v = \sqrt{100 + \frac{100}{3}} = \frac{20}{\sqrt{3}} \) m/s and \( \cos\phi = \frac{10}{v} = \frac{\sqrt{3}}{2} \).
3. Radius \( r = \frac{v^2}{g\cos\phi} = \frac{400/3}{10 \times \sqrt{3}/2} = \frac{80}{3\sqrt{3}} \) m.

Teacher's Note:
1. Radius of curvature uses the component of g perpendicular to the velocity.
2. That component is g cos φ, where φ is the angle of the velocity with the horizontal.

Answer: (a) \( \sin^{-1}\left(\frac{4}{5}\right) \)

Explanation:
1. Horizontal velocity \( u_x = 36 \) m/s.
2. \( y = 48t - 4.9t^2 \), so \( u_y = 48 \) m/s.
3. \( \tan\theta = \frac{48}{36} = \frac{4}{3} \), so \( \sin\theta = \frac{4}{5} \) and \( \theta = \sin^{-1}\left(\frac{4}{5}\right) \).

Teacher's Note:
1. Differentiate or compare coefficients to find the initial velocity components.
2. A 3-4-5 triangle gives the sine directly.

Answer: (c) 8 m/s²

Explanation:
1. \( a_x = \frac{d^2x}{dt^2} = 8 \) m/s².
2. \( a_y = \frac{d^2y}{dt^2} = 0 \).
3. The acceleration is constant at 8 m/s² for all t, including t = 5 s.

Teacher's Note:
1. Acceleration is the second derivative of position with respect to time.
2. A constant result means the time value is not needed.

Answer: (c) 3 min

Explanation:
1. Shortest path means the resultant is perpendicular to the bank, so the crossing speed = \( \sqrt{5^2 - 3^2} = 4 \) km/h.
2. Time = 0.2 km / 4 km/h = 0.05 h = 3 min.

Teacher's Note:
1. Convert the width to kilometres or the speed to m/s before dividing.
2. The component of swimmer's velocity along the stream must cancel the river's speed.

Answer: (c) 3.0 km

Explanation:
1. \( R = \frac{u^2\sin2\theta}{g} \).
2. For 15°: \( 1.5 = \frac{u^2\sin30^\circ}{g} = \frac{u^2}{2g} \), so \( \frac{u^2}{g} = 3 \) km.
3. For 45°: \( R = \frac{u^2}{g} \times 1 = 3.0 \) km.

Teacher's Note:
1. At the same speed, 45° gives the maximum range.
2. Use sin 30° = ½ for the first case.

Answer: (a) \( \tan^{-1}\left(\frac{1}{5}\right) \)

Explanation:
1. Horizontal velocity = 500 m/s and vertical velocity on landing = gt = 10 × 10 = 100 m/s.
2. \( \tan\phi = \frac{100}{500} = \frac{1}{5} \).
3. The angle with the horizontal is \( \tan^{-1}\left(\frac{1}{5}\right) \).

Teacher's Note:
1. The bomb keeps the plane's horizontal velocity.
2. Its vertical velocity grows as gt.

Answer: (c) \( (V_1^2 - V_2^2)^{1/2} \)

Explanation:
1. The package's horizontal velocity on the ground frame is \( V_1\sin\theta \), and since it looks vertical to the pilot, \( V = V_1\sin\theta \).
2. The vertical velocity is the same in both frames: \( V_2 = V_1\cos\theta \).
3. So \( V = \sqrt{V_1^2 - V_2^2} \).

Teacher's Note:
1. A horizontal change of frame does not alter the vertical velocity.
2. The pilot's speed equals the horizontal component of the package's velocity.

Answer: (c) \( 40\sqrt{5} \) m

Explanation:
1. \( u_x = 50\cos37^\circ = 40 \) m/s and \( u_y = 50\sin37^\circ = 30 \) m/s.
2. 80 m horizontally takes t = 2 s, and then \( y = 30 \times 2 - 5 \times 4 = 40 \) m.
3. Distance \( = \sqrt{80^2 + 40^2} = 40\sqrt{5} \) m.

Teacher's Note:
1. Find the time from the horizontal motion first.
2. Then get y and use Pythagoras.

Answer: (a) 1 s

Explanation:
1. Acceleration is g, directed vertically downwards.
2. The velocity is perpendicular to it when the velocity is horizontal, that is at the top of the path.
3. \( t = \frac{u\sin30^\circ}{g} = \frac{10}{10} = 1 \) s.

Teacher's Note:
1. A vertical acceleration is perpendicular only to a horizontal velocity.
2. That occurs at the highest point.

Answer: (d) \( \frac{1}{\sqrt{3}} \)

Explanation:
1. Maximum height \( H = \frac{u^2\sin^2\theta}{2g} \).
2. Equal heights require \( u_1\sin60^\circ = u_2\sin30^\circ \).
3. \( \frac{u_1}{u_2} = \frac{\sin30^\circ}{\sin60^\circ} = \frac{1}{\sqrt{3}} \).

Teacher's Note:
1. Height depends only on the vertical component of the initial velocity.
2. Equate the vertical components.

Answer: (b) 4 : 1

Explanation:
1. Minimum kinetic energy occurs at the top, where only \( u\cos\theta \) remains, so \( (u_1\cos\theta_1)^2 : (u_2\cos\theta_2)^2 = 4 : 1 \).
2. Heights \( \propto (u\sin\theta)^2 \) are also in the ratio 4 : 1.
3. Both components are in the ratio 2 : 1, so \( \tan\theta \) is equal, and the range \( R = \frac{2(u\cos\theta)(u\sin\theta)}{g} \) is in the ratio 2 × 2 = 4 : 1.

Teacher's Note:
1. Minimum KE depends on the horizontal component and height on the vertical component.
2. The range is proportional to their product.