JEE Physics Motion In A Plane MCQs Set 28

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Chapter-wise Objective Questions: Motion In A Plane

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Multiple Choice Questions

Question 1: A hunter aims his gun directly at a monkey and fires. At the instant the bullet leaves the barrel, the monkey drops. Which statement is correct (assume the bullet reaches the monkey's line before hitting the ground)?
(a) The bullet will never hit the monkey
(b) The bullet will always hit the monkey
(c) The bullet may or may not hit the monkey
(d) Cannot be predicted
Show Answer & Explanation

Answer: (b) The bullet will always hit the monkey

Explanation:
1. Both the bullet and the monkey fall the same distance ½gt² below the straight line of sight.
2. So the bullet always meets the monkey.

Teacher's Note:
1. Gravity acts equally on both, so relative motion is along the line of sight.

Question 2: For which angle between two equal vectors A and B will the magnitude of their sum be equal to the magnitude of each vector?
(a) 60°
(b) 120°
(c) 0°
(d) 90°
Show Answer & Explanation

Answer: (b) 120°

Explanation:
1. |A + B|² = A² + A² + 2A² cos θ = A².
2. So cos θ = −1/2 and θ = 120°.

Teacher's Note:
1. Use the cosine rule for vector addition.

Question 3: If a particle moves in a circle describing equal angles in equal intervals of time, its velocity vector
(a) remains constant
(b) changes in magnitude
(c) changes in direction
(d) changes in both magnitude and direction
Show Answer & Explanation

Answer: (c) changes in direction

Explanation:
1. Equal angles in equal times means constant angular speed and constant speed.
2. The direction of the velocity keeps rotating.

Teacher's Note:
1. Speed is constant; velocity direction changes.

Question 4: A small particle of mass m is projected at an angle θ with the x-axis with initial velocity v₀ in the x-y plane (x horizontal, y vertically up). At a time t < v₀ sin θ/g, the angular momentum of the particle about the origin is
(a) \( -mg v_0t^{2}\cos\theta\,\hat j \)
(b) \( mg v_0t\cos\theta\,\hat k \)
(c) \( -\frac12 mg v_0t^{2}\cos\theta\,\hat k \)
(d) \( \frac12 mg v_0t^{2}\cos\theta\,\hat i \)
Show Answer & Explanation

Answer: (c) \( -\frac12 mg v_0t^{2}\cos\theta\,\hat k \)

Explanation:
1. L_z = m(x vy − y vx) with x = v₀ cos θ t, y = v₀ sin θ t − gt²/2, vx = v₀ cos θ and vy = v₀ sin θ − gt.
2. L_z = m v₀ cos θ (−gt² + gt²/2) = −½ m g v₀ t² cos θ.
3. The direction is along −k.

Teacher's Note:
1. Compute x vy − y vx and attach the axis direction.

Question 5: Two balls are projected simultaneously from the same point in the same vertical plane with velocities v1 and v2 at angles θ1 and θ2 with the horizontal. If v1 cos θ1 = v2 cos θ2, the path of one ball as seen from the position of the other is
(a) a parabola
(b) a horizontal straight line
(c) a vertical straight line
(d) a straight line making 45° with the vertical
Show Answer & Explanation

Answer: (c) a vertical straight line

Explanation:
1. The horizontal velocity components are equal, so the horizontal relative velocity is zero.
2. The relative acceleration is zero too, so the relative motion is along a vertical straight line.

Teacher's Note:
1. Equal horizontal components give no relative horizontal motion.

Question 6: A particle describes uniform circular motion in a circle of radius 2 m with angular speed 2 rad/s. The magnitude of the change in its velocity in π/2 s is
(a) 0 m/s
(b) \( 2\sqrt2\ \text{m/s} \)
(c) 8 m/s
(d) 4 m/s
Show Answer & Explanation

Answer: (c) 8 m/s

Explanation:
1. Speed v = ωr = 4 m/s.
2. The angle turned in π/2 s is ωt = π, so the velocity reverses.
3. The change is 4 − (−4) = 8 m/s.

Teacher's Note:
1. Reversal of velocity doubles its magnitude as change.

Question 7: Two stones move with the same angular speed in circular paths of radii 1 m and 2 m. The ratio of their linear speeds is
(a) 2
(b) 1/2
(c) 1/3
(d) 3
Show Answer & Explanation

Answer: (b) 1/2

Explanation:
1. v = ωr with the same ω, so v1/v2 = r1/r2.
2. = 1/2.

Teacher's Note:
1. Linear speed is proportional to radius for the same angular speed.

Question 8:
Assertion (A): The trajectory of a projectile is quadratic in y and linear in x.
Reason (R): The y component of the trajectory is independent of the x component.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. The equation y = x tan θ − gx²/(2u² cos² θ) is quadratic in x and linear in y, so the Assertion is false.
2. The vertical and horizontal motions are independent, so the Reason is true.

Teacher's Note:
1. Check which variable carries the square.

Question 9: Two projectiles A and B thrown with speeds in the ratio 1 : √2 acquire the same heights. If A is thrown at an angle of 45° with the horizontal, the angle of projection of B is
(a) 0°
(b) 60°
(c) 30°
(d) 45°
Show Answer & Explanation

Answer: (c) 30°

Explanation:
1. H ∝ u² sin²θ, so uA sin 45° = uB sin θ.
2. With uB = √2 uA: 1/√2 = √2 sin θ, so sin θ = 1/2 and θ = 30°.

Teacher's Note:
1. Equal heights fix the vertical component.

Question 10: The direction of the angular velocity vector is along
(a) the tangent to the circular path
(b) the inward radius
(c) the outward radius
(d) the axis of rotation
Show Answer & Explanation

Answer: (d) the axis of rotation

Explanation:
1. Angular velocity is an axial vector.
2. It points along the axis of rotation in the direction given by the right-hand rule.

Teacher's Note:
1. Use the right-hand rule on the sense of rotation.

Question 11: Centripetal acceleration is
(a) a constant vector
(b) a constant scalar
(c) a vector of changing magnitude
(d) not a constant vector
Show Answer & Explanation

Answer: (d) not a constant vector

Explanation:
1. Its magnitude v²/r is constant in uniform circular motion, but its direction always points to the centre and keeps changing.
2. So it is not a constant vector.

Teacher's Note:
1. A vector with changing direction is not constant.

Question 12: A cricket ball is hit at an angle of 30° to the horizontal with kinetic energy E. Its kinetic energy when it reaches the highest point is
(a) \( \frac E2 \)
(b) 0
(c) \( \frac{2E}{3} \)
(d) \( \frac{3E}{4} \)
Show Answer & Explanation

Answer: (d) \( \frac{3E}{4} \)

Explanation:
1. At the top only the horizontal component u cos 30° remains.
2. KE = ½m(u cos 30°)² = E cos² 30° = 3E/4.

Teacher's Note:
1. Only the horizontal velocity survives at the top.

Question 13: A ball is thrown from the rear end of a train compartment to the front end, the train moving with a constant horizontal velocity. An observer A sitting in the compartment and an observer B on the ground both draw the trajectory. They will have
(a) equal horizontal and equal vertical ranges
(b) equal vertical ranges but different horizontal ranges
(c) different vertical ranges but equal horizontal ranges
(d) different horizontal and different vertical ranges
Show Answer & Explanation

Answer: (b) equal vertical ranges but different horizontal ranges

Explanation:
1. The vertical motion is the same in both frames, so the heights agree.
2. The ground observer adds the train's speed to the horizontal velocity, so the horizontal range is larger.

Teacher's Note:
1. Frames moving horizontally relative to each other agree on vertical motion.

Question 14: A projectile has maximum range at an angle ...X... and the maximum range is equal to ...Y.... X and Y are
(a) \( \frac{\pi}{4}\ \text{and}\ \frac{v_0^{2}}{2g} \)
(b) \( \frac{\pi}{2}\ \text{and}\ \frac{v_0}{g} \)
(c) \( \frac{\pi}{4}\ \text{and}\ \frac{v_0^{2}}{g} \)
(d) \( \frac{\pi}{2}\ \text{and}\ \frac{v_0^{2}}{g} \)
Show Answer & Explanation

Answer: (c) \( \frac{\pi}{4}\ \text{and}\ \frac{v_0^{2}}{g} \)

Explanation:
1. R = v0² sin 2θ/g is maximum when 2θ = 90°, so θ = π/4.
2. R_max = v0²/g.

Teacher's Note:
1. Maximum range occurs at 45°.

Question 15: When a body moves with a constant speed along a circle
(a) no work is done on it
(b) no acceleration is produced in the body
(c) no force acts on the body
(d) its velocity remains constant
Show Answer & Explanation

Answer: (a) no work is done on it

Explanation:
1. The centripetal force is perpendicular to the displacement at every point.
2. So the work done is zero.
3. The body is accelerated and its velocity changes direction.

Teacher's Note:
1. A force perpendicular to motion does no work.

Question 16: The component of a vector r along the x-axis will have maximum value if
(a) r is along the positive y-axis
(b) r is along the positive x-axis
(c) r makes an angle of 45° with the x-axis
(d) r is along the negative y-axis
Show Answer & Explanation

Answer: (b) r is along the positive x-axis

Explanation:
1. The x-component is r cos θ, where θ is the angle with the x-axis.
2. It is largest when cos θ = 1, that is when r lies along the positive x-axis.

Teacher's Note:
1. The component along an axis is the magnitude times the cosine of the angle with that axis.

Question 17: A stone tied to the end of a string 1 m long is whirled in a horizontal circle with constant speed. If the stone makes 22 revolutions in 44 seconds, the magnitude and direction of the acceleration of the stone are
(a) π² m/s² along the radius towards the centre
(b) π² m/s² along the radius away from the centre
(c) π² m/s² along the tangent to the circle
(d) π²/4 m/s² along the radius towards the centre
Show Answer & Explanation

Answer: (a) π² m/s² along the radius towards the centre

Explanation:
1. Frequency = 22/44 = 0.5 Hz, so ω = π rad/s.
2. a = ω²r = π² m/s², directed towards the centre.

Teacher's Note:
1. Centripetal acceleration is ω²r.

Question 18:
Assertion (A): If A·B = B·C, then A may not always be equal to C.
Reason (R): The dot product of two vectors involves the cosine of the angle between them.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. B·(A − C) = 0 means A − C is perpendicular to B, not that A = C.
2. The cosine in the dot product allows different vectors to give the same product, which explains it.

Teacher's Note:
1. Equal projections on B do not make two vectors equal.

Question 19: The vertical component of velocity of a projectile at its maximum height (u is the velocity of projection and θ the angle of projection) is
(a) \( u\sin\theta \)
(b) \( u\cos\theta \)
(c) \( \frac{u}{\sin\theta} \)
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. The projectile stops rising at the top.
2. So the vertical component is zero there.

Teacher's Note:
1. Zero vertical velocity marks the top.

Question 20:
Assertion (A): When the range is maximum, the angle of projection may be 45° or 135°.
Reason (R): Whether θ is 45° or 135°, the value of the range remains the same; only the sign changes.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. sin 2θ = ±1 at 45° and 135°, so the magnitude of the range is the same, with a reversed direction at 135°.
2. The Reason correctly states this.

Teacher's Note:
1. Projection at 135° throws the body in the opposite horizontal direction.

Question 21: A particle of mass m is projected with velocity v at 30° with the horizontal. The magnitude of its angular momentum about the point of projection when it is at its maximum height h is
(a) \( \frac{\sqrt3}{2}\frac{mv^{2}}{g} \)
(b) zero
(c) \( \frac{mv^{3}}{\sqrt2 g} \)
(d) \( \frac{\sqrt3}{16}\frac{mv^{3}}{g} \)
Show Answer & Explanation

Answer: (d) \( \frac{\sqrt3}{16}\frac{mv^{3}}{g} \)

Explanation:
1. At the top the velocity is v cos 30° (horizontal) and the height is h = v² sin² 30°/2g = v²/8g.
2. L = m(v cos 30°)h = m (√3 v/2)(v²/8g) = √3 m v³/16g.

Teacher's Note:
1. L equals momentum times the perpendicular height.

Question 22: A body travelling in a circle at constant speed
(a) has a constant velocity
(b) is not accelerated
(c) has an inward radial acceleration
(d) has an outward radial acceleration
Show Answer & Explanation

Answer: (c) has an inward radial acceleration

Explanation:
1. The velocity changes direction continuously.
2. The resulting acceleration v²/r is directed towards the centre.

Teacher's Note:
1. Constant speed does not mean zero acceleration.

Question 23: The resultant of \( \vec A\times\vec 0 \) will be equal to
(a) zero
(b) A
(c) a zero vector
(d) a unit vector
Show Answer & Explanation

Answer: (c) a zero vector

Explanation:
1. The cross product of two vectors is a vector.
2. Crossing with the null vector gives the null (zero) vector, not the scalar zero.

Teacher's Note:
1. A cross product is always a vector, so the result is a zero vector.

Question 24:
Assertion (A): If the dot product and the cross product of A and B are both zero, one of the vectors must be a null vector.
Reason (R): A null vector is a vector with zero magnitude.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A·B = 0 needs cos θ = 0 or a zero vector, and A × B = 0 needs sin θ = 0 or a zero vector.
2. Both cannot hold for non-zero vectors, so one vector must be null.
3. The Reason defines the null vector used in this conclusion.

Teacher's Note:
1. cos θ and sin θ cannot vanish together.

Question 25: Select the correct statements about a football thrown in a parabolic path. I. At the highest point the vertical component of velocity is zero. II. At the highest point the velocity of the football acts horizontally. III. At the highest point the acceleration of the ball acts vertically downwards.
(a) I and II
(b) II and III
(c) I and III
(d) I, II and III
Show Answer & Explanation

Answer: (d) I, II and III

Explanation:
1. At the top the vertical velocity is zero, so the velocity is horizontal.
2. Gravity still acts, so the acceleration is vertically downward.

Teacher's Note:
1. Velocity and acceleration need not be in the same direction.

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