JEE Physics Motion In A Plane MCQs Set 27

Answer: (c) \( |\hat i|=|\hat j|=|\hat k|=1 \)

Explanation:
1. A unit vector has magnitude exactly 1.
2. So the three axis unit vectors all have magnitude 1, and magnitudes can never be negative.

Teacher's Note:
1. Unit vectors only indicate direction and carry magnitude 1.

Answer: (a) A-1, B-2, C-4, D-3

Explanation:
1. |A| = 5 and the unit vector is A/5 = 0.6i + 0.8j.
2. A + B = 4i + 2j with magnitude √20.
3. A − B = 2i + 6j.

Teacher's Note:
1. Components make vector arithmetic simple.

Answer: (d) kinetic energy

Explanation:
1. Speed is constant, so kinetic energy is constant.
2. Velocity, momentum and acceleration change direction continuously.

Teacher's Note:
1. Only quantities depending on speed alone are constant.

Answer: (b) II only

Explanation:
1. On a curved path the velocity direction changes, so there is acceleration even at constant speed.
2. For constant speed this acceleration is perpendicular to the velocity.

Teacher's Note:
1. Constant speed with changing direction needs centripetal acceleration.

Answer: (c) an axial vector

Explanation:
1. Angular momentum L = r × p is directed along the axis of rotation.
2. Vectors that point along an axis and follow the right-hand rule are axial vectors.

Teacher's Note:
1. Torque, angular velocity and angular momentum are axial vectors.

Answer: (a) 1 : 1

Explanation:
1. The two angles add up to 90°, so they are complementary.
2. Complementary angles give equal ranges.

Teacher's Note:
1. R is the same for θ and 90° − θ.

Answer: (a) \( 9\hat i+6\hat j \)

Explanation:
1. a = F/m = 2t i + (4t/3) j.
2. v = ∫a dt = t² i + (2t²/3) j.
3. At t = 3 s, v = 9i + 6j m/s.

Teacher's Note:
1. Integrate the acceleration from rest.

Answer: (b) \( 2t\sqrt{a^{2}+b^{2}} \)

Explanation:
1. vx = 2at and vy = 2bt.
2. Speed = 2t√(a² + b²).

Teacher's Note:
1. Speed is the magnitude of the velocity vector.

Answer: (c) negligible

Explanation:
1. Projectile motion theory neglects air resistance.
2. Its effect is taken as negligible.

Teacher's Note:
1. The ideal projectile moves under gravity alone.

Answer: (a) 40 m

Explanation:
1. Maximum range = u²/g.
2. = 400/10 = 40 m.

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

Answer: (a) periodic but not simple harmonic

Explanation:
1. The motion repeats after every revolution, so it is periodic.
2. It is not simple harmonic because the acceleration is not directed along a line in proportion to displacement.

Teacher's Note:
1. Its projection on a diameter is SHM, but the circular motion itself is not.

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

Explanation:
1. Fragments can fly in any direction as long as the total momentum is conserved, for example one fragment up and one down, so the Assertion is false.
2. The Reason is correct.

Teacher's Note:
1. Zero net momentum in a direction allows opposite fragments.

Answer: (a) \( \propto R \)

Explanation:
1. The two angles are θ and 90° − θ.
2. t₁t₂ = (2u sin θ/g)(2u cos θ/g) = 2(u² sin 2θ/g)/g = 2R/g, so t₁t₂ ∝ R.

Teacher's Note:
1. Use complementary angles for equal range.

Answer: (a) perpendicular to each other

Explanation:
1. Velocity is tangential and acceleration is radial.
2. So they are mutually perpendicular.

Teacher's Note:
1. Tangent and radius are perpendicular.

Answer: (c) parabolic path

Explanation:
1. Relative to the ground the stone has a horizontal velocity equal to the train's and falls under gravity.
2. This is projectile motion, a parabola. (The path is a straight line only for an observer in the train.)

Teacher's Note:
1. Choose the frame of the ground observer.

Answer: (d) A·B ≤ 0

Explanation:
1. Squaring: A² + 2A·B + B² = A², so A·B = −B²/2.
2. Since B² ≥ 0, we get A·B ≤ 0 always.
3. B = 0 is only one special case, and the vectors need not be perpendicular or antiparallel.

Teacher's Note:
1. Square both sides and compare with the dot-product expansion.

Answer: (b) −1/2

Explanation:
1. The dot product must be zero: 2(−4) + 3(4) + 8α = 0.
2. So 4 + 8α = 0 and α = −1/2.

Teacher's Note:
1. Perpendicular vectors have zero dot product.

Answer: (b) constant velocity

Explanation:
1. No horizontal force acts, so the horizontal acceleration is zero.
2. The horizontal velocity therefore stays constant.

Teacher's Note:
1. Horizontal and vertical motions are independent.

Answer: (c) 15 km/h

Explanation:
1. For motion in the same direction, relative velocity = 80 − 65.
2. = 15 km/h.

Teacher's Note:
1. Subtract the velocities when both move the same way.

Answer: (a) 3.5

Explanation:
1. The trajectory y = x tan θ − gx²/(2u² cos² θ) is the same if g/u² is the same.
2. g′ = 9.8 × (3/5)² = 9.8 × 0.36 = 3.5 m/s².

Teacher's Note:
1. Equate g/u² for both cases at the same angle.

Answer: (b) radial velocity and tangential acceleration

Explanation:
1. The velocity is entirely tangential, so the radial velocity is zero.
2. The speed is constant, so the tangential acceleration is zero; only radial acceleration exists.

Teacher's Note:
1. Velocity tangential, acceleration radial.

Answer: (a) I and II

Explanation:
1. In circular motion at constant speed the particle returns to its starting point, so the average velocity can be zero.
2. The average acceleration also need not vanish.
3. Constant speed does give equal path lengths in equal intervals, so III is correct.

Teacher's Note:
1. Constant speed does not mean constant velocity.

Answer: (a) 10t

Explanation:
1. vx = 8t and vy = 6t.
2. Speed = √(64t² + 36t²) = 10t.

Teacher's Note:
1. Differentiate each coordinate and combine.

Answer: (c) \( a_r\neq0\ \text{and}\ a_t=0 \)

Explanation:
1. Uniform speed requires a_t = 0.
2. Moving in a circle requires a non-zero radial (centripetal) acceleration.

Teacher's Note:
1. No tangential acceleration, but centripetal acceleration is present.

Answer: (b) \( \theta=\tan^{-1}4 \)

Explanation:
1. R = 2u² sin θ cos θ/g and H = u² sin² θ/(2g).
2. Setting R = H gives 2 cos θ = sin θ/2, so tan θ = 4.

Teacher's Note:
1. Equate the two formulas and simplify.