Motion in a Plane

What to remember

• A vector has magnitude and direction and obeys the triangle (parallelogram) law of addition. The resultant of vectors A and B at angle θ has magnitude R = √(A² + B² + 2AB cos θ).
• Dot product: A·B = AB cos θ (a scalar). Cross product: A × B = AB sin θ n̂ (a vector perpendicular to both, given by the right-hand rule). A × B = −(B × A).
• Projectile motion: the horizontal velocity is constant and the vertical motion has acceleration g downward. Time of flight T = 2u sin θ/g; maximum height H = u² sin²θ/(2g); range R = u² sin 2θ/g. The range is maximum at θ = 45°.
• In uniform circular motion, the speed is constant but the velocity changes. The acceleration is centripetal: a = v²/r = ω²r, directed towards the centre.

Scalars and vectors

A scalar has magnitude only: mass, distance, speed, time, energy, temperature. A vector has both magnitude and direction and obeys the vector rules of addition: displacement, velocity, acceleration, force. Current and pressure have magnitude and direction in some sense but are scalars, because they do not obey the vector law of addition.

A vector is written in bold or with an arrow; in these notes, "vector A" and its magnitude is |A| = A.

• Position vector: the vector from the origin to the point P(x, y). r = x î + y ĵ.
• Displacement vector: the vector from the initial position to the final position, Δr = r₂ − r₁. It depends only on the end points, not on the path.
• Equal vectors: same magnitude and same direction, wherever they are placed.
• Negative of a vector: same magnitude, opposite direction.
• Multiplying vector A by a real number λ gives a vector of magnitude |λ|A. If λ > 0 the direction is the same as A; if λ < 0 it is opposite.
• Unit vector: a vector of magnitude 1 which shows direction only. n̂ = A/|A|. The unit vectors along x, y and z are î, ĵ and k̂.
• Null (zero) vector: has zero magnitude and no definite direction.

Addition and subtraction of vectors

Triangle law. If two vectors are drawn head to tail, the resultant is the vector from the tail of the first to the head of the second. Parallelogram law. If two vectors are drawn from the same point as adjacent sides of a parallelogram, the resultant is the diagonal from that point.

Analytical method. For vectors A and B with an angle θ between them (when drawn from the same point):

• Magnitude of the resultant: R = √(A² + B² + 2AB cos θ).
• Direction: tan α = B sin θ/(A + B cos θ), where α is the angle of R with A.

For subtraction, A − B = A + (−B). Its magnitude is √(A² + B² − 2AB cos θ).

Properties: vector addition is commutative (A + B = B + A) and associative ((A + B) + C = A + (B + C)). The resultant of three vectors can be zero only if the magnitude of each is not greater than the sum of the other two (the largest cannot exceed the sum of the others). The resultant of two vectors lies between |A − B| and A + B.

Resolution of a vector

A vector can be split into components along perpendicular axes. For a vector A making angle θ with the x-axis:

• Ax = A cos θ and Ay = A sin θ.
• A = Ax î + Ay ĵ.
• Magnitude A = √(Ax² + Ay²), direction tan θ = Ay/Ax.

For vectors given in components, A + B = (Ax + Bx) î + (Ay + By) ĵ. Addition and subtraction are done component by component.

Example: A = 3î + 4ĵ has magnitude 5. The unit vector along A is (3î + 4ĵ)/5 = 0.6î + 0.8ĵ.

In three dimensions, A = Ax î + Ay ĵ + Az k̂ and A = √(Ax² + Ay² + Az²).

Scalar (dot) product and vector (cross) product

The direction of A × B is given by the right-hand rule: curl the fingers of the right hand from A to B through the smaller angle; the thumb points along A × B. The magnitude |A × B| equals the area of the parallelogram with sides A and B.

Uses:

• Work W = F·s (dot product). Power P = F·v.
• Torque τ = r × F. Angular momentum L = r × p.
• Angle between vectors: cos θ = A·B/(AB).
• Component of A along B = A·B̂ = (A·B)/B.
• Two vectors are perpendicular if A·B = 0; parallel if A × B = 0.

Example: A = 2î + 3ĵ and B = î − ĵ. A·B = 2 − 3 = −1. A × B = (2)(−1) − (3)(1) = −5 in the k̂ direction, so A × B = −5k̂.

Motion in a plane

For a particle moving in a plane with position r = x î + y ĵ:

• Velocity v = dr/dt = vx î + vy ĵ, where vx = dx/dt and vy = dy/dt. The velocity is always tangent to the path.
• Acceleration a = dv/dt = ax î + ay ĵ.
• The motion along x and along y are independent of each other.

For constant acceleration a in a plane:

• v = v₀ + at
• r = r₀ + v₀t + ½at²

These vector equations are the same as the two sets of component equations: vx = v₀x + axt, x = x₀ + v₀x t + ½ax t², and the same for y.

Projectile motion

A projectile is a body thrown into the air that moves under gravity alone. Let it be launched with speed u at angle θ above the horizontal (air resistance neglected).

• Initial components: ux = u cos θ, uy = u sin θ.
• Horizontal: ax = 0, so x = u cos θ · t and vx = u cos θ is constant.
• Vertical: ay = −g, so y = u sin θ · t − ½gt² and vy = u sin θ − gt.

Facts:

• At the highest point, the velocity is u cos θ horizontal (vertical component is zero). The speed there is the minimum. The acceleration is still g downward.
• Range is maximum when θ = 45°: R_max = u²/g.
• For angles θ and (90° − θ) (complementary angles), the range is the same, because sin 2θ = sin(180° − 2θ). Their heights and times of flight are different.
• At θ = 45°, R = 4H. In general, H/R = tan θ/4, so tan θ = 4H/R.
• R = 4H/tan θ. At θ = 45°, H = R/4.
• A projectile launched from the ground returns with the same speed u, at the same angle below the horizontal.

Horizontal projection from a height. A body is projected horizontally with speed u from height h. The initial vertical velocity is zero.

• Time of flight: T = √(2h/g). It does not depend on u.
• Horizontal range: R = u√(2h/g).
• Trajectory: y = g x²/(2u²) (measured downward from the launch point).
• Final speed: v = √(u² + 2gh). Angle with the horizontal: tan α = vy/u = √(2gh)/u.

Worked example 1. A ball is thrown at 20 m s⁻¹ at 30° above the horizontal (g = 10 m s⁻²). T = 2 × 20 × 0.5/10 = 2 s. H = 400 × 0.25/20 = 5 m. R = 400 × sin 60°/10 = 40 × 0.866 = 34.6 m.

Worked example 2. A stone is thrown horizontally at 15 m s⁻¹ from a cliff 80 m high (g = 10 m s⁻²). T = √(2 × 80/10) = 4 s. Range = 15 × 4 = 60 m. Vertical speed at the ground = 10 × 4 = 40 m s⁻¹.

Uniform circular motion

A particle moves on a circle of radius r with constant speed v. Its velocity changes direction, so it has acceleration.

• The centripetal acceleration points towards the centre. Its magnitude is constant, but its direction changes continuously.
• The velocity is along the tangent. It is perpendicular to the acceleration at every instant.
• In uniform circular motion, the speed is constant, but the velocity and acceleration are not constant (as vectors).
• Change in velocity when the particle turns through angle θ: |Δv| = 2v sin(θ/2). In a half turn, |Δv| = 2v.
• The centripetal force is mv²/r. It is a name for the net force towards the centre; it is not a new kind of force.

Example: A stone is whirled in a circle of radius 0.5 m, making 2 revolutions per second. ω = 2π × 2 = 4π rad s⁻¹, v = ωr = 2π m s⁻¹ ≈ 6.28 m s⁻¹, a = ω²r = 16π² × 0.5 ≈ 79 m s⁻².

Relative velocity in two dimensions

The velocity of A relative to B is v_AB = v_A − v_B (a vector subtraction).

Rain and umbrella. Rain falls vertically at speed v_r. A man walks with speed v_m on level ground. The rain relative to the man has horizontal component v_m (towards the man) and vertical component v_r. Its speed is √(v_r² + v_m²). The man must hold the umbrella tilted forward (towards his direction of motion) at angle θ from the vertical, where tan θ = v_m/v_r.

River and boat. Let v_b be the speed of the boat in still water and v_r the speed of the river current. The boat's velocity relative to the ground is the vector sum of the two.

The crossing time for the shortest-time heading depends only on the perpendicular component of the boat's velocity relative to water, not on the river's speed.

Worked example 3. A boat has a speed of 5 m s⁻¹ in still water. It crosses a river 100 m wide with current 3 m s⁻¹. Shortest time = 100/5 = 20 s, drift = 3 × 20 = 60 m. For zero drift, sin θ = 3/5, so the net speed across is √(25 − 9) = 4 m s⁻¹ and the time is 100/4 = 25 s.

Formula summary

NEET traps

• 1. The angle in R = √(A² + B² + 2AB cos θ) is the angle between the vectors drawn tail to tail; the "+" sign becomes "−" for subtraction.
• 2. Electric current and pressure have direction in a loose sense but are scalars, as they do not follow the vector law of addition.
• 3. A × B = −B × A, but A·B = B·A.
• 4. For complementary launch angles, the range is equal, but the time of flight and the maximum height are different.
• 5. At the top of a projectile path the speed is not zero (it is u cos θ) unless the launch is vertical.
• 6. The time of flight in horizontal projection depends only on the height, not on the launch speed.
• 7. In uniform circular motion, the acceleration is not zero and not constant: its direction changes continuously.
• 8. The centripetal force is not an extra force acting in addition to the others.
• 9. Rain-umbrella: the umbrella is tilted towards the direction of the man's motion, from the vertical, with tan θ = v_man/v_rain.
• 10. A boat reaches the opposite point directly across only if its speed in still water is greater than the river speed; shortest time and shortest path are different headings.

One-liners

• 1. The resultant of two equal vectors of magnitude A at 120° has magnitude A.
• 2. The magnitude of the resultant of two vectors lies between |A − B| and A + B.
• 3. î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ.
• 4. The dot product of perpendicular vectors is zero; the cross product of parallel vectors is zero.
• 5. The area of a parallelogram with sides A and B is |A × B|.
• 6. Work is the dot product of force and displacement.
• 7. Torque is the cross product r × F.
• 8. The path of a projectile is a parabola.
• 9. The maximum range is u²/g at 45°.
• 10. At 45°, the range equals four times the maximum height.
• 11. Complementary angles give equal ranges.
• 12. The horizontal velocity of a projectile stays constant.
• 13. For horizontal projection from height h, the time of flight is √(2h/g).
• 14. Centripetal acceleration = v²/r = ω²r.
• 15. In a half revolution of uniform circular motion, the change in velocity has magnitude 2v.

Practice questions

1. The resultant of two perpendicular vectors of magnitudes 3 N and 4 N has magnitude

   1. 5 N
   2. 7 N
   3. 12 N
   4. 1 N
   Answer

   A. 5 N

   R = √(3² + 4²) = 5 N (7 N is the maximum, at 0°).

2. Which of the following is a scalar quantity?

   1. torque
   2. displacement
   3. work
   4. momentum
   Answer

   C. work

   Work = F·s is the dot product, a scalar. The others are vectors.

3. For A = 2î + 3ĵ + k̂ and B = î − 2ĵ + 4k̂, the value of A·B is

   1. −4
   2. 12
   3. 8
   4. 0
   Answer

   D. 0

   A·B = 2(1) + 3(−2) + 1(4) = 2 − 6 + 4 = 0, so the vectors are perpendicular.

4. Two vectors of magnitudes 4 and 3 units make an angle of 30°. The magnitude of their cross product is

   1. 3√3 units
   2. 6 units
   3. 12 units
   4. 6√3 units
   Answer

   B. 6 units

   |A × B| = AB sin 30° = 4 × 3 × ½ = 6.

5. A projectile is launched at 20 m s⁻¹ at 30° above the horizontal. Taking g = 10 m s⁻², the time of flight is

   1. 1 s
   2. 2 s
   3. 4 s
   4. 2√3 s
   Answer

   B. 2 s

   T = 2u sin θ/g = 2 × 20 × 0.5/10 = 2 s.

6. A ball is thrown at 40 m s⁻¹ at 30° above the horizontal (g = 10 m s⁻²). The maximum height reached is

   1. 10 m
   2. 40 m
   3. 20 m
   4. 80 m
   Answer

   C. 20 m

   H = u² sin²θ/(2g) = 1600 × 0.25/20 = 20 m. Check: uy = 20 m s⁻¹, H = 400/20 = 20 m.

7. A body is projected at 20 m s⁻¹ at 45° to the horizontal (g = 10 m s⁻²). Its horizontal range is

   1. 40 m
   2. 20 m
   3. 10 m
   4. 80 m
   Answer

   A. 40 m

   R = u² sin 90°/g = 400/10 = 40 m.

8. The horizontal range of a projectile launched at 25° with a given speed equals its range when launched at the same speed at

   1. 75°
   2. 50°
   3. 45°
   4. 65°
   Answer

   D. 65°

   Equal ranges occur for complementary angles: 90° − 25° = 65°.

9. A stone is thrown horizontally at 10 m s⁻¹ from the top of a 45 m high tower (g = 10 m s⁻²). It lands at a horizontal distance of

   1. 45 m
   2. 90 m
   3. 30 m
   4. 10 m
   Answer

   C. 30 m

   T = √(2 × 45/10) = 3 s. Range = 10 × 3 = 30 m.

10. A wheel makes 120 revolutions per minute. Its angular velocity is

    1. 120π rad s⁻¹
    2. 240π rad s⁻¹
    3. 2π rad s⁻¹
    4. 4π rad s⁻¹
    Answer

    D. 4π rad s⁻¹

    120 rev/min = 2 rev/s. ω = 2π × 2 = 4π rad s⁻¹.

11. A particle moves in a circle of radius 5 m with a constant speed of 10 m s⁻¹. The magnitude of its acceleration is

    1. 50 m s⁻²
    2. 20 m s⁻²
    3. 0
    4. 2 m s⁻²
    Answer

    B. 20 m s⁻²

    a = v²/r = 100/5 = 20 m s⁻².

12. The unit vector along the vector 6î + 8ĵ is

    1. 0.6î + 0.8ĵ
    2. 0.8î + 0.6ĵ
    3. 6î + 8ĵ
    4. î + ĵ
    Answer

    A. 0.6î + 0.8ĵ

    Magnitude = 10, so the unit vector is (6î + 8ĵ)/10 = 0.6î + 0.8ĵ.

13. Two forces, each of magnitude F, act at an angle of 120° to each other. The magnitude of the resultant is

    1. 2F
    2. F/2
    3. √3 F
    4. F
    Answer

    D. F

    R = √(F² + F² + 2F² cos 120°) = √(2F² − F²) = F.

14. Two forces of 7 N and 4 N act at a point. Which of the following cannot be the magnitude of their resultant?

    1. 11 N
    2. 12 N
    3. 3 N
    4. 6 N
    Answer

    B. 12 N

    The resultant lies between 7 − 4 = 3 N and 7 + 4 = 11 N, so 12 N is impossible.

15. A vector of magnitude 10 units makes an angle of 30° with the x-axis. Its y-component is

    1. 5 units
    2. 10 units
    3. 5√3 units
    4. 10√3 units
    Answer

    A. 5 units

    Ay = A sin 30° = 10 × ½ = 5 units.

16. A particle goes half way round a circle of radius r. The ratio of the magnitude of its displacement to the path length is

    1. 1 : 2
    2. π : 2
    3. 2 : π
    4. 1 : π
    Answer

    C. 2 : π

    Displacement = diameter = 2r; path length = πr. Ratio = 2 : π.

17. Given below are two statements: Statement I: Vector addition is commutative. Statement II: Vector subtraction is commutative.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    A + B = B + A, but A − B = −(B − A), so Statement II is incorrect.

18. Given below are two statements: Statement I: A × B = B × A. Statement II: A·B = B·A.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The cross product changes sign on reversing the order, A × B = −B × A. Statement I is incorrect; the dot product is commutative.

19. Given below are two statements: Statement I: The range of a projectile is maximum when it is launched at 45°. Statement II: Projectiles launched at complementary angles with the same speed have the same range.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    R = u² sin 2θ/g is maximum at θ = 45° and has equal values for θ and 90° − θ. Both are correct.

20. Given below are two statements: Statement I: The acceleration of a projectile at the highest point of its path is zero. Statement II: The horizontal component of velocity of a projectile changes during its flight.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    The acceleration is g throughout, and the horizontal velocity stays constant. Both are incorrect.

21. Given below are two statements: Statement I: In uniform circular motion, the acceleration is constant. Statement II: In uniform circular motion, the speed is constant.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The acceleration has constant magnitude but changing direction, so it is not constant as a vector. Statement II is correct.

22. Which of the following statements are correct? 1. The dot product of two perpendicular vectors is zero. 2. The cross product of two parallel vectors is zero. 3. For any vector A, A·A = 0.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    A·A = A², not zero, so 3 is incorrect.

23. Which of the following statements are correct? 1. In horizontal projection, the time of flight does not depend on the launch speed. 2. In horizontal projection, the range is u√(2h/g). 3. The trajectory of a projectile is a parabola.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    D. 1, 2 and 3

    All three statements are correct.

24. Which of the following statements are correct? 1. A boat can reach the point directly opposite on a river bank only if its speed in still water is less than the river speed. 2. The shortest time to cross is obtained by heading perpendicular to the bank. 3. A man walking in vertical rain holds the umbrella tilted by θ from the vertical with tan θ = v_man/v_rain.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    B. 2 and 3 only

    Statement 1 is wrong: the boat needs a speed greater than the river speed. Statements 2 and 3 are correct.

25. Given below are two statements: Statement I: The displacement vector depends only on the initial and final positions. Statement II: The path length depends only on the initial and final positions.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Path length depends on the actual path. Statement II is incorrect.

26. Given below are two statements: Statement I: Torque is given by τ = r × F. Statement II: Work done by a force is W = F·s.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Torque is a cross product (vector) and work a dot product (scalar). Both are correct.

27. Assertion (A): A projectile launched at 30° and another launched at 60° with the same speed have the same range. Reason (R): The range depends on sin 2θ, which has the same value for θ and (90° − θ).

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    sin 60° = sin 120°, so R is equal; R explains A.

28. Assertion (A): The horizontal velocity of a projectile remains constant throughout its flight. Reason (R): At the highest point, the velocity of the projectile is purely horizontal.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    A is true because there is no horizontal force. R is true but does not explain why horizontal velocity is constant.

29. Assertion (A): A man walking in vertical rain holds his umbrella tilted forward. Reason (R): The rain acquires a horizontal velocity in the direction of the man's motion because he walks.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Relative to the man the rain comes with a horizontal component against his motion. R is false: the rain's own velocity does not change.

30. Assertion (A): A boat with a speed in still water smaller than the river speed can cross to the point directly opposite the starting point. Reason (R): For zero drift, the boat's upstream component must satisfy v_b sin θ = v_r.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    With v_b < v_r, sin θ = v_r/v_b exceeds 1, which is impossible. A is false; R is true.

31. Assertion (A): In horizontal projection from a given height, the time of flight depends on the launch speed. Reason (R): The horizontal velocity of the projectile is constant.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    The time of flight is √(2h/g), independent of the launch speed. A is false; R is true.

32. Assertion (A): In uniform circular motion, the magnitude of the centripetal acceleration is constant. Reason (R): The centripetal acceleration is always directed towards the centre of the circle.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    a = v²/r is constant in magnitude because v and r are constant. The direction statement is also true but does not explain why the magnitude is constant.

33. Match List I with List II. List I: a. Work b. Torque c. Displacement d. Speed List II: i. Scalar from a dot product ii. Vector from a cross product iii. Vector that depends only on end points iv. Scalar with no direction

    1. a-i, b-ii, c-iv, d-iii
    2. a-ii, b-i, c-iii, d-iv
    3. a-i, b-ii, c-iii, d-iv
    4. a-iii, b-ii, c-i, d-iv
    Answer

    C. a-i, b-ii, c-iii, d-iv

    Work is a dot product (scalar), torque a cross product (vector), displacement a vector between end points, speed a scalar.

34. Match List I with List II. A ball is projected at 20 m s⁻¹ at 30° above the horizontal (g = 10 m s⁻²). List I: a. Time of flight b. Maximum height c. Horizontal range d. Speed at the highest point List II: i. 5 m ii. 2 s iii. 20√3 m iv. 10√3 m s⁻¹

    1. a-ii, b-i, c-iii, d-iv
    2. a-i, b-ii, c-iii, d-iv
    3. a-ii, b-iii, c-i, d-iv
    4. a-ii, b-i, c-iv, d-iii
    Answer

    A. a-ii, b-i, c-iii, d-iv

    T = 2 s; H = 5 m; R = 400 × (√3/2)/10 = 20√3 m; speed at top = 20 cos 30° = 10√3 m s⁻¹.

35. Match List I with List II. List I: a. Angular velocity ω b. Linear speed v c. Centripetal acceleration a d. Period T List II: i. 2π/ω ii. ωr iii. ω²r iv. Δθ/Δt

    1. a-iv, b-ii, c-i, d-iii
    2. a-iv, b-ii, c-iii, d-i
    3. a-ii, b-iv, c-iii, d-i
    4. a-iv, b-iii, c-ii, d-i
    Answer

    B. a-iv, b-ii, c-iii, d-i

    ω = Δθ/Δt, v = ωr, a = ω²r, T = 2π/ω.

36. Match List I with List II. Vectors A and B have magnitudes 4 and 5. List I: a. A·B when the angle is 0° b. A·B when the angle is 90° c. Magnitude of A × B when the angle is 30° d. A·B when the angle is 180° List II: i. 0 ii. 10 iii. 20 iv. −20

    1. a-i, b-iii, c-ii, d-iv
    2. a-iii, b-ii, c-i, d-iv
    3. a-iii, b-i, c-ii, d-iv
    4. a-iii, b-i, c-iv, d-ii
    Answer

    C. a-iii, b-i, c-ii, d-iv

    4 × 5 = 20; cos 90° = 0; 20 sin 30° = 10; 20 cos 180° = −20.

37. Match List I with List II. Two vectors, each of magnitude F, make the angle given below. List I: a. 0° b. 60° c. 90° d. 120° List II: i. F ii. √2 F iii. 2F iv. √3 F

    1. a-iii, b-iv, c-ii, d-i
    2. a-ii, b-iv, c-iii, d-i
    3. a-iii, b-iv, c-i, d-ii
    4. a-iv, b-iii, c-ii, d-i
    Answer

    A. a-iii, b-iv, c-ii, d-i

    R = 2F cos(θ/2): 2F; √3 F; √2 F; F.

38. Match List I with List II. A boat has speed 5 m s⁻¹ in still water and crosses a 100 m wide river with current 3 m s⁻¹. List I: a. Shortest crossing time b. Drift in the shortest time c. Speed across the river for zero drift d. Time of crossing for zero drift List II: i. 25 s ii. 60 m iii. 20 s iv. 4 m s⁻¹

    1. a-iii, b-iv, c-ii, d-i
    2. a-iii, b-ii, c-i, d-iv
    3. a-iii, b-ii, c-iv, d-i
    4. a-ii, b-iii, c-iv, d-i
    Answer

    C. a-iii, b-ii, c-iv, d-i

    t = 100/5 = 20 s; drift = 3 × 20 = 60 m; for zero drift √(25 − 9) = 4 m s⁻¹, time = 100/4 = 25 s.

39. A ball is projected at 50 m s⁻¹ at an angle θ above the horizontal with sin θ = 0.6 and cos θ = 0.8 (g = 10 m s⁻²). Its horizontal range is

    1. 120 m
    2. 300 m
    3. 250 m
    4. 240 m
    Answer

    D. 240 m

    T = 2 × 30/10 = 6 s; horizontal speed = 40 m s⁻¹; R = 40 × 6 = 240 m. Check: R = 2 u² sin θ cos θ/g = 2 × 2500 × 0.48/10 = 240 m.

40. A stone is thrown horizontally at 15 m s⁻¹ from the top of a 20 m high cliff (g = 10 m s⁻²). The speed with which it hits the ground is

    1. 20 m s⁻¹
    2. 25 m s⁻¹
    3. 15 m s⁻¹
    4. 35 m s⁻¹
    Answer

    B. 25 m s⁻¹

    T = 2 s; vy = 20 m s⁻¹; v = √(15² + 20²) = 25 m s⁻¹. Check by energy: v² = 225 + 2 × 10 × 20 = 625.

41. Rain falls vertically at 12 m s⁻¹. A man walks on level ground at 5 m s⁻¹. The speed of the rain relative to the man is

    1. 17 m s⁻¹
    2. 7 m s⁻¹
    3. 10 m s⁻¹
    4. 13 m s⁻¹
    Answer

    D. 13 m s⁻¹

    v_rel = √(12² + 5²) = 13 m s⁻¹.

42. A boat has a speed of 10 m s⁻¹ in still water. It is to cross a 200 m wide river flowing at 6 m s⁻¹ and land at the point directly opposite the starting point. The time of crossing is

    1. 12.5 s
    2. 20 s
    3. 25 s
    4. 33 s
    Answer

    C. 25 s

    Net speed across = √(10² − 6²) = 8 m s⁻¹; t = 200/8 = 25 s. Check: sin θ = 6/10, cos θ = 0.8, speed across = 10 × 0.8 = 8 m s⁻¹.

43. A particle moves on a circle of radius 0.4 m with an angular velocity of 5 rad s⁻¹. Its centripetal acceleration is

    1. 12.5 m s⁻²
    2. 10 m s⁻²
    3. 50 m s⁻²
    4. 2 m s⁻²
    Answer

    B. 10 m s⁻²

    a = ω²r = 25 × 0.4 = 10 m s⁻². Check: v = ωr = 2 m s⁻¹ and v²/r = 4/0.4 = 10.

44. A particle moves in a circle at a constant speed of 10 m s⁻¹. The magnitude of the change in its velocity when it moves through one quarter of the circle is

    1. 10√2 m s⁻¹
    2. 0
    3. 10 m s⁻¹
    4. 20 m s⁻¹
    Answer

    A. 10√2 m s⁻¹

    The initial and final velocities are perpendicular, each of magnitude 10 m s⁻¹, so |Δv| = √(10² + 10²) = 10√2 m s⁻¹ (= 2v sin 45°).

45. With a given launch speed, the maximum horizontal range of a projectile on level ground is 80 m. If it is launched at the angle that gives this maximum range, the maximum height reached is

    1. 80 m
    2. 10 m
    3. 40 m
    4. 20 m
    Answer

    D. 20 m

    Maximum range at 45°: R = u²/g = 80, so u² = 800 (g = 10). H = u² sin²45°/(2g) = 800 × 0.5/20 = 20 m. Check: H = R/4.