MATDSC-252: Mathematics FYUG Even Semester Exam, 2025

Programme: FYUG (4th Semester)

Course Code: MATDSC-252

Full Marks: 70 | Pass Marks: 28

Subject Matter: Statics & Dynamics

UNIT I: Statics (Basics & Equilibrium)

2 Marks

Question 1 (a)

State the law of triangle of forces.

If three forces acting at a point can be represented in magnitude and direction by the sides of a triangle taken in order, then they are in equilibrium. Conversely, if three forces acting at a point are in equilibrium, they can be represented by the sides of a triangle.
2 Marks

Question 1 (b)

Define moment of a force.

The moment of a force about a point is a measure of its tendency to rotate a body about that point. Mathematically, it is the product of the magnitude of the force (F) and the perpendicular distance (p) from the point to the line of action of the force.

Formula: M = F * p

6 Marks

Question 2 (a)

State and prove Lami's theorem.

Statement: If three coplanar forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces.

Proof: Let forces P, Q, and R act at point O and be in equilibrium. Let α, β, and γ be the angles between (Q,R), (R,P), and (P,Q) respectively.

By the triangle law of forces, these forces can be represented by the sides of a triangle ABC. By the Sine Rule in triangle ABC:

BC / sin(A) = CA / sin(B) = AB / sin(C)

Since the interior angles of the triangle are related to the angles between the forces (e.g., A = 180° - α), we get:

P / sin α = Q / sin β = R / sin γ

UNIT II: Friction & Centre of Gravity

2 Marks

Question 3 (b)

Write two laws of limiting friction.

  1. The direction of limiting friction is always opposite to the direction in which the body intends to move.
  2. The magnitude of limiting friction (F) is directly proportional to the normal reaction (R) between the surfaces. (F = μR)
6 Marks

Question 4 (a)

Show that the inclination of a uniform ladder resting on rough ground and a rough wall is given by tan θ = (1 - μμ') / 2μ.

Proof: Let the ladder AB have length 2a and weight W acting at its midpoint G. Let θ be the angle with the horizontal.

  • At ground A: Normal reaction R, Friction μR.
  • At wall B: Normal reaction S, Friction μ'S.

Resolving forces horizontally: μR = S

Resolving forces vertically: R + μ'S = W => R + μ'(μR) = W => R(1 + μμ') = W

Taking moments about A: W * a cos θ = S * 2a sin θ + μ'S * 2a cos θ

Divide by a cos θ: W = 2S tan θ + 2μ'S

Substitute W and S: R(1 + μμ') = 2(μR) tan θ + 2μ'(μR)

1 + μμ' = 2μ tan θ + 2μμ' => 1 - μμ' = 2μ tan θ

tan θ = (1 - μμ') / 2μ

UNIT III: Dynamics & SHM

4 Marks

Question 5 (a)

Write expressions for radial and cross-radial components of velocity and acceleration.

Velocity:

  • Radial velocity (vᵣ) = dr/dt = ṙ
  • Cross-radial velocity (v_θ) = r(dθ/dt) = rθ̇

Acceleration:

  • Radial acceleration (aᵣ) = ṙ̈ - rθ̇²
  • Cross-radial acceleration (a_θ) = (1/r) * d/dt(r²θ̇)
2 Marks

Question 5 (b)

Define SHM. Write also the expression for time period.

Simple Harmonic Motion (SHM) is the motion of a particle along a straight line such that its acceleration is always directed towards a fixed point on the line and is proportional to its distance from that point.

T = 2π / √μ

UNIT IV: Projectiles & Central Orbits

5 Marks

Question 7 (c)

Prove that the path of a projectile in vacuum is a parabola.

Proof: Let a particle be projected with velocity u at an angle α.

Horizontal distance: x = (u cos α)t => t = x / (u cos α)

Vertical distance: y = (u sin α)t - (1/2)gt²

Substituting t: y = u sin α * (x / u cos α) - (1/2)g * (x / u cos α)²

y = x tan α - (gx²) / (2u² cos² α)

This is of the form y = ax - bx², which represents a parabola.

UNIT V: Impulse, Momentum & Impact

4 Marks

Question 9 (b)

Show that Impulse = Change in momentum.

From Newton's Second Law: F = dp/dt = d(mv)/dt

F dt = d(mv)

Integrating from t₁ to t₂: ∫ F dt = mv₂ - mv₁

The integral of force over time is Impulse (J). Thus, J = Change in Momentum.

5 Marks

Question 10 (c)

A ball impinges directly upon another at rest and itself reduces to rest. If half of the KE is destroyed, find the coefficient of restitution (e).

Let masses be m₁, m₂ and initial velocity of m₁ be u. Let final velocities be v₁=0 and v₂.

  1. Momentum: m₁u = m₂v₂ => v₂ = (m₁/m₂)u
  2. Restitution: v₂ - v₁ = e(u - 0) => v₂ = eu
  3. Equating v₂: eu = (m₁/m₂)u => e = m₁/m₂
  4. Energy: Initial KE = (1/2)m₁u². Final KE = (1/2)m₂v₂² = (1/2)m₂(eu)².
  5. Given Final KE = (1/2) * Initial KE => (1/2)m₂e²u² = (1/4)m₁u²
  6. 2m₂e² = m₁ => e² = m₁ / (2m₂)
  7. Since e = m₁/m₂, substitute m₁ = em₂: e² = em₂ / (2m₂) => e² = e/2

Final Answer: e = 1/2