Applied Mathematics — Second: Dynamics · Mathematics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 37 questions on this lesson, each with an explanation.
A body of mass 3 kg moves with speed 12 m/sec. Find the magnitude of its momentum.
Correct answer: (B) 36 kg.m/sec
Explanation: Step 1: Momentum p = m v. Step 2: p = 3 × 12 = 36. ∴ Momentum = 36 kg.m/sec
A horizontal force of magnitude 0.6 kg.wt acts on a body of mass 300 gm resting on a smooth horizontal plane. Find the acceleration of the body.
Correct answer: (B) 19.6 m/sec²
Explanation: Step 1: F = 0.6 kg.wt = 0.6 × 9.8 = 5.88 N and m = 300 gm = 0.3 kg. Step 2: a = F ÷ m = 5.88 ÷ 0.3 = 19.6 m/sec².
A body moves in a straight line with a uniform velocity of 54 km/h while its mass increases at a constant rate of 0.4 kg/sec. Find the magnitude of the force acting on it.
Correct answer: (D) 6 N
Explanation: Step 1: F = d(mv)/dt = v dm/dt, because v is constant. Step 2: v = 54 × 5/18 = 15 m/sec. Step 3: F = 15 × 0.4 = 6 N.
A body of mass 3 kg is acted on by a variable force F = 6t newtons, where t is the time in seconds. Find its acceleration at t = 2 s.
Correct answer: (C) 4 m/sec²
Explanation: Step 1: At t = 2: F = 6 × 2 = 12 N. Step 2: From F = m a, a = F ÷ m = 12 ÷ 3. Step 3: a = 4. ∴ Acceleration = 4 m/sec²
A body of mass 2 kg moves under the action of the forces F₁ = a î + 3 ĵ, F₂ = 4 î + b ĵ and F₃ = −î + 2 ĵ (in newtons). Its displacement vector is S = (t² − 3) î + (5t − 2t²) ĵ metres. Find the constants a and b.
Correct answer: (A) a = 1, b = −13
Explanation: Step 1: V = dS/dt = 2t î + (5 − 4t) ĵ and a = dV/dt = 2 î − 4 ĵ. Step 2: F₁ + F₂ + F₃ = (a + 3) î + (b + 5) ĵ. Step 3: By Newton's second law this equals m a = 4 î − 8 ĵ. Step 4: a + 3 = 4 gives a = 1, and b + 5 = −8 gives b = −13.
A force F = (2r + 6) newtons acts on a body of mass 2 kg, initially at rest at the origin O, in the direction of its motion along a straight line, where r is its distance from O in metres. Find r when the velocity of the body becomes 4 m/sec.
Correct answer: (A) 2 m
Explanation: m v dv/dr = F ⇒ 2 ∫₀ᵛ v dv = ∫₀ʳ (2r + 6) dr ⇒ v² = r² + 6r. v = 4 ⇒ r² + 6r − 16 = 0 ⇒ (r + 8)(r − 2) = 0; the body moves from rest in the positive direction, so r = 2 m. (8/3 m comes from treating the acceleration as the constant 3 m/sec², its value at O.)
A body of variable mass m = (t + 3) kg moves in a straight line, and its displacement vector is S = (t² + 4t) c metres, where c is a unit vector in the direction of motion and t is in seconds. Find the magnitude of the force acting on the body when t = 2 sec.
Correct answer: (C) 18 N
Explanation: Step 1: V = dS/dt = (2t + 4) c, so the momentum H = m V = (t + 3)(2t + 4) c = (2t² + 10t + 12) c. Step 2: F = dH/dt = (4t + 10) c. Step 3: When t = 2: F = 8 + 10 = 18 N (m × a = 5 × 2 = 10 N alone forgets that the mass changes).
A metal ball of mass 50 gm moves in a straight line through a dusty medium, and dust sticks to it at a constant rate of 2 gm per second. Its displacement after t seconds is s = (t² + 5t) cm. Find the magnitude of the force acting on the ball when t = 5 sec.
Correct answer: (D) 150 dyne
Explanation: Step 1: The mass after t seconds is m = (50 + 2t) gm and the velocity is v = ds/dt = (2t + 5) cm/sec. Step 2: F = d(mv)/dt = m dv/dt + v dm/dt = (50 + 2t) × 2 + (2t + 5) × 2. Step 3: When t = 5: m = 60 gm, v = 15 cm/sec, so F = 60 × 2 + 15 × 2 = 120 + 30. Step 4: F = 150 dyne.
A body of mass 2 kg starts from rest and moves in a straight line under a force F = (12 − 4t) newtons acting along the line, where t is in seconds. Find the distance it covers until it comes to rest again.
Correct answer: (B) 36 m
Explanation: Step 1: a = F ÷ m = 6 − 2t, so v = ∫₀ᵗ (6 − 2t) dt = 6t − t². Step 2: It is at rest again when 6t − t² = 0, i.e. t = 6 sec (v is positive on ]0, 6[). Step 3: Distance = ∫₀⁶ (6t − t²) dt = [3t² − t³/3]₀⁶ = 108 − 72. Step 4: Distance = 36 m (9 m/sec is its greatest velocity, at t = 3).
Sand falls vertically at a steady rate of 5 kg/sec onto a horizontal belt moving at a constant 2 m/sec. Find the horizontal force needed to keep the belt moving at this speed.
Correct answer: (C) 10 N
Explanation: Step 1: The sand must be accelerated to the belt speed; here mass is variable, so F = v × (dm/dt). Step 2: F = 2 × 5. Step 3: F = 10. ∴ Force = 10 N
The first 10 questions in every subject are free, each with an explanation.