Motion & Force MCQs set 3 for PIEAS Entry Test Physics — 20 solved questions.
Q1. A body moves in a circular path with constant speed. What is its acceleration?
Answer: v² / r
Explanation: Centripetal acceleration = v² / r, directed towards the center.
Q2. A 5 kg block is moving on a surface with friction coefficient 0.2. What is the frictional force?
Answer: 10 N
Explanation: Frictional force = μ * N = 0.2 * 5 kg * 10 m/s² = 10 N.
Q3. A car moves with 30 m/s velocity and is stopped in 6 s. What is its average deceleration?
Answer: 5 m/s²
Explanation: Average deceleration = Δv / Δt = (0 - 30) / 6 = 5 m/s².
Q4. What is the momentum of a 3 kg body moving with 4 m/s velocity?
Answer: 12 kg m/s
Explanation: Momentum = mass * velocity = 3 kg * 4 m/s = 12 kg m/s.
Q5. A body is moving with constant acceleration. What is its velocity-time graph?
Answer: Straight line
Explanation: v = u + at, a straight line with slope a.
Q6. A 2 kg block is attached to a spring with spring constant 100 N/m. What is its frequency?
Answer: √(100 / 2) / (2 * π)
Explanation: Frequency = (1 / (2 * π)) * √(k / m) = √(100 / 2) / (2 * π).
Q7. A body is projected upwards with 45° angle. What is its horizontal velocity component?
Answer: v * cos(45°)
Explanation: Horizontal component = v * cos(θ) = v * cos(45°).
Q8. A 10 N force acts on a 5 kg body. What is its acceleration?
Answer: 2 m/s²
Explanation: Using F = ma, a = F / m = 10 N / 5 kg = 2 m/s².
Q9. A body moves in a straight line with constant velocity. What is its net force?
Answer: 0 N
Explanation: Net force = 0, as acceleration = 0, using F = ma.
Q10. A 20 kg block is pulled upwards with a force of 250 N. What is its acceleration?
Answer: 2.5 m/s²
Explanation: Using F = ma and F - mg = ma, a = (250 - 20 * 10) / 20 = 2.5 m/s².
Q11. A body is thrown upwards. What is its acceleration at the highest point?
Answer: g
Explanation: Acceleration = g, downwards, at all points, including the highest point.
Q12. A 5 kg block is moving with 8 m/s velocity. What is its kinetic energy?
Answer: 160 J
Explanation: Kinetic energy = 0.5 * m * v² = 0.5 * 5 * 8² = 160 J.
Q13. A 5 kg block is moving with 10 m/s on a frictionless surface. What force is required to stop it in 2 s?
Answer: -25 N
Explanation: F = ma = m * (v - u) / t = 5 * (0 - 10) / 2 = -25 N, using Newton's second law of motion.
Q14. A force of 10 N acts on a 2 kg body for 3 s. If the initial velocity is 0, find the final velocity.
Answer: 15 m/s
Explanation: a = F / m = 10 / 2 = 5 m/s², v = u + at = 0 + 5 * 3 = 15 m/s, using F = ma and v = u + at.
Q15. A particle moves in a circular path with constant speed. What is the direction of its acceleration?
Answer: Towards the center
Explanation: The acceleration is directed towards the center of the circle, known as centripetal acceleration, given by a = v² / r.
Q16. A 1000 kg car accelerates from 0 to 20 m/s in 5 s. What is its average power?
Answer: 40000 W
Explanation: P = ΔKE / t = (1/2) * m * (v² - u²) / t = (1/2) * 1000 * (20² - 0²) / 5 = 40000 W.
Q17. A body is projected upwards with 20 m/s. What is its velocity at the highest point?
Answer: 0 m/s
Explanation: At the highest point, the velocity is zero, as the body momentarily comes to rest before falling back.
Q18. A 2 kg block is attached to a spring with k = 100 N/m. If it is displaced by 0.2 m, what is its acceleration?
Answer: 10 m/s²
Explanation: F = -kx = ma, a = -kx / m = -100 * 0.2 / 2 = -10 m/s², using Hooke's law and F = ma.
Q19. A force F = 2t N acts on a 3 kg body. What is its velocity at t = 4 s, if u = 0?
Answer: 16/3 m/s
Explanation: ∫F dt = ∫(2t) dt = t², from 0 to 4 = 16, Δp = 16, mv - mu = 16, v = 16 / 3 m/s.
Q20. A particle is moving with constant velocity. What is its acceleration?
Answer: 0 m/s²
Explanation: Since the velocity is constant, the acceleration is zero, as a = dv / dt = 0.