PU CET Lahore (Engineering & CS) Physics Motion & Force — Set 3

PU CET Lahore (Engineering & CS) Physics Motion & Force — Set 3

1. Question 1

   Q1. A body moves in a circular path with constant speed. What is its acceleration?

   • A) 0 m/s²
   • B) v² / r
   • C) v / r
   • D) r / v

   Answer: v² / r

   Explanation: Centripetal acceleration = v² / r, directed towards the center.

2. Question 2

   Q2. A 5 kg block is moving on a surface with friction coefficient 0.2. What is the frictional force?

   • A) 10 N
   • B) 5 N
   • C) 0 N
   • D) 20 N

   Answer: 10 N

   Explanation: Frictional force = μ * N = 0.2 * 5 kg * 10 m/s² = 10 N.

3. Question 3

   Q3. A car moves with 30 m/s velocity and is stopped in 6 s. What is its average deceleration?

   • A) 5 m/s²
   • B) 10 m/s²
   • C) -5 m/s²
   • D) 0 m/s²

   Answer: 5 m/s²

   Explanation: Average deceleration = Δv / Δt = (0 - 30) / 6 = 5 m/s².

4. Question 4

   Q4. What is the momentum of a 3 kg body moving with 4 m/s velocity?

   • A) 12 kg m/s
   • B) 10 kg m/s
   • C) 7 kg m/s
   • D) 5 kg m/s

   Answer: 12 kg m/s

   Explanation: Momentum = mass * velocity = 3 kg * 4 m/s = 12 kg m/s.

5. Question 5

   Q5. A body is moving with constant acceleration. What is its velocity-time graph?

   • A) Straight line
   • B) Parabola
   • C) Circle
   • D) Ellipse

   Answer: Straight line

   Explanation: v = u + at, a straight line with slope a.

6. Question 6

   Q6. A 2 kg block is attached to a spring with spring constant 100 N/m. What is its frequency?

   • A) √(100 / 2) / (2 * π)
   • B) √(2 / 100) / (2 * π)
   • C) √(100 / 2)
   • D) 100 / 2

   Answer: √(100 / 2) / (2 * π)

   Explanation: Frequency = (1 / (2 * π)) * √(k / m) = √(100 / 2) / (2 * π).

7. Question 7

   Q7. A body is projected upwards with 45° angle. What is its horizontal velocity component?

   • A) v * cos(45°)
   • B) v * sin(45°)
   • C) v
   • D) 0

   Answer: v * cos(45°)

   Explanation: Horizontal component = v * cos(θ) = v * cos(45°).

8. Question 8

   Q8. A 10 N force acts on a 5 kg body. What is its acceleration?

   • A) 2 m/s²
   • B) 1 m/s²
   • C) 5 m/s²
   • D) 0 m/s²

   Answer: 2 m/s²

   Explanation: Using F = ma, a = F / m = 10 N / 5 kg = 2 m/s².

9. Question 9

   Q9. A body moves in a straight line with constant velocity. What is its net force?

   • A) 0 N
   • B) 5 N
   • C) 10 N
   • D) Variable

   Answer: 0 N

   Explanation: Net force = 0, as acceleration = 0, using F = ma.

10. Question 10

    Q10. A 20 kg block is pulled upwards with a force of 250 N. What is its acceleration?

    • A) 2.5 m/s²
    • B) 5 m/s²
    • C) 10 m/s²
    • D) 0 m/s²

    Answer: 2.5 m/s²

    Explanation: Using F = ma and F - mg = ma, a = (250 - 20 * 10) / 20 = 2.5 m/s².

11. Question 11

    Q11. A body is thrown upwards. What is its acceleration at the highest point?

    • A) 0 m/s²
    • B) g
    • C) -g
    • D) 2g

    Answer: g

    Explanation: Acceleration = g, downwards, at all points, including the highest point.

12. Question 12

    Q12. A 5 kg block is moving with 8 m/s velocity. What is its kinetic energy?

    • A) 160 J
    • B) 100 J
    • C) 80 J
    • D) 40 J

    Answer: 160 J

    Explanation: Kinetic energy = 0.5 * m * v² = 0.5 * 5 * 8² = 160 J.

13. Question 13

    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?

    • A) 25 N
    • B) -25 N
    • C) 50 N
    • D) -50 N

    Answer: -25 N

    Explanation: F = ma = m * (v - u) / t = 5 * (0 - 10) / 2 = -25 N, using Newton's second law of motion.

14. Question 14

    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.

    • A) 10 m/s
    • B) 15 m/s
    • C) 20 m/s
    • D) 25 m/s

    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.

15. Question 15

    Q15. A particle moves in a circular path with constant speed. What is the direction of its acceleration?

    • A) Tangential to the circle
    • B) Perpendicular to the circle
    • C) Towards the center
    • D) Away from the center

    Answer: Towards the center

    Explanation: The acceleration is directed towards the center of the circle, known as centripetal acceleration, given by a = v² / r.

16. Question 16

    Q16. A 1000 kg car accelerates from 0 to 20 m/s in 5 s. What is its average power?

    • A) 2000 W
    • B) 4000 W
    • C) 40000 W
    • D) 80000 W

    Answer: 40000 W

    Explanation: P = ΔKE / t = (1/2) * m * (v² - u²) / t = (1/2) * 1000 * (20² - 0²) / 5 = 40000 W.

17. Question 17

    Q17. A body is projected upwards with 20 m/s. What is its velocity at the highest point?

    • A) 0 m/s
    • B) 10 m/s
    • C) 20 m/s
    • D) -20 m/s

    Answer: 0 m/s

    Explanation: At the highest point, the velocity is zero, as the body momentarily comes to rest before falling back.

18. Question 18

    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?

    • A) 5 m/s²
    • B) 10 m/s²
    • C) 15 m/s²
    • D) 20 m/s²

    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.

19. Question 19

    Q19. A force F = 2t N acts on a 3 kg body. What is its velocity at t = 4 s, if u = 0?

    • A) 4 m/s
    • B) 8/3 m/s
    • C) 16/3 m/s
    • D) 32/3 m/s

    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.

20. Question 20

    Q20. A particle is moving with constant velocity. What is its acceleration?

    • A) 0 m/s²
    • B) 1 m/s²
    • C) 2 m/s²
    • D) Variable

    Answer: 0 m/s²

    Explanation: Since the velocity is constant, the acceleration is zero, as a = dv / dt = 0.