FSc Pre-Engineering Physics Motion & Force — Set 2

FSc Pre-Engineering Physics Motion & Force — Set 2

1. Question 1

   Q1. A force of 10 N acts on a 5 kg body. What is its acceleration?

   • A) 1 m/s²
   • B) 2 m/s²
   • C) 0.5 m/s²
   • D) 5 m/s²

   Answer: 2 m/s²

   Explanation: F = ma, so a = F/m = 10/5 = 2 m/s², applying Newton's second law.

2. Question 2

   Q2. A particle moves in a circular path with constant speed. What is its acceleration?

   • A) Zero
   • B) Tangential
   • C) Centripetal
   • D) None of these

   Answer: Centripetal

   Explanation: For uniform circular motion, acceleration is directed towards the center, known as centripetal acceleration.

3. Question 3

   Q3. What is the force required to keep a 1000 kg car moving at constant velocity on a level road?

   • A) 0 N
   • B) 1000 N
   • C) 9800 N
   • D) 10000 N

   Answer: 0 N

   Explanation: At constant velocity, net force is zero, as per Newton's first law.

4. Question 4

   Q4. A 50 N force is applied to a body at an angle of 30° to the horizontal. What is its horizontal component?

   • A) 25 N
   • B) 43.3 N
   • C) 50 N
   • D) 25√3 N

   Answer: 43.3 N

   Explanation: Horizontal component = F cos(θ) = 50 * cos(30°) = 50 * √3 / 2 = 43.3 N, resolving the force.

5. Question 5

   Q5. A 5 kg block is pulled with a force of 20 N on a frictionless surface. What is its acceleration?

   • A) 2 m/s²
   • B) 4 m/s²
   • C) 1 m/s²
   • D) 5 m/s²

   Answer: 4 m/s²

   Explanation: F = ma, so a = F/m = 20/5 = 4 m/s², applying Newton's second law.

6. Question 6

   Q6. A projectile is launched at 45°. What is its range if initial velocity is 20 m/s?

   • A) 20 m
   • B) 40 m
   • C) 40.8 m
   • D) 80 m

   Answer: 40.8 m

   Explanation: Range = (u² sin(2θ)) / g = (20² * sin(90°)) / 9.8 = 400 / 9.8 = 40.8 m, using the range formula.

7. Question 7

   Q7. A body moves with constant acceleration. What is the shape of its velocity-time graph?

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

   Answer: Straight line

   Explanation: For constant acceleration, the velocity-time graph is a straight line, as a = Δv / Δt.

8. Question 8

   Q8. A force of 100 N acts on a body for 2 s. What is the impulse?

   • A) 50 Ns
   • B) 100 Ns
   • C) 200 Ns
   • D) 400 Ns

   Answer: 200 Ns

   Explanation: Impulse = F * Δt = 100 * 2 = 200 Ns, using the impulse formula.

9. Question 9

   Q9. A car rounds a curve with 50 m radius at 20 m/s. What is its centripetal acceleration?

   • A) 4 m/s²
   • B) 8 m/s²
   • C) 10 m/s²
   • D) 40 m/s²

   Answer: 8 m/s²

   Explanation: a = v² / r = 20² / 50 = 8 m/s², using the centripetal acceleration formula.

10. Question 10

    Q10. A body is moving in a straight line with constant speed. What is its acceleration?

    • A) Zero
    • B) Constant
    • C) Increasing
    • D) Decreasing

    Answer: Zero

    Explanation: For constant speed in a straight line, acceleration is zero, as there's no change in velocity.

11. Question 11

    Q11. A 10 kg block is lifted vertically upwards with 50 N force. What is its acceleration?

    • A) 0 m/s²
    • B) 5 m/s²
    • C) -5 m/s² (downwards)
    • D) None of these

    Answer: -5 m/s² (downwards)

    Explanation: Net force = F - mg = 50 - 100 = -50 N. a = F_net / m = -50 / 10 = -5 m/s², applying Newton's second law.

12. Question 12

    Q12. A particle moves in a circular path with 10 m radius at 5 m/s. What is its angular velocity?

    • A) 0.5 rad/s
    • B) 1 rad/s
    • C) 2 rad/s
    • D) 5 rad/s

    Answer: 0.5 rad/s

    Explanation: ω = v / r = 5 / 10 = 0.5 rad/s, using the formula for angular velocity.

13. Question 13

    Q13. A body is projected upwards with 30° angle. What is its time of flight if initial velocity is 20 m/s?

    • A) 1 s
    • B) 2 s
    • C) 2.04 s
    • D) 4 s

    Answer: 2.04 s

    Explanation: T = 2u sin(θ) / g = 2 * 20 * sin(30°) / 9.8 = 2.04 s, using the time of flight formula.

14. Question 14

    Q14. A 20 N force is applied to a 10 kg body on a frictionless surface at 30° to the horizontal. What is its horizontal acceleration?

    • A) 1 m/s²
    • B) √3 m/s²
    • C) 2 m/s²
    • D) 1.732 m/s²

    Answer: 1.732 m/s²

    Explanation: Horizontal component of force = F cos(30°) = 20 * √3 / 2 = 17.32 N. a = F_horizontal / m = 17.32 / 10 = 1.732 m/s².

15. Question 15

    Q15. A car accelerates from rest to 30 m/s in 6 s. What is its average acceleration?

    • A) 2 m/s²
    • B) 5 m/s²
    • C) 3 m/s²
    • D) 6 m/s²

    Answer: 5 m/s²

    Explanation: a = Δv / Δt = 30 / 6 = 5 m/s², calculating average acceleration.

16. Question 16

    Q16. A 5 kg block is moving on a frictionless surface with 10 m/s velocity. What is the force required to stop it in 2 s?

    • A) 25 N
    • B) -25 N
    • C) 50 N
    • D) -50 N (opposite direction)

    Answer: 25 N

    Explanation: F = ma = m * Δv / Δt = 5 * 10 / 2 = 25 N, applying Newton's second law to find the required force.

17. Question 17

    Q17. A 2 kg block is moving with 5 m/s velocity on a frictionless surface. What is its momentum?

    • A) 10 kg m/s
    • B) 5 kg m/s
    • C) 2 kg m/s
    • D) 0 kg m/s

    Answer: 10 kg m/s

    Explanation: Momentum = mass * velocity = 2 kg * 5 m/s = 10 kg m/s, using the formula p = mv.

18. Question 18

    Q18. A body moves with constant velocity. What is its acceleration?

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

    Answer: 0 m/s²

    Explanation: Acceleration = 0, as velocity is constant, using a = dv / dt = 0.

19. Question 19

    Q19. What is the force required to stop a 5 kg body moving with 10 m/s in 2 s?

    • A) 25 N
    • B) 10 N
    • C) -25 N
    • D) 50 N

    Answer: 25 N

    Explanation: Using F = ma and a = Δv / Δt, F = m * (0 - 10) / 2 = 25 N.

20. Question 20

    Q20. A 10 kg block is pulled with a force of 20 N on a frictionless surface. 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 = 20 N / 10 kg = 2 m/s².