Class 9 Science Motion Numerical Worksheet Chapter 7 Motion | Detailed Solutions & Marking Scheme
🏫 VEDANT CLASSES – A Pathway to Success…
📅 Date: ________
📘 Subject: Physics (Class 9)
📊 Chapter: Motion
📝 Topic-wise Practice Worksheet (40 Questions)
🔹 Section A: Distance & Displacement
Q1. A particle is moving along a circular path of diameter 5 m. Calculate:
(a) the total distance covered, and
(b) the net displacement of the particle when it completes 3 full revolutions.
Q2. A body is thrown vertically upwards from the ground and reaches a maximum height ‘h’. It then returns back to the ground. Calculate:
(a) the total distance travelled by the body, and
(b) the net displacement during the entire motion.
Q3. A body moves from point A to point B covering a distance of 15 m. It then moves from point B to point C covering a distance of 20 m in a direction perpendicular to AB. Calculate:
(a) the total distance travelled, and
(b) the magnitude of displacement of the body.
Q4. An object is moving along a circular path of radius ‘r’. Calculate the distance travelled and displacement of the object when:
(a) it completes half a revolution, and
(b) it completes one full revolution.
Q5. In a long-distance race, athletes have to complete four rounds of a circular track such that the finishing point coincides with the starting point. The length of the track is 200 m. Answer the following:
(a) What is the total distance covered by the athlete?
(b) What is the net displacement of the athlete at the end of the race?
(c) Is the motion uniform or non-uniform? Give reason.
(d) Is the distance travelled equal to displacement? Justify your answer.
🔹 Section B: Speed, Velocity & Average Speed
Q6. An object travels a distance of 16 m in 4 seconds and then travels another 16 m in 2 seconds. Calculate the average speed of the object during the entire journey.
Q7. Vishnu swims in a swimming pool of length 90 m. He swims from one end to the other and returns back to the starting point along the same straight path in 1 minute, covering a total distance of 180 m. Calculate:
(a) the average speed, and
(b) the average velocity of Vishnu.
Q8. Amit is driving a car with a constant velocity of 45 km/h. Calculate the distance travelled by the car:
(a) in 1 minute, and
(b) in 1 second.
Q9. The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the total time taken is 8 hours, calculate the average speed of the car in:
(a) km/h, and
(b) m/s.
Q10. An electric train is moving with a constant speed of 120 km/h. Calculate the distance travelled by the train in 30 seconds.
Q11. A body is moving with a uniform velocity of 15 m/s. What will be its velocity after 10 seconds? Give reason for your answer.
Q12. A train travels a certain distance at a speed of 30 km/h and returns over the same distance at a speed of 45 km/h. Calculate the average speed of the train for the entire journey.
Q13. A train of length 100 m is moving on a straight track and passes a stationary pole in 5 seconds. Calculate:
(a) the speed of the train, and
(b) the time taken by the train to cross a bridge of length 500 m.
Q14. A car travels along a straight road such that it moves with a speed of 40 km/h during the first half of the total time and with a speed of 60 km/h during the second half of the time. Calculate the average speed of the car for the entire journey.
Q15. A train travels at a speed of 60 km/h for 0.52 hours, then at 30 km/h for 0.24 hours, and finally at 70 km/h for 0.71 hours. Calculate the average speed of the train for the complete journey.
Q16. A cyclist completes one full round of a circular track of diameter 105 m in 5 minutes. Calculate the speed of the cyclist.
Q17. A cyclist is moving along a circular track of radius 50 m and completes one revolution in 4 minutes. Calculate:
(a) the average speed, and
(b) the average velocity during one complete revolution.
🔹 Section C: Acceleration & Retardation
Q18. Starting from rest, Bhuvan paddles his bicycle and attains a velocity of 6 m/s in 30 seconds. He then applies brakes and the velocity reduces to 4 m/s in the next 5 seconds. Calculate the acceleration in both cases.
Q19. A body starts moving with an initial velocity of 0.5 m/s on a horizontal surface. Due to friction, it experiences a uniform retardation of 0.05 m/s². Calculate the time taken by the body to come to rest.
Q20. A car increases its speed from 36 km/h to 70 km/h in 5 seconds. Calculate the acceleration of the car. If the same car comes to rest in 20 seconds, calculate the retardation.
Q21. A scooter starts from rest and attains a velocity of 36 km/h in 10 seconds. It then comes to rest in 20 seconds. Calculate the acceleration and retardation of the scooter.
Q22. A moving train is brought to rest in 20 seconds by applying brakes. If the retardation produced is 2 m/s², calculate the initial velocity of the train.
Q23. A car accelerates uniformly from 18 km/h to 36 km/h in 5 seconds. Calculate:
(a) the acceleration, and
(b) the distance travelled during this time.
Q24. A body moving on a rough surface slows down uniformly due to friction. Its initial velocity is 0.5 m/s and retardation is 0.05 m/s². Calculate the time taken to stop.
Q25. A bus is moving with a speed of 54 km/h. After applying brakes, it comes to rest in 8 seconds. Calculate:
(a) the acceleration, and
(b) the distance travelled before stopping.
Q26. A motorcycle is moving with a speed of 5 m/s and is subjected to a uniform acceleration of 0.2 m/s². Calculate:
(a) the speed after 10 seconds, and
(b) the distance travelled in this time.
Q27. The brakes applied to a car produce a uniform retardation of 6 m/s². If the car comes to rest in 2 seconds, calculate the distance travelled during this time.
Q28. A train starts from rest and attains a velocity of 72 km/h in 5 minutes with uniform acceleration. Calculate:
(a) the acceleration, and
(b) the distance travelled during this period.
🔹 Section D: Graph-Based Questions
Q29. The position-time graph of a body is given. Calculate the speed of the body as it moves from:
(a) point A to B,
(b) point B to C, and
(c) point C to D.
Q30. The velocity-time graph of an ascending lift is given. Calculate the acceleration of the lift:
(a) during the first 2 seconds,
(b) between 2 s and 10 s, and
(c) during the last 2 seconds.
Q31. A body is moving with a constant velocity of 5 m/s. Using a velocity-time graph, calculate the distance travelled by the body in 5 seconds.
Q32. Study the given speed-time graph and answer the following:
(a) Identify the type of motion represented by segments OA, AB, and BC.
(b) Calculate the acceleration of the body.
(c) Calculate the retardation of the body.
(d) Calculate the distance travelled between A and B.
Q33. Using the graph in the question, calculate:
(a) the distance travelled from O to A,
(b) the distance travelled from B to C, and
(c) the total distance travelled by the body in 16 seconds.
🔹 Section E: Equations of Motion
Q34. A body starts from rest and accelerates uniformly at 10 m/s². Calculate the distance travelled by the body in 2 seconds.
Q35. An object starts from rest and moves with an acceleration of 8 m/s². Calculate the distance travelled in 1 second.
Q36. A train starts from rest and moves with a uniform acceleration of 0.2 m/s² for 5 minutes. Calculate:
(a) the final velocity, and
(b) the distance travelled.
Q37. A car starts from rest and moves with a constant acceleration of 5 m/s² for 8 seconds. It then continues to move with constant velocity. Calculate the total distance covered in 12 seconds.
🔹 Section F: Uniform Circular Motion
Q38. Calculate the speed of the tip of the second hand of a clock if its length is 1.5 cm.
Q39. The length of the minute hand of a clock is 5 cm. Calculate the speed of its tip.
Q40. A train has to cover a total distance of 120 km with an average speed of 60 km/h. If it covers the first 30 km at a speed of 30 km/h, calculate the speed required to cover the remaining 90 km so that the average speed remains 60 km/h.
Perfect 👍 — here is your DETAILED SOLVED ANSWER KEY with STEP-WISE MARKING SCHEME (CBSE/GSEB Style) for all 40 questions.
I’ve kept it exam-ready, step marking, unit marking, formula marking exactly how boards check.
🏫 VEDANT CLASSES – A Pathway to Success…
📘 Chapter: Motion – Answer Key & Marking Scheme
🔹 Section A: Distance & Displacement
Q1.
Diameter = 5 m ⇒ Radius = 2.5 m
Distance = 3 × circumference
= 3 × (π × D) = 3 × (π × 5) = 15π m
Displacement = 0 (same starting & ending point)
✅ Answer: Distance = 15π m, Displacement = 0
Marking (2M):
Formula (1M)
Final answer (1M)
Q2.
Distance = upward + downward = h + h = 2h
Displacement = 0
Marks (2M):
Concept (1M) + Answer (1M)
Q3.
Distance = 15 + 20 = 35 m
Displacement = √(15² + 20²) = √(225 + 400) = √625 = 25 m
Marks (2M):
Formula (1M)
Calculation (1M)
Q4.
Half circle:
Distance = πr
Displacement = diameter = 2r
Full circle:
Distance = 2πr
Displacement = 0
Marks (3M):
Each part 1.5M
Q5.
(a) Distance = 4 × 200 = 800 m
(b) Displacement = 0
(c) Non-uniform (speed may vary)
(d) No, distance ≠ displacement
Marks (4M):
1M each part
🔹 Section B: Speed & Velocity
Q6.
Total distance = 32 m
Total time = 6 s
Avg speed = 32/6 = 5.33 m/s
Marks (2M)
Q7.
Distance = 180 m, Time = 60 s
Avg speed = 180/60 = 3 m/s
Avg velocity = 0
Marks (3M):
Speed (2M) + Velocity (1M)
Q8.
45 km/h = 12.5 m/s
(a) 1 min = 60 s ⇒ distance = 12.5 × 60 = 750 m
(b) 1 s ⇒ 12.5 m
Marks (2M)
Q9.
Distance = 400 km
Time = 8 hr
Speed = 50 km/h
= 13.89 m/s
Marks (2M)
Q10.
120 km/h = 33.33 m/s
Distance = 33.33 × 30 = 1000 m
Marks (2M)
Q11.
Velocity remains 15 m/s
Marks (1M)
Q12.
Avg speed = 2xy/(x+y)
= 2×30×45/(75) = 36 km/h
Marks (2M)
Q13.
Speed = 100/5 = 20 m/s
Bridge total length = 100 + 500 = 600 m
Time = 600/20 = 30 s
Marks (3M)
Q14.
Avg speed = (40+60)/2 = 50 km/h
Marks (1M)
Q15.
Distance = (60×0.52)+(30×0.24)+(70×0.71) = 88.1 km
Time = 1.47 hr
Speed = 88.1/1.47 ≈ 59.9 km/h
Marks (3M)
Q16.
Distance = πD = 105π
Time = 300 s
Speed = 105π / 300 ≈ 1.1 m/s
Marks (2M)
Q17.
Distance = 2πr = 100π
Speed = 100π/240 ≈ 1.31 m/s
Velocity = 0
Marks (3M)
🔹 Section C: Acceleration
Q18.
a₁ = (6−0)/30 = 0.2 m/s²
a₂ = (4−6)/5 = −0.4 m/s²
Marks (3M)
Q19.
v = 0
0 = 0.5 − 0.05t ⇒ t = 10 s
Marks (2M)
Q20.
Convert speeds
a = (19.44−10)/5 ≈ 1.89 m/s²
Retardation = 19.44/20 ≈ 0.97 m/s²
Marks (3M)
Q21.
a = 10/10 = 1 m/s²
Retardation = −10/20 = −0.5 m/s²
Marks (3M)
Q22.
v = 0, a = −2
0 = u − 2×20 ⇒ u = 40 m/s
Marks (2M)
Q23.
a = (10−5)/5 = 1 m/s²
s = ut + ½at² = 5×5 + ½×1×25 = 37.5 m
Marks (3M)
Q24.
Same as Q19 ⇒ t = 10 s
Marks (2M)
Q25.
a = (0−15)/8 = −1.875 m/s²
s = ut + ½at² = 15×8 − ½×1.875×64 = 60 m
Marks (3M)
Q26.
v = 5 + 0.2×10 = 7 m/s
s = 5×10 + ½×0.2×100 = 60 m
Marks (3M)
Q27.
u = 12 m/s
s = ut − ½at² = 12×2 − ½×6×4 = 12 m
Marks (2M)
Q28.
v = 20 m/s
a = 20/300 = 0.067 m/s²
s = ½at² = 3000 m
Marks (3M)
🔹 Section D: Graphs
Q29–Q33
✔ Depends on graph
Marking Pattern (each 3–4M):
Correct slope/area (1M)
Formula (1M)
Calculation (1M)
Units (1M)
🔹 Section E: Equations of Motion
Q34.
s = ½×10×4 = 20 m
(2M)
Q35.
s = ½×8×1 = 4 m
(2M)
Q36.
v = at = 0.2×300 = 60 m/s
s = ½×0.2×90000 = 9000 m
(3M)
Q37.
s₁ = ½×5×64 = 160 m
v = 40 m/s
s₂ = 40×4 = 160 m
Total = 320 m
(3M)
🔹 Section F: Circular Motion
Q38.
Distance = 2πr = 3π cm
Time = 60 s
Speed = 0.157 cm/s
(2M)
Q39.
Distance = 10π cm
Time = 3600 s
Speed ≈ 0.0087 cm/s
(2M)
Q40.
Total time = 120/60 = 2 hr
Time for 30 km = 1 hr
Remaining time = 1 hr
Speed = 90 km/h
(3M)
✅ Final Marking Insight (Teacher Use)
✔ Step marking strictly followed
✔ Units compulsory → deduct 0.5M
✔ Formula missing → deduct 1M
✔ Final answer wrong but steps correct → give 50% marks
Comments
Post a Comment