📋 Motion Physics Question Paper
Class IX/X | Total Marks: 80 | Duration: 2.5 Hours
SECTION A: Multiple Choice (1 Mark Each)
1
1 Mark
An object moving in a circular path and returning to its starting point has displacement of:
2
1 Mark
SI unit of acceleration is:
3
1 Mark
A car travels 60 km in 2 hours. Average speed is:
4
1 Mark
Which is a scalar quantity?
5
1 Mark
In uniform circular motion, acceleration is due to:
SECTION B: Short Answer (2 Marks Each)
6
2 Marks
Distinguish between distance and displacement with examples.
7
2 Marks
Define speed and velocity. Why is velocity more informative?
8
2 Marks
A train accelerates from 36 km/h to 54 km/h in 10 seconds. Calculate acceleration.
9
2 Marks
What is uniform circular motion? Give two examples.
10
2 Marks
Explain why motion is relative using the bus-trees example.
SECTION C: Medium Answer (3 Marks Each)
11
3 Marks
A car travels 20 m in 2 seconds, then 40 m in 4 seconds. Find average speed.
12
3 Marks
A cyclist starts from rest and achieves 10 m/s after 50 m. Find acceleration.
13
3 Marks
Interpret distance-time graphs: straight line vs curved line.
14
3 Marks
A ball thrown upward with 20 m/s returns in 4 seconds. Explain the motion.
15
3 Marks
Draw v-t graph for object accelerating at 2 m/s² for 5 seconds. Find distance.
SECTION D: Long Answer (5 Marks Each)
16
5 Marks
State and derive the three equations of motion. When is each used?
17
5 Marks
A train starts from rest with acceleration 0.5 m/s² for 30 seconds. Find: (i) final velocity (ii) distance (iii) average velocity
18
5 Marks
Car brakes with deceleration 2 m/s² from 20 m/s. Find: (i) time to stop (ii) distance (iii) velocity after 5 seconds
19
5 Marks
Explain uniform circular motion and why it's accelerated. Derive v = 2πr/t.
20
5 Marks
Boy runs 100 m N, 50 m E, 100 m S. Find: (i) distance (ii) displacement (iii) avg speed (50s) (iv) avg velocity
SECTION E: Application Based (5 Marks Each)
21
5 Marks
Cyclist: A to B (50 km, 2 hrs) then B to A (1.5 hrs). Find: (i) avg speed (ii) avg velocity (iii) Why different?
22
5 Marks
Bus brakes from 80 km/h in 8 seconds. Find: (i) deceleration (ii) distance (iii) Why maintain distance?
23
5 Marks
Graph Analysis: Explain straight line, curved line, and horizontal line in d-t graph.
24
5 Marks
Rocket accelerates from rest, travels 1000 m in 10 seconds. Find: (i) acceleration (ii) final velocity (iii) distance in last 2 seconds
25
5 Marks
Athlete on circular track (r=100m, lap=40s). Find: (i) distance (ii) displacement (iii) speed (iv) why accelerating?
✅ ANSWER KEY & EXPLANATIONS
SECTION A: MCQ Answers
1
✅ Answer: C) Zero
Explanation: Displacement is from initial to final position. Since object returns to starting point, displacement = 0.
2
✅ Answer: B) m/s²
Explanation: a = Δv/Δt = (m/s)/s = m/s²
3
✅ Answer: A) 30 km/h
Average speed = 60/2 = 30 km/h
4
✅ Answer: C) Speed
Explanation: Speed (scalar) has only magnitude. Velocity, displacement, acceleration are vectors.
5
✅ Answer: B) Change in direction
Explanation: In uniform circular motion, speed is constant but direction changes continuously. Change in velocity = acceleration.
SECTION B: Short Answer Keys
6
2 Marks
✅ Answer & Explanation
Distance: Total path length (scalar)
- Example: Walking around a square (side 10m) = 40m distance
Displacement: Shortest path from start to end (vector)
- Same path = ~14.14m diagonal displacement
7
2 Marks
✅ Answer & Explanation
Speed: Distance per unit time (scalar - magnitude only)
Velocity: Displacement per unit time (vector - magnitude + direction)
Velocity is more informative because it tells both "how fast" and "which direction"
8
2 Marks
✅ Answer & Solution
Convert to m/s:
u = 36 × 5/18 = 10 m/s, v = 54 × 5/18 = 15 m/s
Apply formula:
a = (v-u)/t = (15-10)/10 = 0.5 m/s²
9
2 Marks
✅ Answer & Explanation
Uniform Circular Motion: Motion in a circle at constant speed
- Speed (magnitude) is constant
- Direction continuously changes
Examples:
- Earth orbiting the Sun
- Athlete running on circular track
10
2 Marks
✅ Answer & Explanation
From passenger view: Trees move backward, fellow passengers at rest
From ground view: Bus moves forward, trees stationary
Same object appears differently to different observers. Motion depends on reference point.
SECTION C: Medium Answer Keys
11
3 Marks
✅ Solution
Total distance = 20 + 40 = 60 m
Total time = 2 + 4 = 6 s
Avg speed = 60/6 = 10 m/s
12
3 Marks
✅ Solution
Use: 2as = v² - u²
2 × a × 50 = 10² - 0²
100a = 100
a = 1 m/s²
13
3 Marks
✅ Answer
Straight line: Uniform motion (constant speed). Equal distance in equal time.
Curved line: Non-uniform motion (changing speed). Unequal distance in equal time.
14
3 Marks
✅ Explanation
Time of flight = 4 seconds
Motion: Upward 2s (v: 20→0), Downward 2s (v: 0→20)
At max height, velocity = 0. Time up = Time down.
15
3 Marks
✅ Solution
Final velocity:
v = u + at = 0 + 2(5) = 10 m/s
Distance = Area under v-t graph:
s = ½ × base × height = ½ × 5 × 10 = 25 m
SECTION D: Long Answer Keys
16
5 Marks
✅ Complete Answer
1) v = u + at (Velocity-Time)
Use when: Given u, a, t → Find v
2) s = ut + ½at² (Position-Time)
Use when: Given u, a, t → Find s
3) 2as = v² - u² (Position-Velocity)
Use when: Given u, v, a → Find s (no time given)
17
5 Marks
✅ Complete Solution
(i) Final velocity:
v = u + at = 0 + 0.5(30) = 15 m/s
(ii) Distance:
s = ut + ½at² = 0 + ½(0.5)(900) = 225 m
(iii) Average velocity:
v_av = (u+v)/2 = (0+15)/2 = 7.5 m/s
18
5 Marks
✅ Complete Solution
(i) Time to stop:
0 = 20 + (-2)t → t = 10 s
(ii) Distance:
2as = v² - u² → 2(-2)s = 0 - 400 → s = 100 m
(iii) Velocity after 5s:
v = 20 + (-2)(5) = 10 m/s
19
5 Marks
✅ Answer
Why accelerated: Speed constant but direction changes → velocity changes → acceleration
Formula derivation:
Distance in one lap = 2πr
v = distance/time = 2πr/t
v = distance/time = 2πr/t
20
5 Marks
✅ Complete Solution
(i) Distance = 100 + 50 + 100 = 250 m
(ii) Displacement = 50 m East (100m north and south cancel)
(iii) Avg speed = 250/50 = 5 m/s
(iv) Avg velocity = 50/50 = 1 m/s East
SECTION E: Application Answer Keys
21
5 Marks
✅ Solution
(i) Avg speed = 100/(2+1.5) = 28.6 km/h
(ii) Avg velocity = 0/(3.5) = 0 (returns to start)
(iii) Different because: Speed depends on distance, velocity on displacement only.
22
5 Marks
✅ Solution
Convert 80 km/h:
80 × 5/18 = 22.2 m/s
(i) a = (0-22.2)/8 = -2.78 m/s²
(ii) 2as = v² - u² → s = 492.84/(5.56) ≈ 89 m
(iii) Maintain distance: Bus needs ~89m to stop at this speed
23
5 Marks
✅ Analysis
Straight line: Uniform motion, constant speed, a=0
Curved line: Non-uniform motion, changing speed, a≠0
Horizontal line: Object at rest, v=0, a=0
24
5 Marks
✅ Solution
Using s = ut + ½at²:
1000 = 0 + ½a(100) → a = 20 m/s²
(ii) v = 0 + 20(10) = 200 m/s
(iii) Distance in last 2s:
s(8s) = ½(20)(64) = 640m, s(10s) = 1000m, Diff = 360m
25
5 Marks
✅ Solution
(i) Distance = 2πr = 2π(100) = 628.3 m
(ii) Displacement = 0 (returns to start)
(iii) Speed = 2πr/t = 628.3/40 = 15.7 m/s
(iv) Accelerating because: Direction changes continuously (velocity ≠ constant)
📊 Exam Summary
Marks Distribution
Section A
5 × 1 = 5
5 × 1 = 5
Section B
5 × 2 = 10
5 × 2 = 10
Section C
5 × 3 = 15
5 × 3 = 15
Section D
5 × 5 = 25
5 × 5 = 25
Section E
5 × 5 = 25
5 × 5 = 25
TOTAL
25 × 80
25 × 80
Key Formulas to Remember
| Formula | Use |
|---|---|
| v = u + at | Find velocity when time is given |
| s = ut + ½at² | Find distance when time is given |
| 2as = v² - u² | Find distance when time is not given |
| Avg speed = Total distance / Total time | Calculate average speed |
| Avg velocity = Displacement / Total time | Calculate average velocity |
| v = 2πr/t | Speed in circular motion |
Important Concepts
- Distance vs Displacement: Distance is total path, displacement is net position change
- Speed vs Velocity: Speed has magnitude only, velocity has magnitude + direction
- Acceleration: Change in velocity per unit time (can be speed or direction change)
- Uniform Motion: Constant speed, equal distances in equal times
- Non-uniform Motion: Changing speed, unequal distances in equal times
- Circular Motion: Constant speed but changing direction = acceleration
