CBSE Class 9 Science Motion Sure Shot Questions

CBSE Class 9 Science Motion Sure Shot Questions. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

ASSIGNMENT QUESTIONS
CLASS 9 SCIENCE : CHAPTER - MOTION

MULTIPLE CHOICE QUESTIONS

1. If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration

Answer: (b) uniform acceleration

1. Displacement s = ½at² for an object with uniform acceleration a, so s is proportional to t².
2. Hence the object moves with uniform acceleration.

Teacher's Note:
a) Any relation of the form s proportional to t² means constant acceleration.
b) Uniform velocity would give s proportional to t, not t².

2. The distance time graph of a body coincides with its time axis. The body must be
(a) in uniform motion
(b) at rest
(c) in uniformly accelerated motion
(d) in zig-zag motion

Answer: (b) at rest

1. The distance from the starting point stays zero at every instant, since the graph lies along the time axis.
2. A body whose distance stays zero is at rest.

Teacher's Note:
a) Distance-time graphs show how far the body has moved, so a zero line means no movement.

3. From the given v - t graph (see below Fig.), it can be inferred that the object is
(a) in uniform motion
(b) at rest
(c) in non-uniform motion
(d) moving with uniform acceleration

[Figure: Velocity (v) against time (t) graph, showing a horizontal straight line at a constant positive velocity.]

Answer: (a) in uniform motion

1. The velocity-time graph is a horizontal straight line.
2. The velocity is constant and not zero, so the object is in uniform motion.

Teacher's Note:
a) A horizontal v-t line means zero acceleration.
b) If the line were on the time axis, the object would be at rest instead.

4. The velocity time graph of a body is parallel to the time axis. The body is
(a) at rest
(b) having uniform acceleration
(c) having zero acceleration
(d) having non-uniform acceleration

Answer: (c) having zero acceleration

1. A graph parallel to the time axis has constant velocity.
2. Acceleration is the slope of the v-t graph, which is zero here.

Teacher's Note:
a) Being parallel to the time axis does not mean the body is at rest, unless the velocity is zero.
b) Slope zero gives zero acceleration.

5. A particle is moving in a circular path of radius r. The displacement after half a circle would be:
(a) Zero
(b) π r
(c) 2 r
(d) 2π r

Answer: (c) 2 r

1. After half a circle, the particle is at the opposite end of the diameter.
2. Displacement is the straight line from start to finish, which equals the diameter 2r.

Teacher's Note:
a) πr is the distance travelled along the half circle, not the displacement.
b) Displacement is measured along the straight line joining the start and end points.

6. A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) u/g (b) u²/2g (c) u²/g (d) u/2g

Answer: (b) u²/2g

1. At the greatest height, the final velocity v = 0.
2. Using v² = u² - 2gh, we get 0 = u² - 2gh, so h = u²/2g.

Teacher's Note:
a) Use the sign convention carefully, since g acts downward and opposes the upward motion.
b) The time to reach the greatest height is u/g, which is a different quantity.

7. The numerical ratio of displacement to distance for a moving object is
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal or less than 1

Answer: (d) equal or less than 1

1. Displacement is the shortest straight-line distance between start and end, while distance is the length of the path.
2. So the magnitude of displacement is never greater than the distance, and the ratio is equal to or less than 1.

Teacher's Note:
a) The ratio equals 1 only when the object moves in a straight line without turning back.
b) For a round trip, the displacement is zero and the ratio is zero.

8. Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m/s. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity

Answer: (c) in accelerated motion

1. The speed is constant, but the direction of motion keeps changing on a circular path.
2. A change in velocity, including its direction, means the boy has acceleration, so the motion is accelerated.

Teacher's Note:
a) Constant speed does not mean constant velocity, because velocity has direction.
b) Option (d) is wrong for the same reason.

9. Area under a v - t graph represents a physical quantity which has the unit
(a) m²
(b) m
(c) m³
(d) m/s

Answer: (b) m

1. Area under a v-t graph = velocity × time, with unit (m/s) × s.
2. This gives metres, which is the unit of displacement.

Teacher's Note:
a) The area under a v-t graph gives the displacement covered in that interval.

10. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in below Fig.. Choose the correct statement
(a) Car A is faster than car D.
(b) Car B is the slowest.
(c) Car D is faster than car C.
(d) Car C is the slowest.

[Figure: Distance (m) against time (s) graph with four straight lines from the origin, labelled C (steepest), D, A and B (least steep).]

Answer: (b) Car B is the slowest.

1. Speed is the slope of a distance-time graph, so the steeper the line, the faster the car.
2. Slope order: C is steepest, then D, then A, then B. So B has the smallest slope and is the slowest.

Teacher's Note:
a) Read the steepness of each line from the origin.
b) Car A is slower than car D, and car C is faster than car D.

11. Slope of a velocity - time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed

Answer: (c) the acceleration

1. Acceleration = change in velocity / time taken.
2. This is exactly the slope of a velocity-time graph.

Teacher's Note:
a) The area under a v-t graph gives displacement, while the slope gives acceleration.

12. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun

Answer: (a) If the car is moving on straight road

1. Distance equals the magnitude of displacement only when the object moves along a straight line without changing direction.
2. A car moving on a straight road, without turning back, satisfies this condition.

Teacher's Note:
a) On a circular path, the displacement is always smaller than the distance travelled.
b) A pendulum returns along the same path, so its displacement and distance are different over a full swing.

13. Which of the following figures (see below Figure) represents uniform motion of a moving object correctly?

(a)

(b)

(c)

(d)

Answer: (a)

1. Uniform motion means equal distances are covered in equal intervals of time.
2. On a distance-time graph, this is a straight line through the origin, which is graph (a).

Teacher's Note:
a) Graph (b) shows increasing speed and graph (c) shows decreasing speed.
b) Graph (d) shows the body at rest, not moving.

SHORT ANSWER QUESTIONS

14. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify you answer.

Answer:
No. Displacement is zero when the object returns to its starting point, but it can still cover a path of non-zero length. The distance travelled is the total length of the path, which is not zero.

Teacher's Note:
a) Example: a runner who completes one full lap of a 400 m track has zero displacement but travels 400 m.
b) Distance is zero only when the object does not move at all.

15. How will the equations of motion for an object moving with a uniform velocity change?

Answer:
1. For uniform velocity, acceleration a = 0.
2. First equation: v = u + at becomes v = u.
3. Second equation: s = ut + ½at² becomes s = ut, or s = vt.
4. Third equation: v² = u² + 2as becomes v² = u², so v = u.

Teacher's Note:
a) Setting a = 0 in each equation is the quickest way to get the result.
b) The displacement is then simply velocity multiplied by time.

16. A car starts from rest and moves along the x-axis with constant acceleration 5 m/s² for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

Answer:
1. Distance in the first 8 s: s₁ = ut + ½at² = 0 + ½ × 5 × 8² = 160 m.
2. Velocity after 8 s: v = u + at = 0 + 5 × 8 = 40 m/s.
3. Distance in the next 4 s at constant velocity: s₂ = 40 × 4 = 160 m.
4. Total distance in 12 s = 160 + 160 = 320 m.

Teacher's Note:
a) The velocity at the end of acceleration becomes the starting velocity for the constant-velocity stage.
b) Add the two parts only after both are found separately.

17. A motorcyclist drives from A to B with a uniform speed of 30 km/h and returns back with a speed of 20 km/h. Find its average speed.

Answer:
1. Let the distance from A to B be d km.
2. Time from A to B = d/30 h, and time from B to A = d/20 h.
3. Total distance = 2d km, and total time = d/30 + d/20 = 5d/60 = d/12 h.
4. Average speed = 2d / (d/12) = 24 km/h.

Teacher's Note:
a) Average speed is total distance divided by total time, not the mean of the two speeds.
b) The mean of 30 and 20 would give 25 km/h, which is wrong here.

18. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.

Answer: Taking upward as positive, with the stone thrown at 20 m/s and g = 10 m/s², the velocity falls in a straight line from +20 m/s at t = 0, through 0 at the top (t = 2 s), to -20 m/s at t = 4 s when the stone is back at the starting point. The graph is one straight line with slope -g.

Teacher's Note:
a) The velocity passes through zero at the maximum height, where the stone changes direction.
b) The slope of the line is -g, which is the constant downward acceleration.

19. The velocity-time graph (see below Figure) shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds.

[Figure: Velocity (m/s) against time (s) graph, showing a horizontal straight line at 20 m/s from 0 s to 25 s.]

Answer:
1. (i) The graph is parallel to the time axis, so acceleration = 0.
2. (ii) Velocity = 20 m/s, read from the graph.
3. (iii) Distance in 15 s = velocity × time = 20 × 15 = 300 m.

Teacher's Note:
a) The area under the graph from 0 s to 15 s is a rectangle of height 20 and width 15.
b) Zero slope means zero acceleration, even though the cyclist is moving.

LONG ANSWER QUESTIONS

20. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in below figure. Plot a velocity-time graph for the same.

[Figure: Displacement (m) against time (s) graph. The displacement rises in a straight line from 0 m at 0 s to 100 m at 50 s, then falls in a straight line back to 0 m at 100 s.]

Answer:
1. From 0 to 50 s: velocity = (100 - 0)/(50 - 0) = +2 m/s.
2. From 50 to 100 s: velocity = (0 - 100)/(100 - 50) = -2 m/s.
3. The velocity-time graph is a line at +2 m/s from 0 s to 50 s, then a line at -2 m/s from 50 s to 100 s.

Teacher's Note:
a) The negative velocity shows that the girl is walking back towards her starting point.
b) Each part of a displacement-time graph gives a constant velocity, equal to its slope.

21. An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start.

Answer:
1. Using s = ut + ½at² with u = 0, in the first 2 s: 20 = ½ × a × 2², so a = 10 m/s².
2. Check: distance in the first 6 s = ½ × 10 × 6² = 180 m, and 180 - 20 = 160 m, which matches the next 4 s.
3. Velocity after 7 s: v = u + at = 0 + 10 × 7 = 70 m/s.

Teacher's Note:
a) The motion starts from rest, so u = 0 in every calculation.
b) Use the check of 160 m to confirm the acceleration before moving on.

22. An electron moving with a velocity of 5 × 10⁴ m/s enters into a uniform electric field and acquires a uniform acceleration of 10⁴ m/s² in the direction of its initial motion.
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time?

Answer:
1. Initial velocity u = 5 × 10⁴ m/s, acceleration a = 10⁴ m/s², final velocity v = 2u = 10⁵ m/s.
2. (i) Time t = (v - u)/a = (10⁵ - 5 × 10⁴)/10⁴ = 5 s.
3. (ii) Distance s = ut + ½at² = (5 × 10⁴ × 5) + ½ × 10⁴ × 5² = 2.5 × 10⁵ + 1.25 × 10⁵ = 3.75 × 10⁵ m.

Teacher's Note:
a) The acceleration acts in the direction of motion, so the velocity increases.
b) Check the answer with v² = u² + 2as, which gives the same distance of 3.75 × 10⁵ m.

23. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.

Answer:
1. Distance travelled in t seconds: s_t = ut + ½at².
2. Distance travelled in the 4th second = s₄ - s₃, and in the 5th second = s₅ - s₄.
3. Distance travelled between the 4th and 5th seconds (the interval from t = 4 s to t = 5 s) = s₅ - s₄ = u(5 - 4) + ½a(5² - 4²).
4. Hence s = u + 4.5a, which is the general relation s_nth = u + ½a(2n - 1) with n = 5.

Teacher's Note:
a) Subtract the distance at the start of the interval from the distance at the end of it.
b) The result is written in the form u + ½a(2n - 1), which is the standard relation for the nth second.

24. Two stones are thrown vertically upwards simultaneously with their initial velocities u₁ and u₂ respectively. Prove that the heights reached by them would be in the ratio of u₁² : u₂² (Assume upward acceleration is -g and downward acceleration to be +g).

Answer:
1. For the first stone, at the greatest height v = 0, so 0² = u₁² + 2(-g)h₁, which gives h₁ = u₁²/2g.
2. For the second stone, similarly h₂ = u₂²/2g.
3. Ratio h₁ : h₂ = (u₁²/2g) : (u₂²/2g) = u₁² : u₂².

Teacher's Note:
a) The result does not depend on the value of g, since it cancels in the ratio.
b) The stones are thrown at the same time, but the time taken is not needed to find the heights.

25. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?

Answer:
1. Both objects start from rest with the same acceleration g = 9.8 m/s².
2. Distance fallen in 2 s by each object = ½ × 9.8 × 2² = 19.6 m.
3. Height of first object after 2 s = 150 - 19.6 = 130.4 m.
4. Height of second object after 2 s = 100 - 19.6 = 80.4 m.
5. Difference in heights = 130.4 - 80.4 = 50 m.
6. At any time t, the difference is (150 - ½gt²) - (100 - ½gt²) = 50 m, so the difference does not change with time.

Teacher's Note:
a) Both objects fall the same distance in the same time, so their separation stays fixed at 50 m.
b) The answer is the same whatever value of g is used.