Science 1: Chapter no 1: Law of motion
Class 9 Science 1 Chapter 1 Exercise Answers
1. Match the First Column with Appropriate Entries in the Second and Third Columns
| S. No. | Column 1 | Column 2 | Column 3 |
|---|---|---|---|
| 1 | Negative acceleration | The velocity of the object decreases. | A vehicle moving with a velocity of 10 m/s stops after 5 seconds. |
| 2 | Positive acceleration | The velocity of the object increases. | A car initially at rest reaches a velocity of 50 km/h in 10 seconds. |
| 3 | Zero acceleration | The velocity of the object remains constant. | A vehicle is moving with a constant velocity of 25 m/s. |
2. Clarify the Differences
A. Distance and Displacement
| Distance | Displacement |
|---|---|
| Distance is the total length of the actual path travelled by an object. | Displacement is the shortest distance between the initial and final positions of an object. |
| Distance is a scalar quantity. | Displacement is a vector quantity. |
| Distance has only magnitude. | Displacement has both magnitude and direction. |
| Distance is always positive. | Displacement may be positive, negative or zero depending on the direction of motion. |
B. Uniform Motion and Non-uniform Motion
| Uniform Motion | Non-uniform Motion |
|---|---|
| An object covers equal distances in equal intervals of time. | An object covers unequal distances in equal intervals of time. |
| The speed of the object remains constant. | The speed of the object changes with time. |
| The acceleration is zero. | The acceleration is not zero. |
| Example: A train moving at a constant speed on a straight track. | Example: A car moving in city traffic. |
Give Scientific Reasons (Answers)
a. When an object falls freely to the ground, its acceleration is uniform.
- A freely falling object is acted upon only by the force of gravity.
- The acceleration due to gravity remains constant near the Earth's surface.
- Therefore, the velocity of the object increases uniformly with time.
- Hence, the acceleration of a freely falling object is uniform.
b. Even though the magnitudes of action and reaction are equal and opposite, their effects do not get cancelled.
- According to Newton's third law, action and reaction forces are equal in magnitude and opposite in direction.
- These two forces always act on two different objects.
- Since they act on different objects, they cannot cancel each other.
- Therefore, the effects of action and reaction forces do not get cancelled.
c. It is easier to stop a tennis ball as compared to a cricket ball when both are travelling with the same velocity.
- A cricket ball has greater mass than a tennis ball.
- Therefore, the cricket ball has greater momentum at the same velocity.
- More force is required to stop an object having greater momentum.
- Hence, it is easier to stop a tennis ball than a cricket ball.
d. The velocity of an object at rest is considered to be uniform.
- An object at rest has zero velocity.
- The velocity of the object remains zero at all times.
- Since there is no change in velocity, it is said to be constant.
- Therefore, the velocity of an object at rest is considered to be uniform.
Q. Examples Based on Newton's Laws of Motion
1. Seat Belt in a Moving Car (Newton's First Law)
- A passenger moves forward when a moving car stops suddenly.
- This happens because the body tends to remain in motion due to inertia.
- The seat belt applies a force to stop the passenger.
- Thus, the seat belt works according to Newton's First Law of Motion.
2. Kicking a Football (Newton's Second Law)
- A football moves when it is kicked with force.
- Greater force produces greater acceleration in the ball.
- The acceleration also depends on the mass of the ball.
- This example explains Newton's Second Law of Motion.
3. Rocket Launch (Newton's Third Law)
- A rocket throws hot gases downward at high speed.
- The gases exert an equal and opposite force on the rocket.
- This force pushes the rocket upward.
- Therefore, the rocket works according to Newton's Third Law of Motion.
4. Pushing a Shopping Trolley (Newton's Second Law)
- An empty trolley moves easily when pushed.
- A loaded trolley requires more force to move.
- This is because a greater mass needs more force for the same acceleration.
- Hence, this example is based on Newton's Second Law of Motion.
5. Walking on the Ground (Newton's Third Law)
- While walking, we push the ground backward with our feet.
- The ground exerts an equal and opposite force on our feet.
- This reaction force pushes us forward.
- Therefore, walking is based on Newton's Third Law of Motion.
Q.3 Complete the Following Table
Table 1 : Using the Formula v = u + at
| u (m/s) | a (m/s²) | t (sec) | v (m/s) |
|---|---|---|---|
| 2 | 4 | 3 | 14 |
| 10 | 5 | 2 | 20 |
Table 2 : Using the Formula s = ut + ½at²
| u (m/s) | a (m/s²) | t (sec) | s (m) |
|---|---|---|---|
| 5 | 12 | 3 | 69 |
| 7 | 8 | 4 | 92 |
Table 3 : Using the Formula v² = u² + 2as
| u (m/s) | a (m/s²) | s (m) | v² (m²/s²) |
|---|---|---|---|
| 4 | 3 | -4 | 8 |
| 4 | 5 | 8.4 | 10 |
Calculations
1. v = u + at = 2 + (4 × 3) = 14 m/s
2. u = v − at = 20 − (5 × 2) = 10 m/s
3. s = ut + ½at² = (5 × 3) + ½ × 12 × 3² = 15 + 54 = 69 m
4. 92 = (7 × 4) + ½a × 4²
92 = 28 + 8a
a = 8 m/s²
5. 8 = 4² + 2 × 3 × s
8 = 16 + 6s
s = -4 m
6. 10 = u² + 2 × 5 × 8.4
10 = u² + 84
u² = -74 (Not Possible)
Note: The last row of the third table appears to contain a printing error in the textbook because the given values do not satisfy the formula v² = u² + 2as.
Q.7 Solve the Following Examples
(a) Calculate the Average Speed
Given:
- Distance covered in first 3 s = 18 m
- Distance covered in next 3 s = 22 m
- Distance covered in last 3 s = 14 m
Formula:
Average Speed = Total Distance ÷ Total Time
Solution:
Total Distance = 18 + 22 + 14 = 54 m
Total Time = 3 + 3 + 3 = 9 s
Average Speed = 54 ÷ 9 = 6 m/s
Answer: Average Speed = 6 m/s
(b) Calculate the Force and Acceleration
Given:
- Mass (m₁) = 16 kg
- Acceleration (a₁) = 3 m/s²
- Mass (m₂) = 24 kg
Formula:
Force = Mass × Acceleration (F = ma)
Solution:
F = 16 × 3
F = 48 N
Now,
a = F ÷ m
a = 48 ÷ 24
a = 2 m/s²
Answer:
- Force = 48 N
- Acceleration = 2 m/s²
(c) Determine the Velocity of Bullet and Wooden Plank
Given:
- Mass of bullet (m₁) = 10 g = 0.01 kg
- Velocity of bullet (u₁) = 1.5 m/s
- Mass of plank (m₂) = 90 g = 0.09 kg
- Velocity of plank (u₂) = 0 m/s
Formula:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
Solution:
(0.01 × 1.5) + (0.09 × 0) = (0.01 + 0.09)v
0.015 = 0.10v
v = 0.015 ÷ 0.10
v = 0.15 m/s
Answer: Velocity = 0.15 m/s
(d) Calculate the Average Speed
Given:
- Distance covered = 100 m, 80 m and 45 m
- Time taken = 40 s, 40 s and 20 s
Formula:
Average Speed = Total Distance ÷ Total Time
Solution:
Total Distance = 100 + 80 + 45 = 225 m
Total Time = 40 + 40 + 20 = 100 s
Average Speed = 225 ÷ 100
Average Speed = 2.25 m/s
Answer: Average Speed = 2.25 m/s
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