Chapter 1 - Laws of Motion
1. Match the first column with appropriate entries in the second and third columns and remake the table.
| Column 1 | Column 2 | Column 3 |
|---|---|---|
Negative acceleration | The velocity of the object remains constant | A car, initially at rest reaches a velocity of 50 km/hr in 10 seconds |
Positive acceleration | The velocity of the object decreases | A vehicle is moving with a velocity of 25 m/s |
Zero acceleration | The velocity of the object increases | A vehicle moving with the velocity of 10 m/s, stops after 5 seconds |
Ans:
| Column 1 | Column 2 | Column 3 |
|---|---|---|
Negative acceleration | The velocity of the object decreases | A vehicle moving with the velocity of 10 m/s, stops after 5 seconds |
Positive acceleration | The velocity of the object increases | A car, initially at rest reaches a velocity of 50 km/hr in 10 seconds |
Zero acceleration | The velocity of the object remains constant | A vehicle is moving with a velocity of 25 m/s |
2. Clarify the differences
A. Distance and displacement
Ans:
| Distance | Displacement |
|---|---|
(1) Distance is the length of the actual path travelled by an object in motion while going from one point to another. | (1) Displacement is the minimum distance between the starting and finishing points. |
(2) It is a scalar quantity. | (2) It is a vector quantity. |
(3) It is either equal to or greater than displacement. | (3) It is either equal to or less than distance. |
(4) Distance travelled is always positive. | (4) Displacement may be positive or negative or zero. |
B. Uniform and non-uniform motion
Ans:
| Distance | Displacement |
|---|---|
(1) If an object covers equal distances in equal intervals of time it is said to be in uniform motion. | (1) If an object moves unequal distances in equal intervals of time, its motion is said to be non-uniform. |
(2) Distance – time graph for uniform motion is a straight line. | (2) Distance – time graph for uniform motion is not a straight line. |
(3) In uniform motion, acceleration is zero. | (3) In non-uniform motion, acceleration is non-zero. |
3. Complete the following table.
Solution:
(i) Given:
u = 2 m/s
a = 3 m/s²
t = 4 s
To find:
v
Solution:
We know that,
v = u + at
∴ v = 2 + (3 × 4)
∴ v = 2 + 12
∴ v = 14 m/s
(ii) Given:
a = 5 m/s²
t = 2 s
v = 20 m/s
To find:
u
Solution:
We know that,
v = u + at
∴ 20 = u + (5 × 2)
∴ 20 = u + 10
∴ u = 20 – 10
∴ u = 10 m/s
(i) Given:
u = 5 m/s
a = 12 m/s²
t = 3 s
To find:
s
Solution:
We know that,
s = ut + \(\large \frac {1}{2}\) at²
∴ s = (5 × 3) + \(\large \frac {1}{2}\) × 12(3)²
∴ s = 15 + (6 × 9)
∴ s = 15 + 54
∴ s = 69 m
(ii) Given:
u = 7 m/s
t = 4 s
s = 92 m
To find:
a
Solution:
We know that,
s = ut + \(\large \frac {1}{2}\) at²
∴ 92 = (7 × 4) + \(\large \frac {1}{2}\) × a(4)²
∴ 92 = 28 + \(\large \frac {1}{2}\) × 16a
∴ 92 = 28 + 8a
∴ 8a = 92 – 28
∴ 8a = 64
∴ a = \(\large \frac {64}{8}\)
∴ a = 8 m/s²
(i) Given:
u = 4 m/s
a = 3 m/s²
v = 8 m/s
To find:
s
Solution:
We know that,
v² = u² + 2as
∴ 8² = 4² + 2 × 3s
∴ 64 = 16 + 6s
∴ 6s = 64 – 16
∴ 6s = 48
∴ s = \(\large \frac {48}{6}\)
∴ s = 8 m
(ii) Given:
a = 5 m/s²
s = 8.4 m
v = 10 m/s
To find:
u
Solution:
We know that,
v² = u² + 2as
∴ 10² = u² + 2 × 5 × 8.4
∴ 100 = u² + 84
∴ u² = 100 – 84
∴ u² = 16
∴ u = \(\sqrt {16}\)
∴ u = 4 m
4. Complete the sentences and explain them.
a. The minimum distance between the start and finish points of the motion of an object is called the ________ of the object.
Ans: Displacement
Explanation: The minimum distance between the start and finish points of the motion of an object is called the displacement of the object.
b. Deceleration is ________ acceleration
Ans: Negative
Explanation: In deceleration, the velocity of the body goes on decreasing; hence, it is called negative acceleration.
c. When an object is in uniform circular motion, its ________ changes at every point.
Ans: Direction
Explanation: In uniform circular motion, the speed is constant along the circumference, but its direction at every point is tangential.
d. During collision ________ remains constant.
Ans: Total momentum Law of conservation of momentum
Explanation: The total final momentum is equal to the total initial momentum.
e. The working of a rocket depends on Newton’s ________ law of motion.
Ans: Third
Explanation: The escaping gases exert an equal and opposite reaction on the rocket so that it gets propelled in the forward direction.
5. Give scientific reasons.
a. When an object falls freely to the ground, its acceleration is uniform.
Ans:
(i) When the body falls freely on the ground, there are equal changes in velocity at equal intervals of time.
(ii) Thus, the acceleration of the body is constant.
(iii) Hence, it possesses uniform acceleration.
b. Even though the magnitudes of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled.
Ans:
(i) Action and reaction forces act on different bodies.
(ii) They don’t act on the same body; hence, they cannot cancel each other’s effects.
(iii) Hence, even though the magnitudes of action force and reaction force are equal, they do not cancel each other.
c. It is easier to stop a tennis ball as compared to a cricket ball, when both are travelling with the same velocity.
Ans:
(i) The momentum of an object depends on its mass as well as its velocity.
(ii) A cricket ball is heavier than a tennis ball.
(iii) Although they are thrown with the same velocity, a cricket ball has more momentum than a tennis ball.
(iv) Hence, the force required to stop a cricket ball is greater than that required to stop a tennis ball.
(v) Hence, it is easier to stop a tennis ball than a cricket ball moving with the same velocity.
d. The velocity of an object at rest is considered to be uniform.
Ans:
(i) When a body is in a state of rest, it attains a constant velocity.
(ii) A body with constant velocity is said to be in uniform motion.
(iii) Hence, the state of rest is an example of uniform motion.
6. Take 5 examples from your surroundings and give explanations based on Newton’s laws of motion.
Ans:
(i) An electric fan keeps on rotating for some time after being switched off due to inertia. (Newton’s first law)
(ii) Fruits on a tree fall down when the branches are shaken. (Newton’s first law)
(iii) Athletes jump on a bed of sand during a long jump. (Newton’s second law).
(iv) A fielder moves his hands backward while catching a fast-moving ball. (Newton’s second law).
(v) While firing a gun, the gun recoils backwards. (Newton’s third law)
7. Solve the following examples.
a) An object moves 18 m in the first 3 s, 22 m in the next 3 s and 14 m in the last 3 s. What is its average speed?
Given :
Total distance (d) = 18 + 22 + 14 = 54 m
Total time taken (t) = 3 + 3 + 3 = 9 sec
To find :
Average speed
Solution :
We know that,
Average speed = \(\large \frac {Total\, distance\, covered}{Total \, time \, taken}\)
∴ Average speed = \(\large \frac {54}{9}\)
∴ Average speed = 6 m/s
Ans: The object moves with an average speed of 6 m/s.
b) An object of mass 16 kg is moving with an acceleration of 3 m/s². Calculate the applied force. If the same force is applied on an object of mass 24 kg, how much will be the acceleration?
Given :
Mass of 1st body (m\(_1\)) = 16 kg
Acceleration of 1st body (a\(_1\) = 3 m/s²
Mass of 2nd body (m\(_2\)) = 24 kg
To find :
Force on 1st body (F\(_1\)) = ?
Acceleration of 2nd body (a\(_2\)) = ?
Solution :
We know that,
F = m × a
∴ F\(_1\) = m\(_1\) × a\(_1\)
∴ F\(_1\) = 16 × 3
∴ F\(_1\) = 48 N
And,
F\(_2\) = m\(_2\) × a\(_2\)
∴ a\(_2\) = \(\large \frac {F_2}{m_2}\)
∴ a\(_2\) = \(\large \frac {48}{24}\)
∴ a\(_2\) = 2 m/s²
Ans: The force acting on the 1st body is 48 N and the acceleration of the 2nd body is 2 m/s².
c) A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 900 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.
Given :
Mass of bullet (m\(_1\)) = 10 g = \(\large \frac {10}{1000}\) kg = 0.010 kg
Mass of plank (m\(_2\)) = 90 g \(\large \frac {90}{1000}\) kg = 0.090 kg
Initial velocity of bullet (u\(_1\)) = 1.5 m/s
Initial velocity of plank (u\(_2\)) = 0 m/s
To find :
Common velocity (v) = ?
Solution :
Let v\(_1\) and v\(_2\) be the common velocities of the bullet and plank
∴ v\(_1\) = v\(_2\) = v
We know that,
m\(_1\) u\(_1\)+ m\(_2\) u\(_2\) = m\(_1\) v\(_1\) + m\(_2\) v\(_2\)
∴ (0.01 × 1.5) + (0.09 × 0) = (0.01 × v) + (0.09 × v)
∴ 0.015 + 0 = v (0.01 + 0.09)
∴ 0.015 = 0.1v
∴ v = \(\large \frac {0.015}{0.1}\)
∴ v = 0.15 m/s
Ans: The plank moves with a velocity of 0.15 m/s.
d) A person swims 100 m in the first 40 s, 80 m in the next 40 s and 45 m in the last 20 s. What is the average speed?
Given :
Total distance (d) = 100 + 80 + 45 = 225 m
Total time taken (t) = 40 + 40 + 20 = 100 s
To find :
Average speed
Solution :
We know that,
Average speed = \(\large \frac {Total\, distance \, covered}{Total\, time \, taken}\)
∴ Average speed = \(\large \frac {225}{100}\)
∴ Average speed = 2.25 m/s
Ans: The person swims with an average speed of 2.25 m/s.