Newton’s Laws of Motion: Understanding Inertia and Force
Numerical – 3 (C)
1. Linear Momentum Calculation
Problem: A body of mass 5 kg is moving with a velocity of 2 m/s. Calculate its linear momentum.
Answer:
Mass of the body (m) = 5 kg
Velocity (v) = 2 m/s
Linear momentum = mv = (5)(2) kgm/s
= 10 kgm/s
2. Velocity Calculation from Linear Momentum
Problem: The linear momentum of a ball of mass 50 g is 0.5 kg m/s. Find its velocity.
Answer:
Linear momentum = 0.5 kgm/s
Mass (m) = 50 g = 0.05 kg
Velocity = Linear momentum / mass = 0.5 / 0.05 m/s = 10 m/s
3. Acceleration Calculation from Force and Mass
Problem: A force of 15 N acts on a body of mass 2 kg. Calculate the acceleration produced.
Answer:
Force (F) = 15 N
Mass (m) = 2 kg
Acceleration (a) = F / m [From Newton’s second law]
Or, a = (15 / 2) m/s²
Or, a = 7.5 m/s²
5. Force Calculation from Mass and Acceleration
Problem: Calculate the magnitude of force which when applied on a body of mass 0.5 kg produces an acceleration of 5 m/s².
Answer:
Mass (m) = 0.5 kg
Acceleration (a) = 5 m/s²
Force (F) = ma [From Newton’s second law]
Or, F = (0.5)(5) N = 2.5 N
6. Velocity and Momentum Change Calculation
Problem: A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate:
(i) The velocity acquired by the body and
(ii) Change in momentum of the body.
Answer:
Force (F) = 10 N
Mass (m) = 2 kg
Time (t) = 3 s
Initial velocity (u) = 0 m/s
(i) Let v be the final velocity acquired.
From Newton’s second law, F = ma.
Or, a = F / m = 10 / 2 = 5 m/s².
From the 1st equation of motion, a = (v – u) / t
Or, v = at + u
Or, v = (5)(3) + 0 = 15 m/s
(ii) Change in momentum = Final momentum – initial momentum
p = mv – mu
Or, p = m(v – u)
Or, p = 2(15 – 0) = 30 kgm/s
7. Velocity, Acceleration, and Force Calculation
Problem: A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate:
(i) The velocity acquired by the body,
(ii) The acceleration produced by the force, and
(iii) The magnitude of the force.
Answer:
Mass (m) = 100 kg
Distance moved (s) = 100 m
Initial velocity (u) = 0
(i) Velocity of the body = Distance moved / time taken
Velocity = (100 / 5) = 20 m/s
(ii) From Newton’s third equation of motion, v² – u² = 2as.
Or, a = (v² – u²) / 2s
and, a = [(20² – 0²) / 2(100)] m/s²
hence, a = 2 m/s²
(iii) Force (F) = ma
Or, F = (100)(2) N
therefore, F = 200 N
8. Force Calculation from Velocity-Time Graph
Problem: Fig 3.11 shows the velocity-time graph of a particle of mass 100 g moving in a straight line. Calculate the force acting on the particle.
(Hint: Acceleration = Slope of the v-t graph)
Answer:
Slope of a velocity-time graph gives the value of acceleration.
Here, slope = 20 / 5 = 4 m/s².
Or, acceleration (a) = 4 m/s².
Force = Mass x Acceleration.
Given mass (m) = 100 g = 0.1 kg.
Force = (0.1)(4) = 0.4 N.
9. Acceleration Calculation with Varying Mass
Problem: A force causes an acceleration of 10 m/s² in a body of mass 500 g. What acceleration will be caused by the same force in a body of mass 5 kg?
Answer:
Let the force be F.
Force F causes an acceleration, a = 10 m/s² in a body of mass, m = 500 g or 0.5 kg
Thus, F = ma
Or, F = (0.5)(10) = 5 N
Let a’ be the acceleration which force F (= 5 N) causes on a body of mass, m’ = 5 kg.
Then, a’ = F / m’.
Or, a’ = (5 / 5) m/s².
Or, a’ = 1 m/s²
10. Force Calculation from Impulse
Problem: A force acts for 0.1 s on a body of mass 2.0 kg initially at rest. The force is then withdrawn, and the body moves with a velocity of 2 m/s. Find the magnitude of the force.
Answer:
Mass (m) = 2 kg
Initial velocity (u) = 0
Final velocity (v) = 2 m/s
Time (t) = 0.1 s
Acceleration = Change in velocity / time
Or, a = (v – u) / t
Or, a = (2 – 0) / 0.1 = 20 m/s².
Force = Mass x Acceleration
Or, F = (2)(20) = 40 N
11. Force Calculation from Displacement and Time
Problem: A body of mass 500 g, initially at rest, is acted upon by a force which causes it to move a distance of 4 m in 2 s. Calculate the force applied.
Answer:
Mass of an object = 500 g = 0.5 kg
The second equation of motion is s = ut + (1/2)at².
As the body is at rest, u = 0, t = 2 s
Displacement of an object = 4 m
4 = (0)(2) + (1/2)a(2)².
⇒ a = 2 m/s²
F = ma = 0.5 * 2 = 1 Newton
Thus, the force applied to move the body is 1 Newton.
12. Braking Force Calculation
Problem: A car of mass 480 kg moving at a speed of 54 km/h is stopped in 10 s. Calculate the force applied by the brakes.
Answer:
Mass (m) = 480 kg.
Initial velocity (u) = 54 km/h = 15 m/s
Final velocity (v) = 0.
Time (t) = 10 s.
Acceleration = Change in velocity / time.
Or, a = (v – u) / t
Or, a = (0 – 15) / 10 = -1.5 m/s²
Here, the negative sign indicates retardation.
Now, Force = Mass x Acceleration
Or, F = (480)(1.5) = 720 N.
13. Momentum Change, Retardation, and Mass Calculation
Problem: A car is moving with a uniform velocity of 30 m/s. It is stopped in 2 s by applying a force of 1500 N through its brakes.
Calculate the following values:
(a) The change in momentum of the car.
(b) The retardation produced in the car.
(c) The mass of the car.
Answer:
Initial velocity (u) = 30 m/s
Final velocity (v) = 0
Time (t) = 2 s
Force (F) = 1500 N
Here, a = (v – u) / t = (0 – 30) / 2 = -15 m/s².
Here, the negative sign indicates retardation.
Now, F = ma.
Or, m = F / a = (1500 / 15) = 100 kg
(a) Change in momentum = Final momentum – Initial momentum
Or, p = m(v – u)
and, p = 100(0 – 30)
Hence, p = -3000 kgm/s
(b) Acceleration (a) = (v – u) / t
Or, a = (0 – 30) / 2 = -15 m/s²
Here, the negative sign indicates retardation.
Thus, retardation = 15 m/s²
(c) From Newton’s second law of motion, F = ma
Or, m = F / a = (1500 / 15) = 100 kg.
14. Bullet Momentum, Retardation, and Resistive Force
Problem: A bullet of mass 50 g moving with an initial velocity of 100 m/s strikes a wooden block and comes to rest after penetrating a distance of 2 cm in it.
Calculate:
(i) Initial momentum of the bullet,
(ii) Final momentum of the bullet,
(iii) Retardation caused by the wooden block, and
(iv) Resistive force exerted by the wooden block.
Answer:
Mass (m) = 50 g = 0.05 kg
Initial velocity (u) = 100 m/s
Final velocity (v) = 0
Distance (s) = 2 cm = 0.02 m.
(i) Initial momentum = mu = (0.05)(100) = 5 kgm/s
(ii) Final momentum = mv = (0.05)(0) = 0 kgm/s.
(iii) Acceleration (a) = (v² – u²) / 2s.
Or, a = (0² – 100²) / 2(0.02).
Or, a = -2.5 * 10⁵ m/s².
Therefore, retardation is 2.5 * 10⁵ m/s².
(iv) Force (F) = ma
Or, F = (0.05 kg)(2.5 * 10⁵ m/s²)
Or, F = 12500 N
Ex 3D
1. Action-Reaction Forces
Problem: A boy pushes a wall with a force of 10 N towards the east. What force is exerted by the wall on the boy?
Answer:
The wall exerts an equal force of 10 N on the boy in the opposite direction, i.e., west.
2. Forces in a Hanging Block System
Problem: In the Fig., a block of weight 15 N is hanging from a rigid support by a string. What force is exerted by
(a) the block on the string and
(b) the string on the block? Name and show them in the diagram.
Answer:
(a) The block exerts a 15 N force (weight) on the string downwards.
(b) The string exerts an equal force of 15 N on the block in the opposite direction, i.e., upward direction (tension).
Understanding Inertia
Examples of Inertia
(b) Train Doors: When a train suddenly starts, the sliding doors of some compartments may open because the frame of the sliding door, being in contact with the floor of the train, also comes in motion on the start of the train, but the sliding door remains in its position due to inertia. Thus, the frame moves ahead with the train, while the door slides opposite to the direction of motion of the train. Thus, the door may shut or open accordingly.
(c) Shaking Fruit Trees: People shake branches of a tree in an attempt to cause the fruits to fall because when the branches of the tree are shaken, they come in motion, while the fruits, due to inertia, remain in a state of rest. Thus, the larger and weakly attached fruits get detached from the branches and fall down due to the pull of gravity.
(d) Running After Alighting from a Bus: When alighting from a moving bus, one has to run for some distance because when people alight from a moving bus, they continue to run alongside the bus to avoid falling. If they were to stop at once, the feet would come to rest suddenly, but the upper part of the body would still be in motion, and they would tend to fall forward.
(e) Removing Dust from a Carpet: Dust particles are removed from a carpet by beating it because the part of the carpet where the stick strikes comes in motion at once, while the dust particles settled on it remain in the state of rest due to inertia of rest. Thus, a part of the carpet moves ahead with the stick, leaving behind the dust particles that fall down due to gravity.
Galileo’s Law of Inertia
Problem: Explain Galileo’s law of inertia.
Answer:
Galileo’s law of inertia states that a body continues to be in its state of rest or of uniform motion unless an external force is applied on it.
Newton’s First Law of Motion
Statement of Newton’s First Law
Problem: State Newton’s first law of motion.
Answer:
According to Newton’s first law of motion, if a body is in a state of rest, it will remain in the state of rest, and if the body is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.
Explanation of Newton’s First Law
Problem: State and explain the law of inertia (or Newton’s first law of motion).
Answer:
Statement of Newton’s first law: If a body is in a state of rest, it will remain in the state of rest, and if the body is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.
Types of Inertia
Problem: Name the two kinds of inertia.
Answer:
Two kinds of inertia are as listed below:
(i) Inertia of rest.
(ii) Inertia of motion.
Examples of Inertia
Problem: Give one example of each of the following cases of inertia:
(a) Inertia of rest, and
(b) Inertia of motion.
Answer:
Example of inertia of rest: A coin placed on top of a card remains in place when the card is slightly and quickly jerked horizontally.
Example of inertia of motion: A ball thrown vertically upwards by a person in a moving train comes back to his hand.
Newton’s Second Law of Motion
Difference Between Newton’s First and Second Laws
Problem: How does Newton’s second law of motion differ from the first law of motion?
Answer:
Newton’s first law of motion gives the qualitative definition of force. It explains the force as the cause of acceleration only qualitatively, but Newton’s second law of motion gives the quantitative value of force. It states force as the product of mass and acceleration. Thus, it relates force to the measurable quantities such as acceleration and mass.
Mathematical Form of Newton’s Second Law
Problem: Write the mathematical form of Newton’s second law of motion. State the conditions, if any.
Answer:
Mathematical expression of Newton’s second law of motion is as shown below:
Force = Mass x Acceleration
Above relation holds for the following conditions:
- When the velocity of the body is much smaller than the velocity of light.
- When the mass remains constant.
Statement and Conditions of Newton’s Second Law
Problem: State Newton’s second law of motion. Under what condition does it take the form F = ma?
Answer:
According to Newton’s second law of motion, the rate of change of momentum is directly proportional to the force applied on it, and the change of momentum takes place in the direction in which the force is applied.
The relation F = ma holds for the following conditions:
- When the velocity of the body is much smaller than the velocity of light.
- When the mass remains constant.
Deriving Newton’s First Law from the Second Law
Problem: How can Newton’s first law of motion be obtained from the second law of motion?
Answer:
From Newton’s second law of motion, F = ma. If F = 0, then a = 0.
This means that if no force is applied on the body, its acceleration will be zero. If the body is at rest, then it will remain in the state of rest, and if it is moving, then it will remain moving in the same direction with the same speed. Thus, a body not acted upon by an external force does not change its state of rest or motion. This is the statement of Newton’s first law of motion.