Answer :
Alright, let's address each part of the question step by step.
### Question Q2
#### (i) Unit of Acceleration
Acceleration is defined as the rate of change of velocity with respect to time.
Velocity = Distance / Time
Acceleration = Velocity / Time
Using dimensional analysis,
- Distance has the dimension [tex]\([L]\)[/tex]
- Time has the dimension [tex]\([T]\)[/tex]
Thus,
- Velocity [tex]\(v = \frac{L}{T}\)[/tex]
- Acceleration [tex]\(a = \frac{v}{T} = \frac{L}{T^2}\)[/tex]
Therefore, the unit of acceleration is [tex]\( \left[ \frac{L}{T^2} \right] \)[/tex], which in SI units translates to meters per second squared (m/s²).
#### (ii) Unit of Force by Using Dimension
Force is calculated using Newton’s second law of motion:
Force ([tex]\(F\)[/tex]) = Mass ([tex]\(m\)[/tex]) × Acceleration ([tex]\(a\)[/tex])
Using dimensional analysis,
- Mass has the dimension [tex]\([M]\)[/tex]
- Acceleration has the dimension [tex]\(\left[\frac{L}{T^2}\right]\)[/tex]
Thus,
[tex]\( F = m \times a = [M] \times \left[ \frac{L}{T^2} \right] = \left[ \frac{ML}{T^2} \right] \)[/tex]
Therefore, the unit of force is [tex]\(\left[\frac{ML}{T^2}\right]\)[/tex], which in SI units translates to kilograms meters per second squared (kg·m/s²), also known as a Newton (N).
### Question Q3
#### (a) Statement of Newton's Second Law of Motion
Newton's Second Law of Motion states that the rate of change of momentum of an object is directly proportional to the applied force acting on it and takes place in the direction of the applied force. Mathematically, it can be expressed as:
[tex]\[ F = \frac{dp}{dt} \][/tex]
where [tex]\( p \)[/tex] is the momentum of the object.
#### (b) Verification that [tex]\( F = ma \)[/tex]
Considering Newton's Second Law:
[tex]\[ F = \frac{dp}{dt} \][/tex]
Momentum [tex]\( p \)[/tex] is the product of mass [tex]\( m \)[/tex] and velocity [tex]\( v \)[/tex]:
[tex]\[ p = m \cdot v \][/tex]
Thus,
[tex]\[ \frac{dp}{dt} = \frac{d(m \cdot v)}{dt} \][/tex]
Since mass [tex]\( m \)[/tex] is constant,
[tex]\[ \frac{d(m \cdot v)}{dt} = m \frac{dv}{dt} \][/tex]
[tex]\[ \frac{dv}{dt} \][/tex] is the acceleration [tex]\( a \)[/tex], therefore:
[tex]\[ F = m \cdot a \][/tex]
This completes the proof that [tex]\( F = ma \)[/tex].
### Question Q4
Using Newton's Third Law of Motion:
Newton's Third Law states that for every action, there is an equal and opposite reaction:
[tex]\[ F_{n} = -f_{1n} \][/tex]
We need to show:
[tex]\[ M_n U_n + M_{n1} U_{m1} = m_n U_n + M_m V_n \][/tex]
Let's consider a system where two bodies are interacting:
1. Initial momentum of body 1: [tex]\( M_n U_n \)[/tex]
2. Initial momentum of body 2: [tex]\( M_{n1} U_{m1} \)[/tex]
3. Final momentum of body 1: [tex]\( m_n U_n \)[/tex]
4. Final momentum of body 2: [tex]\( M_m V_n \)[/tex]
According to the law of conservation of momentum, the total momentum before interaction is equal to the total momentum after interaction:
[tex]\[ \text{Initial total momentum} = \text{Final total momentum} \][/tex]
This translates to:
[tex]\[ M_n U_n + M_{n1} U_{m1} = m_n U_n + M_m V_n \][/tex]
This equation states that the sum of the initial momenta of the two bodies must be equal to the sum of their final momenta, assuming no external forces act on them.
These are the detailed step-by-step solutions to the questions presented. If you have any more questions or need further clarification, feel free to ask!
### Question Q2
#### (i) Unit of Acceleration
Acceleration is defined as the rate of change of velocity with respect to time.
Velocity = Distance / Time
Acceleration = Velocity / Time
Using dimensional analysis,
- Distance has the dimension [tex]\([L]\)[/tex]
- Time has the dimension [tex]\([T]\)[/tex]
Thus,
- Velocity [tex]\(v = \frac{L}{T}\)[/tex]
- Acceleration [tex]\(a = \frac{v}{T} = \frac{L}{T^2}\)[/tex]
Therefore, the unit of acceleration is [tex]\( \left[ \frac{L}{T^2} \right] \)[/tex], which in SI units translates to meters per second squared (m/s²).
#### (ii) Unit of Force by Using Dimension
Force is calculated using Newton’s second law of motion:
Force ([tex]\(F\)[/tex]) = Mass ([tex]\(m\)[/tex]) × Acceleration ([tex]\(a\)[/tex])
Using dimensional analysis,
- Mass has the dimension [tex]\([M]\)[/tex]
- Acceleration has the dimension [tex]\(\left[\frac{L}{T^2}\right]\)[/tex]
Thus,
[tex]\( F = m \times a = [M] \times \left[ \frac{L}{T^2} \right] = \left[ \frac{ML}{T^2} \right] \)[/tex]
Therefore, the unit of force is [tex]\(\left[\frac{ML}{T^2}\right]\)[/tex], which in SI units translates to kilograms meters per second squared (kg·m/s²), also known as a Newton (N).
### Question Q3
#### (a) Statement of Newton's Second Law of Motion
Newton's Second Law of Motion states that the rate of change of momentum of an object is directly proportional to the applied force acting on it and takes place in the direction of the applied force. Mathematically, it can be expressed as:
[tex]\[ F = \frac{dp}{dt} \][/tex]
where [tex]\( p \)[/tex] is the momentum of the object.
#### (b) Verification that [tex]\( F = ma \)[/tex]
Considering Newton's Second Law:
[tex]\[ F = \frac{dp}{dt} \][/tex]
Momentum [tex]\( p \)[/tex] is the product of mass [tex]\( m \)[/tex] and velocity [tex]\( v \)[/tex]:
[tex]\[ p = m \cdot v \][/tex]
Thus,
[tex]\[ \frac{dp}{dt} = \frac{d(m \cdot v)}{dt} \][/tex]
Since mass [tex]\( m \)[/tex] is constant,
[tex]\[ \frac{d(m \cdot v)}{dt} = m \frac{dv}{dt} \][/tex]
[tex]\[ \frac{dv}{dt} \][/tex] is the acceleration [tex]\( a \)[/tex], therefore:
[tex]\[ F = m \cdot a \][/tex]
This completes the proof that [tex]\( F = ma \)[/tex].
### Question Q4
Using Newton's Third Law of Motion:
Newton's Third Law states that for every action, there is an equal and opposite reaction:
[tex]\[ F_{n} = -f_{1n} \][/tex]
We need to show:
[tex]\[ M_n U_n + M_{n1} U_{m1} = m_n U_n + M_m V_n \][/tex]
Let's consider a system where two bodies are interacting:
1. Initial momentum of body 1: [tex]\( M_n U_n \)[/tex]
2. Initial momentum of body 2: [tex]\( M_{n1} U_{m1} \)[/tex]
3. Final momentum of body 1: [tex]\( m_n U_n \)[/tex]
4. Final momentum of body 2: [tex]\( M_m V_n \)[/tex]
According to the law of conservation of momentum, the total momentum before interaction is equal to the total momentum after interaction:
[tex]\[ \text{Initial total momentum} = \text{Final total momentum} \][/tex]
This translates to:
[tex]\[ M_n U_n + M_{n1} U_{m1} = m_n U_n + M_m V_n \][/tex]
This equation states that the sum of the initial momenta of the two bodies must be equal to the sum of their final momenta, assuming no external forces act on them.
These are the detailed step-by-step solutions to the questions presented. If you have any more questions or need further clarification, feel free to ask!
