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Class 9, Science, Chapter-8, Lecture-2, Newton's Second Law (Notes)

MOMENTUM:

The quantity of motion of an object measured as the product of its mass and velocity is termed as momentum of the object. 
Mathematically: 
${{\rm{Momentum}}\;{\rm{ }} = {\rm{ }}\;{\rm{mass\; }} \times {\rm{ velocity}}}$  
${p\;\; = \;\;mv}$

Momentum is a VECTOR quantity

SI Unit: kg ms– 1

NEWTON’S SECOND LAW OF MOTION (in terms of momentum):

The rate of change of momentum of a body is directly proportional to the force and the change is in the direction of force. 
Mathematically

$$\left( {{{{m_2}{v_2} - {m_1}{v_1}} \over {{t_2} - {t_1}}}} \right) \propto F$$

$$ \Rightarrow F \propto \left( {{{{m_2}{v_2} - {m_1}{v_1}} \over {{t_2} - {t_1}}}} \right)$$
$$ \Rightarrow F = k \cdot \left( {{{{m_2}{v_2} - {m_1}{v_1}} \over {{t_2} - {t_1}}}} \right){\rm{  (~k ~is ~a ~constant ~of ~proportionality~)}} -  -  - {\rm{(i)}}$$
Unit of force is chosen in such a way that a unit force produces a unit change of momentum in unit time. 
$${\rm{For~ F = 1,~ }}\left( {{m_2}{v_2} - {m_1}{v_1}} \right) = 1{\rm{ ~and~ }}\left( {{t_2} - {t_1}} \right) = 1{\rm{ ~in ~equation ~(i)}}$$
$$1 = k \cdot 1 \cdot 1$$

$$ \Rightarrow k = 1$$
putting $k = 1$ in equation $(i)$
$$F = \left( {{{{m_2}{v_2} - {m_1}{v_1}} \over {{t_2} - {t_1}}}} \right)$$

NEWTON’S SECOND LAW OF MOTION (in terms of acceleration): 

“The acceleration produced by an unbalanced force acting on a body is directly proportional to the magnitude of force and is inversely proportional to the mass of the object, the acceleration taking place in the direction of the force.” 
Mathematically

$a \propto F$  and $a \propto {1 \over m}$

$ \Rightarrow a \propto {F \over m}$

$ \Rightarrow F \propto m \cdot a$

$ \Rightarrow {\rm{ }}F = k \cdot m \cdot a~{\rm{  (~k ~is ~a ~constant ~of ~proportionality~) }} -  -  - {\rm{(ii)}}$

Unit of force is chosen in such a way that a unit force produces a unit acceleration in a unit mass.
$${\rm{For~ F = 1,~ }}m = 1{\rm{ ~and~ }}a = 1{\rm{ ~in ~equation ~(ii)}}$$
$$1 = k \cdot 1 \cdot 1$$

$$ \Rightarrow k = 1$$
putting $k = 1$ in equation $(ii)$
$$F = m \cdot a$$

  • Newton’s second Law gives us the measurement of force.
  • Relation Between Force and Momentum: ${{\rm{Force}} = {\rm{Rate ~of ~change ~of ~momentum}}}$

The First Law is a special case of the Second Law.

Justification: 

When no force acts on a moving object of mass m

$F = 0$

$$ \Rightarrow a = {F \over m} = {0 \over m} = 0$$

$$ \Rightarrow a = 0$$

$${\therefore}{\rm{~when~ }}F = 0{\rm{,~ }}a = 0$$

This shows that when no force acts on an object, the velocity remains constant. This is Newton’s First Law.

UNITS OF FORCE:

  • SI Unit – newton ( N )
  • G.S. unit – dyne ( dyn )

DEFINITION OF ‘ONE NEWTON’  
The force which when acted upon an object of mass 1 kg produces an acceleration of 1 m/s2 is one newton ( 1 N ).

$$1{\rm{ ~newton}} = 1{\rm{ ~kg}} \times 1{\rm{ ~m}}{{\rm{s}}^{ - 2}}$$

$$ \Rightarrow 1{\rm{ ~N}} = 1{\rm{ ~kgm}}{{\rm{s}}^{ - 2}}$$

DEFINITION OF ‘ONE DYNE’  
The force which when acted upon an object of mass 1 gram produces an acceleration of 1 cm/s2 is one dyne.

$$1{\rm{ ~dyne}} = 1{\rm{ ~g}} \times 1{\rm{ ~cm}}{{\rm{s}}^{ - 2}}$$

$$ \Rightarrow 1{\rm{ ~dyn}} = 1{\rm{ ~gcm}}{{\rm{s}}^{ - 2}}$$

Relation between 1 N and 1 dyn:

$$1{\rm{ ~N}} = 1{\rm{ ~kg}} \times {\rm{ 1 ~m}}{{\rm{s}}^{ - 2}}$$

$$\Rightarrow 1{\rm{ ~N}} = 1000{\rm{ ~g}} \times {\rm{ 100 ~cm}}{{\rm{s}}^{ - 2}}$$

$$\Rightarrow 1{\rm{ ~N}} = {10^5}{\rm{ ~gcm}}{{\rm{s}}^{ - 2}}$$

$$\Rightarrow 1{\rm{ ~N}} = {10^5}{\rm{ ~dyn}}$$

Relation Between Force and Momentum

$F = m \cdot a$   (By Newton's Second Law)

$$ \Rightarrow F = m \cdot \left( {{{v - u} \over t}} \right) {\rm{~~~~~}}\left( {\because ~a = {{v - u} \over t}} \right)$$

$$ \Rightarrow F = \left( {{{mv - mu} \over t}} \right)$$

$$ \Rightarrow F = {{{p_2} - {p_1}} \over t}{\rm{ ~~~~  }}\left[ \matrix{
 {\rm{~Where,~ }}{p_2} = mv = {\rm{ ~final ~momentum}} \hfill \cr 
 {\rm{ ~~~~~~~~~~ }}{p_1} = mu = {\rm{ ~initial ~mometum}} \hfill \cr}  \right]$$

$$ \Rightarrow {\rm{~Force}} = {{{\rm{~Change ~in ~Momentum}}} \over {{\rm{~Time ~taken}}}}$$

$$\Rightarrow {{\rm{~Force = ~Rate ~of ~change ~of ~momentum}}}$$