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)
$$ \Rightarrow F = m \cdot \left( {{{v - u} \over t}} \right) {\rm{~~~~~}}\left( {\because ~a = {{v - u} \over t}} \right)$$
$$ \Rightarrow F = \left( {{{mv - mu} \over t}} \right)$$
$$ \Rightarrow F = {{{p_2} - {p_1}} \over t}{\rm{ ~~~~ }}\left[ \matrix{
{\rm{~Where,~ }}{p_2} = mv = {\rm{ ~final ~momentum}} \hfill \cr
{\rm{ ~~~~~~~~~~ }}{p_1} = mu = {\rm{ ~initial ~mometum}} \hfill \cr} \right]$$
$$ \Rightarrow {\rm{~Force}} = {{{\rm{~Change ~in ~Momentum}}} \over {{\rm{~Time ~taken}}}}$$
$$\Rightarrow {{\rm{~Force = ~Rate ~of ~change ~of ~momentum}}}$$