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The momentum **p** of a body is the product of its mass m and velocity **v: ** **p = **m**v**. A body of lesser mass moves with greater velocity compared to a body of greater mass but the product of the mass and its velocity remains the same, for a given force applied. Force, momentum and time are inter-related. The greater the change in the momentum in a given time, the greater is the force that needs to be applied. A force has to be applied on a body for some length of time, during which a change in momentum occurs. The momentum of an object increases when it accelerates and vice versa.

Therefore, the rate of change of momentum depends on the acceleration of the object. Newton’s Second Law of Motion relates the net external force to the acceleration or rate of change of momentum of the body. Newton’s Second Law of Motion states that: “The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.”

Consider a force, F, being applied on a body of mass, m, for a time interval of ‘t’ seconds. This results in change of velocity and momentum of the body.

Force = F

Mass of the body = m

Time interval = Δt

Initial velocity = v

Final velocity = v + Δv

Initial momentum = p = mv

Final momentum = (p + Δp) = m(v + Δv)

Change in momentum = (p + Δp) – p

Δp = m(v + Δv) – mv

Δp = mΔv

According to Newton's Second Law of Motion

F ∝ $\frac{\u2206\text{p}}{\u2206\text{t}}$

F = K( $\frac{\u2206\text{p}}{\u2206\text{t}}$)

Where K is the constant of proportionality

Δt → 0; $\frac{\u2206\text{p}}{\u2206\text{t}}$ = $\frac{\text{dp}}{\text{dt}}$

F = K $\frac{\text{dp}}{\text{dt}}$

$\frac{\text{dp}}{\text{dt}}$ = $\frac{\text{d}}{\text{dt}}$(mv) = m$\frac{\text{dv}}{\text{dt}}$

$\frac{\text{dp}}{\text{dt}}$ = ma [since a = $\frac{\text{dv}}{\text{dt}}$ ]

F = Kma

Taking K = 1

F = ma

__Units of Force__:

m → 1 Kg; a → 1 ms^{–2}

F → kgms^{–2} or N

Special Applications:

1) F = 0, a = 0 as m ≠ 0

2) F_{x} = $\left(\frac{\text{d}{\text{p}}_{\text{x}}}{\text{dt}}\right)$ = ma_{x}

F_{y} = $\left(\frac{\text{d}{\text{p}}_{\text{y}}}{\text{dt}}\right)$ = ma_{y}

F_{z} = $\left(\frac{\text{d}{\text{p}}_{\text{z}}}{\text{dt}}\right)$ = ma_{z}

3) In case of more than one external force: ∑F = ma.

Newton’s Second Law applies to instantaneous force and instantaneous acceleration.It can also be written as: Change in momentum is equal to the product of force applied and the length of time for which it is applied. If a very large force is applied for a very short duration of time, it results in a large change in momentum of the body. This large amount of force acting on an object for a very short duration of time is called as an impulsive force or impulse.

Therefore, the rate of change of momentum depends on the acceleration of the object. Newton’s Second Law of Motion relates the net external force to the acceleration or rate of change of momentum of the body. Newton’s Second Law of Motion states that: “The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.”

Consider a force, F, being applied on a body of mass, m, for a time interval of ‘t’ seconds. This results in change of velocity and momentum of the body.

Force = F

Mass of the body = m

Time interval = Δt

Initial velocity = v

Final velocity = v + Δv

Initial momentum = p = mv

Final momentum = (p + Δp) = m(v + Δv)

Change in momentum = (p + Δp) – p

Δp = m(v + Δv) – mv

Δp = mΔv

According to Newton's Second Law of Motion

F ∝ $\frac{\u2206\text{p}}{\u2206\text{t}}$

F = K( $\frac{\u2206\text{p}}{\u2206\text{t}}$)

Where K is the constant of proportionality

Δt → 0; $\frac{\u2206\text{p}}{\u2206\text{t}}$ = $\frac{\text{dp}}{\text{dt}}$

F = K $\frac{\text{dp}}{\text{dt}}$

$\frac{\text{dp}}{\text{dt}}$ = $\frac{\text{d}}{\text{dt}}$(mv) = m$\frac{\text{dv}}{\text{dt}}$

$\frac{\text{dp}}{\text{dt}}$ = ma [since a = $\frac{\text{dv}}{\text{dt}}$ ]

F = Kma

Taking K = 1

F = ma

m → 1 Kg; a → 1 ms

F → kgms

Special Applications:

1) F = 0, a = 0 as m ≠ 0

2) F

F

F

3) In case of more than one external force: ∑F = ma.

Newton’s Second Law applies to instantaneous force and instantaneous acceleration.It can also be written as: Change in momentum is equal to the product of force applied and the length of time for which it is applied. If a very large force is applied for a very short duration of time, it results in a large change in momentum of the body. This large amount of force acting on an object for a very short duration of time is called as an impulsive force or impulse.

The momentum **p** of a body is the product of its mass m and velocity **v: ** **p = **m**v**. A body of lesser mass moves with greater velocity compared to a body of greater mass but the product of the mass and its velocity remains the same, for a given force applied. Force, momentum and time are inter-related. The greater the change in the momentum in a given time, the greater is the force that needs to be applied. A force has to be applied on a body for some length of time, during which a change in momentum occurs. The momentum of an object increases when it accelerates and vice versa.

Therefore, the rate of change of momentum depends on the acceleration of the object. Newton’s Second Law of Motion relates the net external force to the acceleration or rate of change of momentum of the body. Newton’s Second Law of Motion states that: “The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.”

Consider a force, F, being applied on a body of mass, m, for a time interval of ‘t’ seconds. This results in change of velocity and momentum of the body.

Force = F

Mass of the body = m

Time interval = Δt

Initial velocity = v

Final velocity = v + Δv

Initial momentum = p = mv

Final momentum = (p + Δp) = m(v + Δv)

Change in momentum = (p + Δp) – p

Δp = m(v + Δv) – mv

Δp = mΔv

According to Newton's Second Law of Motion

F ∝ $\frac{\u2206\text{p}}{\u2206\text{t}}$

F = K( $\frac{\u2206\text{p}}{\u2206\text{t}}$)

Where K is the constant of proportionality

Δt → 0; $\frac{\u2206\text{p}}{\u2206\text{t}}$ = $\frac{\text{dp}}{\text{dt}}$

F = K $\frac{\text{dp}}{\text{dt}}$

$\frac{\text{dp}}{\text{dt}}$ = $\frac{\text{d}}{\text{dt}}$(mv) = m$\frac{\text{dv}}{\text{dt}}$

$\frac{\text{dp}}{\text{dt}}$ = ma [since a = $\frac{\text{dv}}{\text{dt}}$ ]

F = Kma

Taking K = 1

F = ma

__Units of Force__:

m → 1 Kg; a → 1 ms^{–2}

F → kgms^{–2} or N

Special Applications:

1) F = 0, a = 0 as m ≠ 0

2) F_{x} = $\left(\frac{\text{d}{\text{p}}_{\text{x}}}{\text{dt}}\right)$ = ma_{x}

F_{y} = $\left(\frac{\text{d}{\text{p}}_{\text{y}}}{\text{dt}}\right)$ = ma_{y}

F_{z} = $\left(\frac{\text{d}{\text{p}}_{\text{z}}}{\text{dt}}\right)$ = ma_{z}

3) In case of more than one external force: ∑F = ma.

Newton’s Second Law applies to instantaneous force and instantaneous acceleration.It can also be written as: Change in momentum is equal to the product of force applied and the length of time for which it is applied. If a very large force is applied for a very short duration of time, it results in a large change in momentum of the body. This large amount of force acting on an object for a very short duration of time is called as an impulsive force or impulse.

Therefore, the rate of change of momentum depends on the acceleration of the object. Newton’s Second Law of Motion relates the net external force to the acceleration or rate of change of momentum of the body. Newton’s Second Law of Motion states that: “The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.”

Consider a force, F, being applied on a body of mass, m, for a time interval of ‘t’ seconds. This results in change of velocity and momentum of the body.

Force = F

Mass of the body = m

Time interval = Δt

Initial velocity = v

Final velocity = v + Δv

Initial momentum = p = mv

Final momentum = (p + Δp) = m(v + Δv)

Change in momentum = (p + Δp) – p

Δp = m(v + Δv) – mv

Δp = mΔv

According to Newton's Second Law of Motion

F ∝ $\frac{\u2206\text{p}}{\u2206\text{t}}$

F = K( $\frac{\u2206\text{p}}{\u2206\text{t}}$)

Where K is the constant of proportionality

Δt → 0; $\frac{\u2206\text{p}}{\u2206\text{t}}$ = $\frac{\text{dp}}{\text{dt}}$

F = K $\frac{\text{dp}}{\text{dt}}$

$\frac{\text{dp}}{\text{dt}}$ = $\frac{\text{d}}{\text{dt}}$(mv) = m$\frac{\text{dv}}{\text{dt}}$

$\frac{\text{dp}}{\text{dt}}$ = ma [since a = $\frac{\text{dv}}{\text{dt}}$ ]

F = Kma

Taking K = 1

F = ma

m → 1 Kg; a → 1 ms

F → kgms

Special Applications:

1) F = 0, a = 0 as m ≠ 0

2) F

F

F

3) In case of more than one external force: ∑F = ma.

Newton’s Second Law applies to instantaneous force and instantaneous acceleration.It can also be written as: Change in momentum is equal to the product of force applied and the length of time for which it is applied. If a very large force is applied for a very short duration of time, it results in a large change in momentum of the body. This large amount of force acting on an object for a very short duration of time is called as an impulsive force or impulse.