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**Newton's law of motion**

•__ First law__: “A body continues in its state of rest or uniform motion in a straight line unless acted by external force”

-It gives the idea of inertia and gives the qualitative definition of force. It is the condition for equillibirium of body. eg A book lying on table, moving fan etc.

•__ Second law__: “The rate of change of momentum is directly proportional to the force applied and change takes place on direction of force.”

**Momentum**: It is the measure of quantity of motion contained on body. Let p be the momentum and v be the velocity then,

$p =mv$

i.e, $F=dtdP =dtd(mv) =mdtdv +vdtdm $

-It gives the quantitative definition of force.

•__ Third law__: “To every action there is equal and opposite reaction.”

$F_{12} =−F_{21} $

It states that force exists on pair.

**Motion of lift and apparent weight**

Let, M be mas of lift ,m=mass of body ,T=tension on rope and R=apparent weight.R

(i) When lift moves up or down with constant Velocity(a=0)

mg-R=ma

so R=mg

∴T=(M+m)g

(ii)When lift moves upward with constant acceleration(a)

R-mg=ma

so, R=mg+ma

∴T=(M+m)(g+a)

$∴$net acceleration =(g+a)

(iii)When lift moves downward with constant acceleration(a)

mg-R=ma

so, R=mg-ma

∴T=(M+m)(g-a)

$∴$net acceleration=(g-a)

(iv) If a=g(free fall) R=0 weightlessness.

(v)If a>g, R is negative i.e acts upward person will rise from floor.

**System of variable mass **

Example:

What is the force needed to hold pipe in which water is flowing with velocity v?

We have,

$F=v.dtdm $

or, $F=v.dtd(ρAl) =v∗ρA∗dtdl =ρAv_{2}$ (away from himself)

__Case I__: When water rebounds with equal speed

F= ρAv(v-(-v))=2ρAv^{2}

__Case II__:When directed at angle θ with horizontal and rebounding is neglected,

F= ρAv(vcosθ -0)= ρAv^{2}cosθ

__Case III__:When directed at angle θ with horizontal and rebounds at same angle.

F= ρAv[vcosθ –(-vcosθ)] = 2ρAv^{2}cosθ

**Mass pulley system**

(i)If m_{2}>m_{1} Net acceleration=$Total massnet force $

=$m_{2}+m_{1}m_{2}g−m_{1}g =(m_{2}+m_{1})(m_{2}−m_{1})g $

Also
T-m_{1}g=m_{1}a and m_{2}g-T=m_{2}a

Solving,
$T=m_{1}+m_{2}2m_{2}m_{1} g$_{
}

Note, the net thrust on pulley is T+T=2T

(ii) T=m_{2}a

M_{1}g-T=m_{1}a

Also net acceleration=$Total massnet force $

=$m_{2}+m_{1}m_{1}g $

$T=m_{1}+m_{2}2m_{2}m_{1} g$

Note, the net thrust on pulley is , $T_{2}+T_{2} =2 T$

If m_{2}gsinβ

Net acceleration=$Total massnet force $

=$m_{2}+m_{1}−m_{2}gsinβ+m_{1}gsinα $

Also, m_{1}gsinα-T=m_{1}a and T -m_{2}gsinβ =m_{2}a

Solving,

$T=m_{2}+m_{1}m_{2}m_{1} g(sinα+sinβ)$

Note, the net thrust on pulley can be found vectorically.

**When moving carriage is given horizontal force**

•If a mass suspended on moving carriage is given horizontal force:

Tcosθ=mg and Tsinθ=ma

So, Tanθ=$ga $

And Tension on rope is

$T=ma_{2}+g_{2} $

**Rocket Motion**

•It based on conservation of linear momentum states that, total momentum remain constant.

∑mv=constant

For two bodies,

$m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}$

Rocket motion,

V=u*$lnmm_{o} $

Where, V=velocity, u=exhaust speed of gas

m_{0}=initial mass, m=residual mass

**Real weight on pan**

If an object is placed on one pan its weight is ‘w1’, in another pan its weight is w2, then true weight of object:

•If balance has equal arms, true weight is $w=2w1+w2 $

•If balance has unequal arms, true weight is $w=w_{1}.w_{2} $

**Contact forces on smooth surfaces**

The forces developed on surface of contact between two bodies is called Contact force.

•Acceleration of system is

$a=m_{1}+m_{2}F $

And
F-F’=m_{1}a, F’=m_{2}a

Solving, contact force is

$F=m_{1}+m_{2}m_{2}F $

If force was applied on block m_{2} ,

contact force is $F=m_{1}+m_{2}m_{1}F $

**Tension on connected systems**

Acceleration of system is $a=(m_{1}+m_{2}+m_{3})F $

$T_{1}=(m_{1}+m_{2}+m_{3})F(m_{2}+m_{3}) $

$T_{2}=(m_{1}+m_{2}+m_{3})Fm_{3} $

Generally, tension on any part of string is = $total sum of massesforce∗(sum of masses behind it) $

**Reaction on different surfaces**

Rxn always lies perpendicular to contact plane.

(i) __On simple plane:__

Reaction =weight(R=mg)

(ii)__On inclined plane:__

R=mgcosθ

__(iii)Pulled by force F__

R=mg-Fsinθ

And ma=Fcosθ

__(iv)Pushed by force F__

R=mg+Fsinθ>mg

ma=Fcosθ

Hence, pulling is easier than pushing because It reduces normal reaction.

**Problem on chain or rope**

•At a distance x from force applied end. (M is total mass of rope)

-Mass of remaining portion is ,$m=LL−x ∗M$

Acceleration of each part is, $a=MF $

So, Tension at C is $T_{c}=m∗a$

or, $T_{c}=LL−x ∗F$

Similarly, for suspended vertically tension at x distance from free end (force applied end) is,

$T_{c}=LL−x ∗F+Lxg ∗M$

**Example**

•If a stone of mass m is dropped from height h and it penetrates s distance on ground, then,

average retardation is $a=sgh $ (*as,energy conservation*, *mgh=mas*)

-Time of penetration is$t=gh2s_{2} $ $(0=2gh −at)$

*-Average resis*tance offered by ground is,

R=mg+ma=$mg(1+ga )$= $mg(1+sh )$

•*No*te that, if a man takes a quick step on platform his weight first decreases then increases.

**Impulse and Momentum**

•It is the product of force and time.If I denotes impulse, F is force and dt be small interval of time then∫$F.dt$

$I=F.dt$=$∫dtdP $. dt =$dp=p_{f}−p_{i}$

Momentum: It is the product of mass and velocity.

It is vector and denoted by $p $ and given by

$p =mv$ where, m is mass of body and v is velocity.