Momentum (p) - class-XI

Momentum of an object is defined as the product of its mass and the velocity with which the object is moving.

Using the work-energy theorem: Work = Change in Kinetic Energy. F * d = 1/2 * m * v^2. F * 0.6 = 1/2 * 0.04 * (90)^2. F * 0.6 = 0.02 * 8100 = 162. F = 162 / 0.6 = 270 N.

Momentum is mass* velocity.

While running a long distance, jumper tries to gain more velocity that increases his momentum. When he jumps, due this momentum, his body covers longer distance.

Second law of motion yields $F = \Delta p/\Delta t$

Momentum combines both mass and velocity. Ant has a greater momentum because of it is moving. Momentum of the elephant would be zero because it is at rest.

$SI$ unit of momentum $= (SI, \text{unit of}, m).(SI,\text{ unit of}, v) = kg ,m/s$

Momentum = mv

According to Newton's second law,

For motion there are two factors, How much is moving? and how fast is moving?

momentum is P = mv

A car and a lorry are moving with same momentum, if same breaking force is applied, then lorry comes to rest in shorter distance as it has greater mass so moves with less velocity.

P = mv

P= mv

$\Delta p = m(v-u) = 3 (10-4)$
$=18 kg  m  s^{-1}$

Option A is correct .

Momentum, P = Mv

$\displaystyle F = \frac{m(v-u)}{t}$

$F=\displaystyle \frac{\Delta P}{t}$
$2 = \displaystyle \frac{0.4}{t} $ (or) $t = 0.2 s$

\begin{array}{l} The, , kineticenergyisE=\frac { 1 }{ 2 } m{ v^{ 2 } } \ The, , , momentum, , is, , p=mv \ v=\sqrt { \frac { { 2E } }{ m }  }  \ therefore, \ \Rightarrow p=mv=m\sqrt { \frac { { 2E } }{ m }  } =\sqrt { 2mE }  \ taking, \log  s, \ \Rightarrow \log  , p=\frac { 1 }{ 2 } \left( { \log  2+\log  m+\log  , E } \right)  \ differentiating: \ \Rightarrow \frac { { \Delta , p } }{ p } =\frac { 1 }{ 2 } \left( { \frac { { \Delta m } }{ m }  } \right) +\frac { 1 }{ 2 } \left( { \frac { { \Delta E } }{ E }  } \right)  \ but, \ \Rightarrow \frac { { \Delta E } }{ E } =0.1 \ so., therefore, \ \Rightarrow \frac { { \Delta p } }{ p } =\frac { 1 }{ 2 } \times 0.1=0.05 \ so, the, correct, option, is, A. \end{array}

Momentum of an object is due to its mass and velocity. m x v = momentum. Mass remains constant for any object. The value of momentum mostly depends on the value of velocity. Thus, velocity out of the four options is closely related to momentum.

Rubber ball has got more elasticity and the change in velocity is more in case of rubber ball which proves that change in momentum of rubber ball is more.

$P = m\times v$

$\because$ $p = mv$ unit of $m = kg$
unit of $v = m$$s^{-1}$ 

$ P = m\times v$
$ 3000 = 60 \times v$
$ v = 50 m/sec $
Hence velocity becomes $50 m/sec $.

(i) Inertia depends on mass, and since the mass of the two balls are $m$ and $2m$, So the ratio of inertia of the balls $1 : 2$

(ii)Momentum , $ p = mv$
So, as per momentum equation, it depends on both mass and velocity.
So, their ratio $= m \times  2v : 2m \times  v = 1: 1$

As we know that momentum (p) =MAASS$\times$VELOCITY(vector quantity)
hence chance in momentum($\Delta$p)=${{m} _{2}{v} _{2}}-{{m} _{1}{v} _{1}}$=$\Delta$(mv)
where ${m} _{1}$=${m} _{2}$=m
$\Delta$v=${v} _{2}-{v} _{1}$

We know that $ p= mv$
where $p =0.5 kg ms^{-1}$ linear momentum
$m=50g = 0.05kg$ mass, $v =$ velocity
$\therefore v = \dfrac{p}{m}$
$\Rightarrow =\dfrac{0.5}{0.05} = 10 m/s$

Linear momentum, $p=mv$
Unit of mass $=kg$
Unit of velocity $=m/s$
Hence, Unit of linear momentum $=kg m/s = kg m/s^2\times s = Ns$

We know that $ p= mv$

The momentum is given as $M= mv$

The momentum is given by mass times it's velocity, and it is a vector quantity with direction same as velocity. The question states momentum is mass times it's speed, which is incorrect since it gives a scalar value. Hence the statement is false and correct answer is B.

We know, p$=$mv and the mass of car is less than truck.
However momentum are same, that shows  velocity of car is greater than the velocity of truck.

Linear momentum is always in the direction of Velocity.
Linear momentum is a vector quantity defined as the product of an object's mass, m, and its velocity, v. Linear momentum is denoted by the letter p and is called momentum for short: Note that a body's momentum is always in the same direction as its velocity vector. .

The initial momentum of ball $P _1 = m\vec{v} $, after batsman hits ball and reverses its direction, momentum of ball $P _2= -m\vec{v} $
Magnitude of change in momentum $\Delta p = \left | P _2-P _1\right | =\left | - 2m \vec{v}\right | =2m \left | \vec{v}\right | $ 
from data given in question $  m=0.15 kg ;;  \left | \vec{ v} \right |= 12m/ s$
$\Delta P=2\times 0.15\times 12= 3. 6 kg m/ s $

Momentum of truck $ P _{truck}= m _{truck}v _{truck}$
from given data in question $ m _{truck} = 38000 ; kg;; v _{truck}= 1.2; m/s$
$P _{truck}= 38000\times 1.2= 45600 ; kg m/ s$
the momentum of car is equal to momentum of truck $P _{car}= m _{car}v _{car}\Rightarrow v _{car}= 45600/ 1400= 32.57 m/s$

Momentum is  the quantity of motion of a moving body, measured as a product of its mass and velocity.

While opening the door the force applied is tangential to the motion of the door.
While applying brakes the motion is on a different axis to the force applied,
When a batsman hits a ball, both the ball and the force applied are in the same direction
While drawing water from the well the motion and the force are in different directions (mostly tangential)

$\Delta P = F\Delta t = 10 \times 10 = 100 kg m/s$

Mass of a body is the quantity of inertia possessed by it, or in simpler words, it is the measure of resistance that a body can provide to change in its motion.
Speed is the measure of how fast a body is moving.
Velocity is the measure of how fast a body is moving along with its direction.
Linear momentum is the quantity which denoted the motion of body along with its direction.

If the collision of the ball and the wall is elastic one then the change in the momentum of the ball after collision is $= mu-(-mu)=2mu$ away from the wall.

Given :   $m = 10$ kg             $v = 36$ km/hr $ = 36\times \dfrac{5}{18}  = 10$ m/s

Given :   $m =20$ kg        

Mass ad velocity both are considered to be quantities which defines motion.

For metallic ball,

Momentum is product of mass and velocity.

Initial momentum of the ball $=P$

$\Delta P={ P } _{ 1 }-{ P } _{ 2 }$ $[$ change in momentum$]$

Forces that do not store energy are called nonconservative or dissipative forces. Friction is a nonconservative force, and there are others. Any friction-type force, like air resistance, is a nonconservative force.

to gain momentum so that the range that he jumps is long

change in momentum = 5 (40 - 30) = $50kgm/s$

The force is $F=\dfrac{\Delta P}{dt}$

$momentum=mass*\delta v=5kg*(20-0.20)=99kgm/s$

P =mv = 0.1 x 15 $kgm/s$ = 1.5$kgm/s$

Let J be the impulse associated with the hammer

P = mv

Impulse applied to an object produces an equivalent vector change in its linear momentum, also in the same direction. The SI unit of impulse is the newton second $(N⋅s)$, and the dimensionally equivalent unit of momentum is the kilogram meter per second $(kg⋅m/s).$

quantity of motion in a body, the relative amount of its motion, as measured by its momentum, varying as the product of mass and velocity.which is define as momentum.

Momentum of the body is defined as the product of mass of the body and its velocity.

Momentum is the quantity of motion of a body which is defined as the product of mass of body and its velocity.

Given :    $P = 50$  $kg$ $m/s$                    $m = 10$  kg

Given :    $m = 10$ kg                   $v =2$ m/s

Given :    $m =25$ g $= 0.025$ kg                   $v =100$ m/s

Given :    $P = 20$  $kg$ $m/s$                    $v = 1000$  $m/s$

Given :    $m = 1000$ kg                   $v =0.5$ m/s

Given :    $P = 1$ $kg$ $m/s$                  $v =2$ m/s

Given :    $m _1 = 4000$ kg           $m _2  =20000$ kg             $v =2$ m/s

Given :   $m =0.01$ kg           $v = 5$ m/s

The principle of conservation of linear momentum states that the linear momentum of a system can be changed only if external forces acts on it.

A force acts on a body of mass 3 kg such that its velocity changes from 4 m/s to 10 m/s

Let A and B be two bodies and F be the force for t time duration.

The momentum of a body is given by

Momentum is defined as the product of the body's mass and its velocity.

Initial velocity while dropping on the ground u=2m/s.
Parachutist was brought to rest within s=0.25 m
Therefore retardation f : $u^2=2fs; ; f=\frac{u^2}{2s}=8m/s^2$
Averarage force from ground on her close is F=mf=600 N

In order to stop a moving body we require a force and depends on the velocity as well as its mass. It was Newton in his second law who defined this new quantity as momentum.
Momentum is the product of mass and velocity of the body or in other words momentum is the measure of quantity of motion.

Momentum = mass $\times$ velocity and its unit is $Kgms^{-1}$
Impulse of the force(J) =force $\times$ time and 
$=ma \times t$
$=m\left(\dfrac{u}{t}\right) \times t$
Thus, impulse (J) is the change of momentum and its unit is $Kgms^{-1}$

Force = $\dfrac{\triangle P}{\triangle t}$ (Newtons 2nd law)

Let velocity of gun is $v$. 

Given,

The sum of the magnitude of the momenta of the individual particles in the system is equal to the total mass times the velocity of centre of mass.
Velocity of COM depends upon the net external force, and as nothing is given about it we can't tell anything strictly about it.
Hence It could be non zero, and it might not be a constant

$Ft = mV _2 - mV _1 \Longrightarrow 10 \times 0.25 = P _t - P _i, P _t = 2.5$

As the rocket is moving upward by ejecting gases, its mass is getting decreased gradually. As momentum is the product of the mass and velocity that is $p = m\times v$. Hence momentum decreases as mass is decreasing.

As we know that momentum is nothing but product of mass and velocity. Let us substitute the given values.

In first case:
$ m=85kg $
$v =60 m/s$
$ p =mv = 85 \times 60 = 5100\ kg m/s$
In second  case:
$ m=? $
$v =2.5 m/s$
$ p =mv = 6.25 $
$\therefore m = \dfrac{p}{v} = \dfrac{6.25}{2.5} = 2.5\ kg$

Mass, $m=50\ g=0.05\ kg$

Mass of ball(m)$=$200g$=$0.2kg
Initial velocity of ball(u)$=$10m${s}^{-1}$
Final velocity of ball(v)$=$5m${s}^{-1}$
Therefore, change in momentum ($\triangle$p) $=$ Final momentum - initial momentum $=m(v-u)=0.2\times(-5-10)=-3\ kgm{s}^{-1}=-3\ Ns$

Given :   $m = 2000$ kg            $u = 36$ km/h $ = 36 \times \dfrac{5}{18}  = 10$ m/s                $v= 72$ km/h $ = 72 \times \dfrac{5}{18}  = 20$ m/s             

Initial momentum  $=mv$

Final momentum  $=-mv$

Change in momentum  $=-mv-(mv)=-2mv$

Initially the velocity of the ball in horizontal direction is $1m/s$.

Impulse $I=$ change in momentum $=mv _2-mv _1$

$v=u+at=1.5\times 4=6ms^{-1}$

Given, $m=3 kg$, $v _1=4 m/s$ and $v _2=10 m/s$

Impulse $I=Ft$

Impulse $=$ change in momentum

We know that,
Momentum = mass $ \times $ velocity
So, we know that when a graph is plotted, momentum against velocity, it is a straight line passing through the origin because momentum varies linearly with respect to velocity.