Tag: force and newton's laws of motion

Questions Related to force and newton's laws of motion

two bodies having $K.E 4 J $ and $9 J$ have equal mass. And their momentum are $p _1$ and $p _2$ respectively then $p _1/p _2$

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. $4:9$

  2. $9:4$

  3. $3:2$

  4. $2:3$

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Correct Option: A

The changes in a momentum of a particle,when the kinetic energy is increased by 0.1% will be:-

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. 0.05%

  2. 0.1%

  3. 1.0%

  4. 10%

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Correct Option: A
Explanation:

\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 is closely related to:

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. force

  2. impulse

  3. velocity

  4. kinetic energy

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Correct Option: C
Explanation:

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.

A metal ball does not rebound when struck on a wall, whereas a rubber ball of same mass when thrown with the same velocity on the wall rebounds. From this it is inferred that:

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. change in momentum is same in both.

  2. change in momentum in rubber ball is more.

  3. change in momentum in metal ball is more.

  4. initial momentum of metal ball is more than that of rubber ball.

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Correct Option: B
Explanation:

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.

 A cannon ball of mass $500 g$ is fired with a speed of $15 m s^{-1}$. Find its momentum.

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. $7500 kg$ $m$ $s^{-1}$

  2. $3750 kg$ $m$ $s^{-1}$

  3. $7.5kg$ $m$ $s^{-1}$

  4. $3.75kg$ $m$ $s^{-1}$

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Correct Option: C
Explanation:

$P = m\times v$


Given m = 500 gram which is also 0.5 Kg after conversion$(500/1000 = 0.5 kg);\ v= 15 \ m/sec$

$P= 0.5\times 15 = 7.5 kg m/sec$

Hence momentum is equal to $7.5 kg\ m/sec$.

The SI unit of momentum is :

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. $kg ms^{-1}$

  2. $kg m s^{-2}$

  3. $kg m s^{2}$

  4. $kg m^{-1}s^{-1}$

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Correct Option: A
Explanation:

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

$\therefore$ unit of momentum 
$P = mV$ is $kg m$$s^{-1}$.
Hence option A is the correct answer.

A body of mass $60 kg$ has the momentum $3000 kg m s^{-1}$. Calculate the speed of the body.

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. $5ms^{-1}$

  2. $0.05ms^{-1}$

  3. $50ms^{-1}$

  4. $500ms^{-1}$

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Correct Option: C
Explanation:

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

Two balls A and B of masses $m$ and $2m$ are in motion with velocities $2v$ and $v$ respectively. What is the ratio of their momentum ?

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. 1:2

  2. 2:1

  3. 1:1

  4. 4:1

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Correct Option: C
Explanation:

(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 $= m \times  2v : 2m \times  v = 1: 1$


A body of mass m moving with a velocity v is acted upon by a force. Write expression for change in momentum  when $v\rightarrow c$.

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. $m \Delta v$

  2. $\Delta \left( mv \right)$

  3. $v \Delta\left(m \right)$

  4. $mv$

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Correct Option: B
Explanation:

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=${v} _{2}-{v} _{1}$

The linear momentum of a ball of mass $50  g$ is $0.5  kg  { m  s }^{ -1 }$. Find its velocity.

physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

  1. $0.01$ ${ m s }^{ -1 }$

  2. $1$ ${ m s }^{ -1 }$

  3. $5$ ${ m s }^{ -1 }$

  4. $10$ ${ m s }^{ -1 }$

Show answer
Correct Option: D
Explanation:

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