Physics · Science General

Collisions, Momentum and Kinetic Energy

385 Questions

Collisions, momentum, and kinetic energy questions analyze the principles of elastic and inelastic impacts. They require calculating mass, velocity, and conserved energy during physical interactions. These foundational physics topics are essential for most government engineering and general science examinations.

Elastic collisionsInelastic collisionsMomentum calculationKinetic energy principlesVelocity after impact

Collisions, Momentum and Kinetic Energy Questions

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

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

  1. 0.05%

  2. 0.1%

  3. 1.0%

  4. 10%
Reveal answer Fill a bubble to check yourself
A Correct answer
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}

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

  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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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 ?

  1. 1:2

  2. 2:1

  3. 1:1

  4. 4:1
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $= m \times  2v : 2m \times  v = 1: 1$


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

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

  1. $m \Delta v$
  2. $\Delta \left( mv \right)$
  3. $v \Delta\left(m \right)$
  4. $mv$
Reveal answer Fill a bubble to check yourself
B Correct answer
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=${v} _{2}-{v} _{1}$

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

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

  1. $0.01$ ${ m s }^{ -1 }$
  2. $1$ ${ m s }^{ -1 }$
  3. $5$ ${ m s }^{ -1 }$
  4. $10$ ${ m s }^{ -1 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

A body of mass $5   kg$ is moving with velocity $2  { ms }^{ -1 }$. Calculate its linear momentum. 

  1. $2.5$ $kg$ ${ ms }^{ -1 }$
  2. $10$ $kg$ ${ ms }^{ -1 }$
  3. $5$ $kg$ ${ ms }^{ -1 }$
  4. $20$ $kg$ ${ ms }^{ -1 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that $ p= mv$


where $p =$ linear momentum

$m=5kg$ mass, $v =2 m/s$ velocity

$ p = 5 \times 2 = 10 kg ms^{-1}$

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

A body P has mass 2 m and velocity 5 v. Another body Q has mass 8 m and velocity 1.25 v. Find out the ratio of momentum of P and Q.

  1. 2:1

  2. 1:1

  3. 1:2

  4. 3:2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The momentum is given as $M= mv$

Where, $m= $ mass of object, 
              $v= $ velocity of object
Momentum of first object $= 2m\times 5v = 10mv$
Momentum of second object $= 8m\times 1.25v=10mv$, taking the ratio of the magnitude of momentum of Ist and IInd objects $ |p _1|:|p _2|= 1:1$
Hence, correct answer is B.

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

Momentum of a body is defined to be the product of its mass and:

  1. speed

  2. velocity

  3. acceleration

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

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

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of $12 ms^{-1}$. If the mass of the ball is $0.15\ kg$, determine the change in momentum of the ball. (Assume linear motion of the ball)

  1. $3.6 kg m/s$
  2. $5 kg m/s$
  3. $3.5 kg m/s$
  4. $7 kg m/s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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 $

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

A force of $10 N$ acts on a body of mass $20 kg$ for $10 s$ .Change in its momentum is:

  1. $5 kg m/s$
  2. $100 kg m/s$
  3. $200 kg m/s$
  4. $1000 kg m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Where,
     $\Delta$ P = Change in momentum,
          F = Applied force
     $\Delta$ t  = time for which force has been applied on the body
                   
                          

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

A ball of mass 'm' moves normal to a wall with a velocity 'u' and rebounds with the same speed. The change in momentum of the ball during the rebounding is

  1. 2mu towards the wall

  2. 2mu away from the wall

  3. zero

  4. mu away from the wall
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.