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

A force acts on a body of mass 3 kg such that its velocity changes from 4 m $s^{-1}$ to 10 m $s^{-1}$.Calculate the change in the momentum of the body.

  1. 2 kg m $s^{-1}$
  2. 14 kg m $s^{-1}$
  3. 18 kg m $s^{-1}$
  4. 42 kg m $s^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

change in momentum 
$\Delta p=m(v _1-v _2)=3*(10-4)=18kg m/s$

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

To reduce the momentum of a given body to half its original value, its velocity must be ___________.

  1. reduced to half

  2. doubled

  3. quadrupled

  4. kept unchanged
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The momentum of a body is given by

          $=MV$
where M is mass and V is velocity
           $P=MV$
So if M is constant and ${P}'=\frac{P}{2}$
      $\frac{P}{2}=M{V}'$
       ${V}'=\frac{V}{2}$

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

A football player kicks a $0.25kg$ ball and imparts it a velocity of $10m/s$. The contact between foot and the ball is only $\cfrac{1}{50}$th of a second. The kicking force is

  1. $250N$
  2. $125N$
  3. $0N$
  4. $3.78N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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


$F=\dfrac{m\triangle V}{\triangle f}=\dfrac{m(V _f-V _i)}{\triangle f}$

Force, $m=0.25kg$

and $V _f=10m/s,V _i=0m/s$

$\therefore F=\dfrac{(0.25)(10-0)}{\dfrac{1}{50}}$

$F=0.25(500)$

$F=25\times 5$

$\boxed{F=125N}$

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

A force of 100 dynes acts on mass of 5 gm for 10 sec. The velocity produced is:

  1. 2 cm/sec

  2. 20 cm/sec

  3. 200 cm/sec

  4. 2000 cm/sec
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that,

Force $F=100\,dyn$

Mass $m=5\,g$

Time $t=10\,s$

Initial velocity $u=0$

Now, the acceleration is

  $ F=ma $

 $ a=\dfrac{F}{m} $

 $ a=\dfrac{100\times {{10}^{-5}}}{5\times {{10}^{-3}}} $

 $ a=0.20\,m/{{s}^{2}} $

Now, from equation of motion

  $ v=u+at $

 $ v=0+0.20\times 10 $

 $ v=2\,m/s $

 $ v=200\,cm/s $

Hence, the velocity produced is $200\ cm/s$

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

Jahnvi is walking at $1.63 m/s$. If she weighs $583 N$, what is the magnitude of her momentum?

  1. <span class="MathJax_Preview"><span class="MathJax"><span class="math"><span class="mrow"><span class="mn">951<span class="mtext"> <span class="mi">k<span class="mi">g<span class="mtext"> <span class="mi">m<span class="texatom"><span class="mrow"><span class="mo">/<span class="mi">s<span class="MJX_Assistive_MathML">951 kg m/s
  2. $68.8 \ kg \ m/s$
  3. $137 \ kg \ m/s$
  4. $672 \ kg \ m/s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$P = mv$
$ = 1.63 \times 583$
$ = 951\,kg\,m/s$
Hence, 
No any option is match so answer is $ 951\,kg\,m/s.$
Multiple choice physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

What is the value of $p _1$ and $m _2$ ?

m (kg) v (m/s) p = mv (kgm/s)
85 60 $p _1$
$m _2$ 2.5 6.25

  1. 5100, 2.5

  2. 5.1 , 2.5

  3. 2500, 5

  4. 55, 2.6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
m (kg) v (m/s) p = mv (kgm/s)
85 60 5100
2.5 2.5 6.25

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 $
$\therefore m = \dfrac{p}{v} = \dfrac{6.25}{2.5} = 2.5\ kg$

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

A hockey ball of mass 200 g travelling at $10\ ms^{-1}$ is struck by a hockey stick so as to return it along its original path with a velocity of $5\ ms^{-1}$. Calculate the change of momentum which occurred in the motion of the hockey ball by the force applied by the hockey stick:

  1. $-3\ Ns$
  2. $3\ Ns$
  3. $6\ Ns$
  4. $-6\ Ns$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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 $=m(v-u)=0.2\times(-5-10)=-3\ kgm{s}^{-1}=-3\ Ns$

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

The change in momentum of a vehicle weighing 2000 kg when its speed increases from 36 km/h to 72 km/h uniformly is :

  1. 15000 kg.m/s

  2. 4000 kg.m/s

  3. 20000 kg.m/s

  4. 7200 kg.m/s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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             

Change in momentum       $\Delta P = m(v-u) = 2000\times (20-10)  =20000$  kg.m/s

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

A 4-kilograms ball slides over horizontal friction less surface with velocity of 1 meter per second ball strike the post and rebounds upward perpendicular to the initial direction with same speed.
Calculate the change of momentum of the ball in initial direction.

  1. -4 kg.m/s

  2. -1 kg.m/s

  3. 0 kg.m/s

  4. 1 kg.m/s

  5. 4 kg.m/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Finally the velocity of ball remains in vertical direction. Thus velocity in initial direction(horizontal)=$0m/s$
Thus the change in momentum=$\vec{p _2}-\vec{p _1}=m(\vec{v _2}-\vec{v _1})=4\times (0-1)kgm/s=-4kgm/s$

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

A force acts on a body of mass 3 kg such that its velocity changes from $4 \,m\, s^{-1}$ to $10 \,m\, s^{-1}$. The change in momentum of the body is:

  1. $42 \,kg\, m\, s^{-1}$
  2. $2 \,kg\, m\, s^{-1}$
  3. $18 \,kg\, m\, s^{-1}$
  4. $14 \,kg\, m\, s^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Change in momentum $=p _2-p _1=mv _2-mv _2=m(v _2-v _1)=3(10-4)=18 $ $kg $ $ms^{-1}$