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 collisions in one dimension collisions work, energy and power mechanics physics

Two solid balls of rubber $A$ and $B$ whose masses are $200\ gm$ and $400\ gm$ respectively, are moving in mutually opposite directions. if the velocity of ball A is $0.3\ m/s$ and both the ball come to rest after collision, then the velocity of ball $B$ is :

  1. $0.15\ m/s$
  2. $-0.15\ m/s$
  3. $1.5\ m/s$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Initial linear momentum of system 

$= [{{m} _{A}}{{\vec{v}} _{A}}+{{m} _{B}}{{\vec{v}} _{B}}]$

$= 0.2 \times  0.3 + 0.4 \times  vB$

Finally both balls come to rest \ final linear momentum = 0 

By the law of conservation of linear momenum        

$0.2 \times  0.3 + 0.4 \times  vB = 0$

$[{{v} _{B}}=-\dfrac{0.2\times 0.3}{0.4}=-0.15\ m/s]$
Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed $v$ toward ball $2$ which is at rest. Immediately after collision, ball 1 travels at a speed of $v/3$ in the same direction. What type of collision has occured?

  1. inelastic

  2. elastic

  3. completely inelastic

  4. cannot be determined from the information given
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solving the equation of conservation of momentum give us that the relative velocity of approach is equal to the relative velocity of separation. Hence coefficient of restitution is 1.
Which means collision is elastic. 

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

The collision of two balls of equal mass takes place at the origin of coordinates. Before collision, the components of velocities are $(V _x = 50 c m s^{-1},  V _{y} = 0)$ and $(V _{x} = -40 c m s^{-1}$ and $V _{y} = 30 c m s^{-1})$. The first ball comes to rest after collision. The velocity (components $V _{x}$ and $V _{y}$ respectively) of the second ball are

  1. 10 and 30 $c m s^{-1}$
  2. 30 and 10 $c m s^{-1}$
  3. 5 and 15 $c m s^{-1}$
  4. 15 and $5 c m s^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

Which of the following does not hold when two particles of masses $m _1$ and $m _2$ undergo elastic collision?

  1. When $m _1 =m _2$ and $m _2$ is stationary, there is maximum transfer of kinetic energy in head an collision
  2. When $m _1=m _2$ and $m _2$ is stationary , there is maximum transfer of momentum in head on collision
  3. when $m _1 >> m _2$ and $m _2$ is stationary, after head on collision $m _2$ moves with twice the velocity of $m _1$
  4. When the collision is oblique and $m _1=m _2$ with $m _2$ stationary, after the collision the particle move in opposite directions.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When m1 = m2 and m2 is stationary, the first particle transfers all its kinetic energy to the second particle in a head-on collision. This is the maximum possible transfer.

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

Two identical balls A and B collide head on elastically. If velocities of A and B, before the collision are +0.5 m/s and -0.3 m/s respectively, then their velocities, after the collision, are respectively

  1. -0.5 m/s and +0.3 m/s

  2. +0.5 m/s and +0.3 m/s

  3. +0.3 m/s and -0.5 m/s

  4. -0.3 m/s and +0.5 m/s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When identical balls collide $elastically $ they just $exchange$ their $SPEEDS$ and get reversed.

This can be verified by applying the $momentum$ conservation and $energy$ conservation, yes energy remains
 conserved for elastic collisions.
so the exchanged velocities will be following $v _1=0.3m/s$ and $v _2=-0.5m/s$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

In a one-dimensional collision between two particles, their relative velocity is $\bar{v _1}$ before the collision and $\bar{v _2}$ and the collision.

  1. $\bar{v _1} = \bar{v _2}$ if the collision is elastic.
  2. $\bar{v _1} = - \bar{v _2}$ if the collision is elastic.
  3. $|\bar{v _2}| = |\bar{v _1}|$ in all cases.
  4. $\bar{v _1} = - k \bar{v _2}$ in all cases, where k $\geq$ 1.
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
Explanation

If $ { v } _{ 1 }$ is relative velocity before collision.
if ${ v } _{ 2 } $ is relative velocity before collision.
$e\le 1\$
$ e=\dfrac { { v } _{ 1 } }{ { -v } _{ 2 } } $


so $\left| { v } _{ 1 } \right| \ge \left| { v } _{ 2 } \right| $
also due to impact the ratios of velocity get changed,relative velocities D is correct. also, for elastic collision e$=$1.
So, option B is correct only if both particles have equal masses,not in general.

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

In a one-dimensional collision between two identical particles $A$  and $B,\  B$ is stationary and  $A$ has momentum $p$  before impact. During impact, $B$  gives impulse $J$ to $A$.

  1. The total momentum of the '$A\ plus\ B$' system is $p$ before and after the impact, and $(p - J)$ during the impact.
  2. During the impact, $A$ gives impulse $J$ to $B$.
  3. The coefficient of restitution is $\displaystyle \dfrac{2 J}{p} - 1$
  4. The coefficient of restitution is $\displaystyle \dfrac{ J}{p} + 1$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

Let  $u=$ speed of A before impact. Thus,  $p=mu$.
Let $v _1, v _2 = $ speeds of  $A$ and $B$ after impact.
$u = v _1 + v _2 $ and $v _1 - v _2 = - eu$
$u = v _1 + v _2$ and $v _1 - v _2 = - eu$


$\therefore v _1 = \dfrac{1}{2} u (1-e)$ and $v _2 = \dfrac{1}{2} u (1 + e)$

$J = mv _2 = m \displaystyle \left [ \dfrac{1}{2} u (1 + e) \right ] = \dfrac{1}{2} p (1 + e)$

$\Rightarrow e=\dfrac{2J}{p}-1$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A sphere of mass m moving with a constant velocity collides with another stationary sphere of same mass. The ratio of velocities of two spheres after collision will be, if the co-efficient of restitution is e:

  1. $\displaystyle \frac{1 - e}{1 + e}$
  2. $\displaystyle \frac{e - 1}{e + 1}$
  3. $\displaystyle \frac{1 + e}{1 - e}$
  4. $\displaystyle \frac{e + 1}{e - 1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The law of conservation  of linear momentum tells us that the overall momentum before the collision must be equal to the overall momentum after a collision.

Since the spheres have identical masses, we can write

$mu + m\times 0 = mv _A + mv _B$

$u = v _A+v _B$

From the definition of the coefficient of restitution, we know that

$e = \dfrac{v _B - v _A}{u}$

solving above two equations

$e \times ( v _A+v _B) = v _B - v _A$

$v _B(1-e) = v _A (1+e)$

$\dfrac{v _A}{v _B} = \dfrac{1-e}{1+e}$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

In head on elastic collision of two bodies of equal masses:

  1. the velocities are interchanged

  2. the speeds are interchanged

  3. the momentum are interchanged

  4. the faster body slows down and the slower body speeds up
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

For a head-on collision with a stationary object of equal mass, the projectile will come to rest and the target will move off with equal velocity. Hence, the velocities are interchanged i.e. the speeds are interchanged which in turn interchanges the momentum. Also, if target have some velocity then the faster body slows down and the slower body speed up.

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A steel ball moving with a velocity $\overline{v}$ collides with an identical ball originally at  rest. The velocity of the first ball after the collision is :

  1. $\left(-\dfrac{1}{2}\right)\overline{v}$
  2. $-\overline{v}$
  3. $\overline{v}$

  4. zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, a steel ball moving with a velocity $\bar v$ collides with an identical ball originally at  rest. hence, masses of two steel balls are equal. For a head-on collision with a stationary object of equal mass, the projectile will come to rest and the target will move off with equal velocity, thus, the velocity of the first ball after the collision is zero.

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

In the elastic collision of heavy vehicle moving with a velocity 10 ms$^{-1}$ and a small stone at rest, the stone will fly away with a velocity equal to : 

  1. 40 ms$^{-1}$
  2. 20 ms$^{-1}$
  3. 10 ms$^{-1}$
  4. 5 ms$^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the elastic collision between a heavy object and a very light object at rest, the velocity of particles after collision is 
for heavy particle, $v _1 = u _1$
for light particle, $v _2 = 2u _1 - u _2$
since, $u _2 = 0$ hence, 
$v _2 = 2u _1$
Therefore, the stone will fly away with a velocity equal to 
$v _2 = 2u _1 = 2(10) = 20 ms^{-1}$

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

Two particles of masses $ {m} _{1}, {m} _{2} $ movie with initial velocities $ u _{1} \text { and } u _{2} $.On collision, one of the particles get excited to higher level, after absorbing energy If final velocities of particles be $  v _{1}  $ and $  v _{2}  $ then we must have :

  1. $
    \dfrac{1}{2} m _{1} u _{1}^{2}+\dfrac{1}{2} m _{2} u _{2}^{2}=\dfrac{1}{2} m _{1} v _{1}^{2}+\dfrac{1}{2} m _{2} v _{2}^{2}-\varepsilon
    $
  2. $
    \dfrac{1}{2} m _{1} u _{1}^{2}+\dfrac{1}{2} m _{2} u _{2}^{2}+\varepsilon=\dfrac{1}{2} m _{1} v _{1}^{2}+\dfrac{1}{2} m _{2} v _{2}^{2}
    $
  3. $
    \dfrac{1}{2} m _{1}^{2} u _{1}^{2}+\dfrac{1}{2} m _{2}^{2} u _{2}^{2}-\varepsilon=\dfrac{1}{2} m _{1}^{2} v _{1}^{2}+\dfrac{1}{2} m _{2}^{2} v _{2}^{2}
    $
  4. $
    m _{1}^{2} u _{1}+m _{2}^{2} u _{2}-\varepsilon=m _{1}^{2} v _{1}+m _{2}^{2} v _{2}
    $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} Total\, \, initial\, \, energy\, \, of\, \, two\, \, particles \ =\frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 } \ Total\, \, final\, \, energy\, \, of\, \, two\, particles \ =\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 }+\in  \ U\sin  g\, \, energy\, \, conservation\, \, principle, \ \frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 }=\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 }+\in  \ \therefore \frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 }-\in =\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 } \end{array}$

Hence,
option $(C)$ is correct answer.

Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

A moving sphere of mass m suffer a perfect elastic collision (not head on) with an  equally massive stationary sphere. after collision both fly off at angle $\theta $ value of which is :

  1. 0
  2. $\pi $

  3. indeterminate
  4. $\pi /2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For an elastic collision between two equal masses where one is initially at rest, the angle between the final velocity vectors is always 90 degrees (pi/2).