Physics · Science General

Collisions, Momentum and Kinetic Energy

331 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 work, energy and power collision of two rigid bodies energy and collisions understanding collisions

a body of mass m falls from height h on ground. If e be the coefficient of restitution of collision betwwen the body and ground then the distance travelled by body before it comes to rest is

  1. $h\left\{ {\dfrac{{1\, + {e^2}}}{{1 - {e^2}}}} \right\}$
  2. $h\left\{ {\dfrac{{1\, - {e^2}}}{{1 + {e^2}}}} \right\}$
  3. ${\dfrac{{2eh}}{{1 + {e^2}}}}$
  4. ${\dfrac{{2eh}}{{1 - {e^2}}}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

A solid spherical ball of radius R collides with a rough horizontal surface as shown in figure. At the time of collision its velocity is $v _{0}$ at an angle $\theta$ to the horizontal and angular velocity $\omega _{0}$ as shown. After collision, angular velocity of ball may

  1. decrease

  2. increase

  3. remains constant

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics systems of particles and rotational motion collision of two rigid bodies energy and collisions understanding collisions

A solid sphere rolls without slipping on a rough horizontal floor, moving with a speed $v$. It makes an elastic collision with a smooth vertical wall. After impact,

  1. it will move with a speed $v$ initially.

  2. its motion will be rolling without slipping.

  3. its motion will be rolling with slipping initially and its rotational motion will stop momentarily at some instant.

  4. its motion will be rolling without slipping only after some time.
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

The velocity would abruptly change and would cause the solid sphere to slide at first and after some time it would attain a constant angular velocity where it would roll without slipping.

Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

In an elastic collision, kinetic energy of the relative motion is converted into the ____ energies of two momentarily compressed bodies, and then is converted back into the _____ energy. Fill in the blanks. 

  1. kinetic,kinetic

  2. elastic,kinetic

  3. elastic,elastic

  4. kinetic,elastic
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

During elastic collision, at first both the bodies get deformed, so total kinetic energy is converted into theie elastic energies and then they regain their original shape (when moving apart with different velocities) suggesting that the elastic energy is converted back into the kinetic energy.

Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision?

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

A. $\dfrac{1-e}{1+e}$


Given,


$m _1=m _2=m$ (say)

$u _1=u $, $u _2=0$

let, $v _1=$ velocity of ball 1 after collision

      $v _2=$ velocity of ball 2 after collision

The coefficient of restitution,

$e=\dfrac{v _2-v _1}{u _1-u _2}$

$eu=v _2-v _1$. . . . . . .(1)

By the conservation of Linear momentum,

$m _1u _1+m _2u _2=m _1v _1+m _2v _2$ 

$u=v _1+v _2$. . . . . . . . .(1)

By solving equation (1) and (2), we get

$v _1=\dfrac{(1-e)u}{2}$

$v _2=\dfrac{(1+e)u}{2}$

The ratio of velocity of two ball,

$\dfrac{v _1}{v _2}=\dfrac{1-e}{1+e}$

Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

A sphere $A$ moving with speed $u$ and rotating with an angular velocity $\omega$ makes a head-on elastic collision with an identical stationary sphere $B$. There is no friction between the surfaces of $A$ and $B$. Choose the correct alternative(s). Discard gravity.

  1. $A$ will stop moving but continue to rotate with an angular velocity $\omega$
  2. $A$ will come to rest and stop rotating
  3. $B$ will move with speed $u$ without rotating
  4. $B$ will move with speed $u$ and rotate with an angular velocity $\omega$.
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
Let $m$ be the mass of sphere and $v _1$ and $v _2$ be the velocities of the spheres A and B respectively after the collision.
Applying conservation of linear momentum:        $P _i = P _f$
$m  u + 0 = mv _1  + mv _2           \implies v _1 + v _2 = u$      ........(1)
Also $\dfrac{v _2- v _1}{ u-0} = -1         \implies v _1 - v _2 = -u$  
On solving we get,       $v _1 = 0 $   and $v _2 = u$
Thus A will stop and B will move with $u$
Also as there is no friction between the A and B, thus there will be no torque. Hence angular velocities of the respective spheres must remains the same as it was initially.
So, A will continue to rotate with $w$ whereas B will not rotate.
Multiple choice physics units and measurement: error analysis accuracy of measurement accuracy and precision accuracy, precision and uncertainty in measurement

Find the maximum possible percentage error in the measurement of force on an object(on mass m) travelling at velocity v in a circle of radius r, if m=(4.0 plus minus 0.1)kg, v=(10 plus minus 0.1)m/s and r=(8.0 plus minus 0.2)m

  1. ( 50 ± 1) N

  2. ( 50 ± 2.5) N

  3. ( 50 ± 1.5) N

  4. ( 50 ± 3.5) N
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Force$(f)=\cfrac{mv^2}{r}=\cfrac{4\times(10)^2}{8}=50\ \therefore\cfrac{\Delta F}{F}=\cfrac{\Delta m}{m}+2(\cfrac{\Delta V}{V})+(\cfrac{\Delta r}{r})\ \quad=\cfrac{0.1}{4}+2\cfrac{(0.1)}{10}+\cfrac{0.8}{8}\ \therefore \cfrac{\Delta F}{F}=0.07$

Percentage error in $F=25\%+2+2.5\%$
$\therefore$ Percentage error in $F=7\%$
Eror in $\Delta F$ in $F=F\times0.07\ \quad=0\times0.07$
$\therefore$ Error in $\Delta F$ in $F=3.5N$
$\therefore F=(50\pm3.5)N$

Multiple choice physics types of energy law of conservation of energy the law of conservation of energy work, energy and machines

Before a rubber ball bounces off from the floor, the ball is in contact with the floor for a fraction of second. Which of the following statements are correct?

  1. Conservation of energy is not valid during this period

  2. Conservation of energy is valid during this period

  3. As ball is compressed, kinetic energy is converted to compressed potential energy

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

The law of conservation of energy is valid at any instant & in all circumstances.

Multiple choice chemistry chemical kinetics collision theory collision theory of chemical reactions rate of chemical reaction

Two identical balls A & B having velocity of 0.5 m/s & -0.3 m/s respectively. colloid elastically in 1-0. The velocity of A & B after collision will be?

  1. -0.5 m/s, 0.3 m/s

  2. -0.3 m/s,0.5 m/s

  3. 0.5 m/s, 0.3 m/s

  4. 0.3 m/s,0.5 m/s
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In elastic collision of identical balls, velocities are exchanged. A(0.5) gives 0.3 to B, B(-0.3) gives -0.5 to A. Final: A=-0.3 m/s, B=0.5 m/s. Mass cancels out.

Multiple choice chemistry how far? how fast? collision theory collision theory of chemical reactions rate of chemical reaction

What does it mean when a collision is elastic?

  1. No energy is gained or lost.

  2. Energy is gained.

  3. Energy is lost.

  4. The particles can stretch out.

  5. The particles slow down.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(A) , No energy is gained or lost.
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms

Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

A molecule of gas in a container hits one wall (1) normally and rebounds back. It suffers no collision and hits the opposite wall (2) which is at an angle of $30^o$ with wall 1.
Assuming the collisions to be elastic and the small collision time to be the same for both the walls, the magnitude of average force by wall 2. $(F _2)$ provided the molecule during collision satisfy

  1. $F _1 > F _2$
  2. $F _1 < F _2$
  3. $F _1=F _2$, both non-zero
  4. $F _1=F _2=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial momentum, $P _1=mvcos 30$
and final momentum, $P _2 = mvcos30$
change in momentum
$\Delta P = -2mv cos30$
$\Delta P =-\sqrt 3 mv$
Force on wall-1
$F _1=\frac {2mv}{\Delta t}$
Force on wall-2
$F _2=\frac {\sqrt 3mv}{\Delta t}$, so $F _1 > F _2$

Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

A hydrogen atom having kinetic energy $E$ collides with a stationary hydrogen atom. Assume all motions are taking place along the line of motion of the moving hydrogen atom. For this situation, mark out the correct statement(s)

  1. For $E\ge20.4\space eV$ only, collision would be elastic
  2. For $E\ge20.4\space eV$ only, collision would be inelastic
  3. For $E = 2.4\space eV$, collision would be perfectly inelastic
  4. For $E = 18\space eV$, the $KE$ of initially moving hydrogen atom after collision is zero
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

K.E=2P.E
For electron in hydrogen to excite, a minimum of 10.2eV energy is required. Therefore, minimum 20.4eV K.E is required for inelastic collision otherwise, electron would not accept energy. And if E=20.4eV, collision would be perfectly inelastic.
If E is less than 20.4eV, collision is elastic and the two hydrogen atoms exchange velocities.
Therefore, B,D are the correct answers.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

Two bodies of masses 10 kg and 2 kg are moving with velocities $2\hat { i } -7\hat { k } +3\hat { j }\ m{ s }^{ -1 }$ and $-10\hat { i } +35\hat { k } -3\hat { j }\ m{ s }^{ -1 }$ respectively. The velocity of their centre of mass is

  1. $2\hat { i }\ m{ s }^{ -1 }$
  2. $2\hat { j }\ m{ s }^{ -1 }$
  3. $\left( 2\hat { j } +2\hat { k } \right) m{ s }^{ -1 }$
  4. $\left( 2\hat { i } +2\hat { j } +2\hat { k } \right) m{ s }^{ -1 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$m _1 = 10\ kg\quad \vec {v _1}=2\hat i -7\hat k+3\hat j\ m/s$
$m _2=2\ kg \quad \vec {v _2}=-10\hat i +35\hat k-3\hat j\ m/s$
velocity of centre of mass should be
$\vec {v}=\dfrac {m _1 \vec {v _1}+m _2 \vec {v _2}}{m _1 +m _2}$
or $\vec {v}=\dfrac {20\hat i-70\hat k+30\hat j+(-20\hat i+70\hat k-6\hat j)}{10+2}$
$\Rightarrow \ \boxed {\vec {v}=\dfrac {24\ \hat j}{12}=2\hat j\ m/s}$