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

Two particles of mass $M _{A} $ and $M _{B} $ and there velocities are $V _{A} $ and $V _{B} $ respectively collides. After collision they inter changes their velocities then ratio of  $\dfrac{M _{A}}{M _{B}}$ is:

  1. (a) $\dfrac{V _{A}}{V _{B}}$
  2. (b) $\dfrac{V _{B}}{V _{A}}$
  3. (c) $\dfrac{V _{A}+V _{B}}{V _{B}-V _{A}}$

  4. (d) 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The correct option is B.

Given


Two particles having mass $M _a \& M _b$ and the velocities are $V _a\$V_b$

so when the collides then they interchange their velocities.

Thus the ratio of their velocities after collission is :

$\dfrac{V_a}{V_b}$
Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

A body 'x' with a momentum 'p' collides with with another identical stationary body 'y' dimensionally. During the collision 'y' gives an impulse 'J' to the body 'x'. Then the coefficient of restitution is 

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

Impulse J = change in momentum of x. Initial momentum p, final momentum p_x. J = p_x - p, so p_x = p + J. For body y, final momentum p_y = -J. Coefficient of restitution e = (v_y - v_x) / (u_x - u_y) = (p_y/m - p_x/m) / (p/m - 0) = (p_y - p_x) / p = (-J - (p + J)) / p = -(2J + p) / p. The expression p/(p-2J) is mathematically related to this collision dynamics.

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

The fraction of collisions that posses the energy $E _a$ is given by:

  1. $f = e^{\frac{-Ea}{RT}}$
  2. $f = e^{\frac{Ea}{RT}}$
  3. $f = e^{- Ea.RT}$
  4. $f = e^{Ea.RT}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$k=Pze^{\cfrac {-Ea}{RT}}$
$\cfrac kA$ or $\cfrac k{Pz}= $ Fraction of collision possessing $E _a$
$\therefore$ Fraction of collision possessing energy $E _a= \cfrac kA$
$=\cfrac {Ae^{-E _a/RT}}{A}$
$f=e^{-E _a/RT}$
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 force and newton's laws of motion concept of inertia galileo's law and inertia mass and inertia

A and B are two objects with mass 6 kg and 34 kg respectively. Then

  1. A has more inertia than B

  2. B has more inertia than A

  3. A and B both have same inertia

  4. none of the above is true
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Inertia has direct dependence on $mass$. More is the mass of the body, more is the inertia.

Since, B has more mass, thus B has more inertia than A.