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 force exerted by collision collisions work, energy and power mechanics physics

Two bodies of equal masses moving with equal velocities in opposite direction collide the resultant velocity of combination is

  1. v

  2. 2v

  3. -v

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

When two equal masses moving with equal but opposite velocities collies. Their velocities cancel each other and net velocity becomes zero.

Multiple choice force exerted by collision collisions work, energy and power mechanics physics

All collision conserves

  1. Kinetic energy

  2. Potential energy

  3. Momentum

  4. All of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
According to the collision theory, all collisions are conserved in momentum in both elastic and inelastic collision. KE is conserved in elastic collision but not in inelastic collision. 
Multiple choice force exerted by collision collisions work, energy and power mechanics physics

A billiard ball moving with a speed of $5  {m}/{s}$ collides with an identical ball originally at rest. If the first ball stops after collision then the second ball will move forward with a speed 

  1. $10 {m}/{s}$
  2. $5 {m}/{s}$
  3. $2.5 {m}/{s}$
  4. $1 {m}/{s}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In elastic head collision velocities get interchanged.

Multiple choice force exerted by collision collisions work, energy and power mechanics physics

In elastic collision, $A$ is conserved while in inelastic collision $B$ is conserved.
I.Momentum
II.Kinetic Energy
III.Potential Energy

  1. $A$ = I, II
    $B$ = I, III
  2. $A$ = I, III
    $B$ = I
  3. $A$ = III
    $B$ = II, III
  4. $A$ = II
    $B$ = I, III
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Collision can be elastic, which means they conserve  both K.E. and momentum. In inelastic collision, they only conserve momentum but not K.E. 
Multiple choice force exerted by collision collisions work, energy and power mechanics physics

In an elastic collision of two particles the following is conserved.

  1. Momentum of each particle

  2. Speed of each particle

  3. Kinetic energy of each particle

  4. Total kinetic energy of both the particles
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total kinetic energy, i.e., kinetic energy of the system is conserved.

Multiple choice force exerted by collision collisions work, energy and power mechanics physics

A ball of  mass $0.2kg$ is thrown against the wall$,$ the ball strikes the wall normally with velocity of $30m/s$ sand rebounds with velocity of $20m/s.$ Calculate the impulse of the force exerted by the ball on the wall

  1. $2N$
  2. $-10N$
  3. $20N$
  4. $40N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $, u = 30 m/s$ $, v = 20 m/s$ 

$m =  0.2 kg$
impulse $-$ change in momentum of the body$,$
$= mv - mu = m (v - u )$
$= 0.2 ( -20 -30 ) = -10 N$
Hence,
option $(B)$ is correct answer.

Multiple choice force exerted by collision collisions work, energy and power mechanics physics

Two solid rubber balls $A$ and $B$ having masses $200$ grams and $400$ grams are moving in opposite directions with velocity of $A$ equal to $0.3   {m}/{s}$. After collision the two balls come to rest, then the velocity of $B$ is

  1. $0.15 {m}/{s}$
  2. $1.5 {m}/{s}$
  3. $- 0.15 {m}/{s}$

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial linear momentum of system $= {m} _{A} \bar { { V } _{ A } } + {m} _{B} \bar { { V } _{ B } }$
                                                      $ = 0.2 \times 0.3 + 0.4 \times {V} _{B}$
Finally both balls come to rest.
Finally linear momentum $= 0$
By the law of conservation of linear momentum.
$0.2 \times 0.3 + 0.4 \times {V} _{B} = 0$
${V} _{B} = -\dfrac{0.2 \times 0.3}{0.4} = -0.15  {m}/{s}$

Multiple choice force exerted by collision collisions work, energy and power mechanics physics

A shell of mass $m$ moving with velocity $V$ suddenly breaks into $2$ pieces. The part having mass ${m}/{4}$ remains stationary. The velocity of the other shell will be:

  1. $V$
  2. $2V$
  3. ${3}/{4} V$
  4. ${4}/{3} V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Mass of the one part which remains at rest is $\dfrac{m}{4}$.
Thus mass of other shell is $\dfrac{3m}{4}$.
Using conservation of linear momentum :  $P _{initial} = P _{final}$
$m V = \dfrac{m}{4} \times 0 + \dfrac{3m}{4} V _2$
Or  $m V =   \dfrac{3m}{4} V _2$
$\implies$ $V _2 = \dfrac{4V}{3}$
Multiple choice force exerted by collision collisions work, energy and power mechanics physics

Two bodies having same mass $40   kg$ are moving in opposite directions, one with a velocity of $10  {m}/{s}$ and the other with $7   {m}/{s}$. If they collide and move as one body, the velocity of the combination is

  1. $10 {m}/{s}$
  2. $7 {m}/{s}$
  3. $3 {m}/{s}$
  4. $1.5 {m}/{s}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By the conservation of momentum
   $40 \times 10 + \left(40\right) \times \left(-7\right) = 80 \times V$
   $V = 1.5   {m}/{s}$

Multiple choice force exerted by collision collisions work, energy and power mechanics physics

Two equal masses ${m} _{1}$ and ${m} _{2}$ moving along the same straight line with velocities $+3  {m}/{s}$ and $-5  {m}/{s}$ respectively collide elastically. Their velocities after the collision will be respectively :

  1. $+4 {m}/{s}$ for both
  2. $-3 {m}/{s}$ and $+5 {m}/{s}$
  3. $-4 {m}/{s}$ and $+4 {m}/{s}$
  4. $-5 {m}/{s}$ and $+3 {m}/{s}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As ${m} _{1} = {m} _{2}$, therefore after elastic collision velocities of masses gets interchanged.

Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

Two spheres $A$ and $B$ of masses $m _1$ and $m _2$ respectively collide. $A$ is at rest initially and $B$ is moving with velocity $v$ along x-axis. After collision $B$ has a velocity $\cfrac{v}{2}$ in a direction perpendicular to the original direction. The mass $A$ moves after collision in the direction

  1. Same as that of $B$
  2. Opposite to that of $B$
  3. $\theta=\tan^{-1}{(1/2)}$ to the x-axis
  4. $\theta=\tan^{-1}{(-1/2)}$ to the x-axis
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By conservation of momentum in the x and y directions: m2*v = m2*(v/2)sin(theta) + m1*v1_x and 0 = m2(v/2)*cos(theta) + m1*v1_y. Since the collision is elastic or specific conditions are implied, the direction of A is determined by the vector sum of momenta. Given the options, the direction is consistent with the conservation laws.

Multiple choice physics momentum collision of two rigid bodies energy and collisions understanding collisions

In a collision between two solid spheres. velocity of separation along the line of impact (assume no external forces act on the system of two spheres during impact):

  1. Cannot be greater than velocity of approach

  2. Cannot be less than velocity of approach

  3. Cannot be equal to velocity of approach

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

$\begin{array}{l} e=\dfrac { { volume\, \, of\, \, sep } }{ { volume\, \, of\, \, app } }  \ 0<e<1 \ \Rightarrow Volume\, \, \, of\, \, sep<volume\, \, of\, \, app \end{array}$

$\therefore $ Option $A$ is correct.

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.