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 modelling collisions collisions momentum work, energy and power physics

A $90\ gm$ ball moving at $100 \ cm/s$ collide head on with a stationary $10\ gm$ ball. The coefficient of restitution is $0.5$. The collision is :

  1. elastic

  2. inelastic

  3. perfect inelastic

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $e = 1$, then the collision is called perfectly elastic.
If $0 < e <1$, the collision is called inelastic.
If $e = 0$,  the collision is called perfectly inelastic.

Multiple choice modelling collisions collisions momentum work, energy and power physics

A body dropped freely from a height h on to a horizontal plane, bounces up and down and finally comes to rest.The coefficient of restitution is e. The ratio of velocities at the beginning and after two rebounds is 

  1. 1 : e

  2. e : 1
  3. $1 : e^3$
  4. $e^2 : 1 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let initial velocity is v at time of collision. $v = \sqrt { 2gh } $

after first re bound velocity ${v} _{1} = ev$
after second rebound velocity ${v} _{2} = e{v} _{1} = {e}^{2}v$
ratio $=\dfrac { { v } _{ 2 } }{ v } =\dfrac { { e }^{ 2 }v }{ v } $
$ ={ e }^{ 2 }:1$

Multiple choice modelling collisions collisions momentum work, energy and power physics

Two bodies of equal masses moving with equal speeds makes a perfectly inelastic collision. If the speed after the collision is reduced to half, the velocities of approach is 

  1. $30 ^ { \circ }$
  2. $60 ^ { \circ }$
  3. $90 ^ { \circ }$
  4. $120 ^ { \circ }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a perfectly inelastic collision of equal masses m, m*v1 + m*v2 = 2*m*v_final. If v_final = v/2, then v1 + v2 = v. This implies the angle between the initial velocity vectors must be 90 degrees.

Multiple choice modelling collisions collisions momentum work, energy and power physics

Two small spheres of equal mass, and heading towards each other with equal speeds, undergo a headon collision (no external force acts on system of two spheres). Then which of the following statement is correct?

  1. Their final velocities must be zero

  2. Their final velocities may be zero

  3. Each must have a final velocity equal to the others initial velocity

  4. Their velocities must be reduced in magnitude
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Nothing is mentioned about coefficient of restitution. 

Hence the only true statement is 'their final velocities may be zero.'

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

A body of mass 2 $ \mathrm{kg}  $ is thrown up vertically with $ \mathrm{K.E.}  $ of 490 Joules. If the acceleration due to gravity is 9.8 $ \mathrm{m} / \mathrm{s}^{2}  $ , then the height at which the K.E. of the body becomes half its originalvalue is given by

  1. $50m$
  2. $12.5m$
  3. $25m$
  4. $10m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial K.E. = 490 J. At height h, K.E. = 245 J. By conservation of energy, initial K.E. = final K.E. + potential energy at height h. 490 = 245 + mgh. 245 = 2 * 9.8 * h. h = 245 / 19.6 = 12.5 m.

Multiple choice physics force and newton's laws of motion first law of motion newton's first law of motion momentum and newton's laws

A force vector applied on a mass is represented by $\vec {F} = 6\hat {i} - 8\hat {j} + 10\hat {k}$ and accelerates with $1\ m/s^{2}$. What is the mass of the body?

  1. $14\ kg$
  2. $10\sqrt {2}kg$
  3. $2\sqrt {10}kg$
  4. $20\ kg$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
To find mass of the body we will use magnitude of force formula $,$ 
According to question$:-$
$\overrightarrow { F } =6-8\widehat { j } +10\widehat { k } $
Magnitude of Force $=\sqrt {\left( {{6^2} + {8^2} + {{10}^2}} \right)} $
                                  $= 14.14 N$
$m = F / a$
     $= 14.14 N / 1 m/s^2$
     $= 14.14 kg$
Hence,
option $(A)$ is correct answer.
Multiple choice physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

A body of mass $0.5$ kg is performing S.H.M. with a time period $\pi /2$ seconds. If its velocity at mean position is $1$ m/s, the restoring force acts on the body at a phase angle $60^o$ from extreme position is

  1. 0.5 N

  2. 1 N

  3. 2 N

  4. 4 N
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$T=\dfrac{ \pi}{2}$
$V _{max}=1 m/sec$
$\omega =\dfrac{2\pi}{T}$
$V=4  rad/sec$
$V _{max=}A\omega$
$A\times 4=1$
$A=\dfrac {1}{4}$
$a=\omega^{2}x$
$F= m\omega^{2}x$
$x=A   cos   60^o$
$\therefore x=\dfrac {A}{2}$
$F=0.5\times (4)^{2}\times \dfrac {1}{4}\times \dfrac{1}{2}$
$F=1N$

Multiple choice physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

A planck with a body of mass m placed on to it starts moving straight up with the law $y=a(1-\cos{\omega t})$ where $\omega$ is displacement. Find the time dependent force:

  1. $-ma\omega^2\cos{\omega t}$
  2. $ma\omega^2\cos{\omega t}$
  3. $ma\omega^2\sin{\omega t}$
  4. $mg+ma\omega^2\cos{\omega t}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total force on the particle will be
$F=mg+m\dfrac { d^{ 2 }y }{ dt^{ 2 } } $
since $y=a(1-\cos { \omega t } )\\ \Rightarrow \dfrac { dy }{ dt } =a\omega \sin { \omega t } \\ \Rightarrow \dfrac { d^{ 2 }y }{ dt^{ 2 } } =a\omega ^{ 2 }\cos { \omega t } $
$\Rightarrow F=mg+ma\omega ^{ 2 }\cos { \omega t } $

Multiple choice physics types of energy renewable and non-renewable resources renewable and non-renewable sources of energy substances, objects and energy

A body of mass 5 kg falls from a height of
30 metre. If its all mechanical energy is changed into heat, then heat produced
will be:-

  1. 350cal

  2. 150 cal

  3. 60cal

  4. 6cal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
work done $=mgn$
$= 5 \times 9.8 \times 30 = 14705$
Heat produce $=\dfrac{1470}{4.2}= \boxed{350\ cal}$
$\boxed {Answer\ is\ A}$



























Multiple choice evs - i substances, objects and energy renewable resources alternative fuels and energy sources alternative sources of energy

If a man speed up 1m/s his KE increases by 44% His original speed in m/s is 

  1. 1

  2. 2

  3. 5

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

Let initial KE be K = 1/2mv^2. New KE is 1.44K = 1/2m(v+1)^2. Taking the ratio: 1.44 = (v+1)^2 / v^2. Taking the square root: 1.2 = (v+1)/v. Thus, 1.2v = v + 1, so 0.2v = 1, which means v = 5 m/s.

Multiple choice evs - i substances, objects and energy renewable resources alternative fuels and energy sources alternative sources of energy

A 200 gm mass has velocity of $(3\hat i+4\hat j)$m/s at certain instant. Find its kinetic energy. 

  1. $2.5 J$
  2. $0.5 J$
  3. $0.8 J$
  4. $1.5 J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First we will calculate the Magnitude of velocity  so $v=\sqrt{3^2+4^2}=5\ m/s$

Now m=200 gm or 0.2 kg

So $KE=\dfrac{mv^2}{2}=\dfrac{0.2\times 5^2}{2}$

so $KE=2.5\ J$

Multiple choice physics along with motion what forces can do? force and it's unit force and its effects

Choose the wrong statement:

  1. 1 kg wt = 9.8 N

  2. Momentum is a vector quantity.

  3. Force is always conserved.

  4. Momentum is conserved in the absence of an external force.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ 1kg-wt = 1 \times 9.8 N $ 

$ momentum = mass \times velocity $ As velocity is vector, momentum is vector.
Friction is non conservative force.
According to Newton's second law, rate of change of momentum is directly proportional to applied force.
So,if force is zero, momentum is conserved.
So wrong statement is option C.