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

 Assertion (A) : A body of "$m _{1}$" collides another body of mass "$m _{2}$" at rest elastically. The fraction of energy transferred to the second body is$\dfrac{m _{1}}{m _1+m _{2}}$
Reason (R) : In an "inelastic collision" only linear momentum is conserved

  1. Both Assertion (A) and Reason (R) are correct

    and R is the correct explanation

  2. Both Assertion (A) and Reason (R) are correct but

    the reason does not give the correct explanation

  3. A is true but R is false

  4. A is false but R is true
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In elastic collision no energy loss takes place that is total energy is transferred.
However, in inelastic collision, energy loss takes place and only linear momentum is conserved.

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

During inelastic collision between two bodies, which of the following quantities always remain conserved?

  1. Total kinetic energy.

  2. Total mechanical energy.

  3. Total linear momentum.

  4. Speed of each body
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since no external forces are acting on the colliding bodies during collision, thus total linear momentum is always conserved in all type of collisions but kinetic energy in not conserved in all collisions.

Kinetic energy is conserved in perfectly elastic collision only but some kinetic energy is lost in inelastic collisions. So, total kinetic energy is not conserved in inelastic collision.

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

In an inelastic collision, the kinetic energy after collision

  1. is same as before collision

  2. is always less than that before collision

  3. is always greater than that before collision

  4. may be less or greater than that before collision
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In inelastic collisions , the kinetic energy is used for deforming the the bodies . In such collisions , the reformation of the bodies is partial. Therefore, the potential energy stored in the deformation is lost as heat and the kinetic energy of the system after collision is less than that of before collisions. However, in some cases it may be greater ( such collisions are called super elastic collisions).

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

A ball hits the floor and rebounds after an inelastic collision. In this case

  1. the momentum of the ball just after the collision is same as that just before the collision

  2. The mechanical energy of the ball remains the same in the collision

  3. the total momentum of the ball and the earth is conserved

  4. the total energy of the ball and the earth remains the same
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

1) K.E will be less since collision is inelastic
2) No impulsive external force on system (earth + ball)

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

Choose the false statement

  1. In a perfect elastic collision the relative velocity of approach is equal to the relative velocity of separation

  2. In an inelastic collision the relative velocity of approach is less than the relative velocity of separation

  3. In an inelastic collision the relative velocity of separation is less than the relative velocity of approach

  4. In perfect inellastic collision relative velocity of separation is zero.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an inelastic collision the relative velocity of approach is more than the relative velocity of separation. Hence B is wrong

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

Two pendulum bobs of mass $m$ and  $2\ m$ collide elastically at the lowest point in their motion. If both the balls are released from height $H$ above the lowest point. The velocity of the bob of mass $m$ just after collision is :

  1. $\sqrt {\dfrac{{2gH}}{3}} $
  2. $\dfrac{5}{3}\sqrt {2gH} $
  3. $\sqrt {2gH} $

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

Lets consider first pendulum bob of mass $m _1$ and second  pendulum bob of mass $m _2$ collide elastically at the lower point in their motion. If both are released  from the height $H$ above the lower point.


At height $H$, Kinetic energy is equal to potential energy.


$ \dfrac{1}{2}mu^2=mgH$

$u=\sqrt{2gH}$

Given ,

$m _1=m$

$m _2=2m$

$u _1=-\sqrt{2gH}$

$u _2=\sqrt{2gH}$

In elastic collision,

The velocity of first $v _1$ after collision at the lower point,

$v _1=(\dfrac{m _1-m _2}{m _1+m _2})u _1+(\dfrac{2m _2}{m _1+m _2})u _2$

By putting the given value of $m _1$, $m _2$, $u _1$, $u _2$ in the above equation,

$v _1=(\dfrac{m-2m}{m+2m})(-\sqrt{2gH})+(\dfrac{4m}{m+2m})\sqrt{2gH}$

$v _1=\dfrac{5}{3}\sqrt{2gH}$

Thus, the correct option is B.

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

Two particles A and B, move with constant velocities $\vec{v _1}$ and $\vec{v _2}$. At the initial moment their position vectors are $\vec{r _1}$ and $\vec{r _2}$ respectively. The condition for particle A and B's collision is:

  1. $\vec{r _1}-\vec{r _2}=\vec{v _1}-\vec{v _2}$
  2. $\dfrac{\vec{r _1}-\vec{r _2}}{|\vec{r _1}-\vec{r _2}|}=\dfrac{\vec{v _2}-\vec{v _1}}{|\vec{v _2}-\vec{v _1}|}$
  3. $\vec{r _1}\cdot \vec{v _1}=\vec{r _2}\cdot \vec{v _2}$
  4. $\vec{r _1}\times \vec{v _1}=\vec{r _2}\times \vec{v _2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice modelling collisions collisions momentum work, energy and power physics

A body of mass 2kg is projected upward from the surface of the ground at t$=$0 with a velocity of 20 m/s. One second later a body B, also of mass 2 kg, is dropped from a height of 20 m. If they collide elastically, then velocities just after collision are 

  1. V$ _A = $ -5 m/s downward , V$ _B = $ 5 m/s upward
  2. V$ _A = $ 10 m/s downward , V$ _B = $ 5 m/s upward
  3. V$ _A = $ 10 m/s upward , V$ _B = $ 10 m/s downward

  4. both move downward with velocity 5 m/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At t=1s, A is at y = 20(1) - 0.5(10)(1)^2 = 15m, velocity vA = 20 - 10(1) = 10 m/s. B is at y = 20 - 0.5(10)(0)^2 = 20m, velocity vB = 0 m/s. They collide at some point. Since masses are equal and collision is elastic, they exchange velocities. The calculation of the exact collision point and post-collision velocities confirms option A.

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

In an inelastic collision-

  1. Momentum of the system is always conserved.

  2. Velocity of separation is less than the velocity of approach.

  3. The coefficient of restitution can be zero.

  4. All of the above.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In inelastic collision momentum of the system is always conserved if $F _{ext}=0$.
Velocity of separation is less than the velocity of approach since co-efficient of restitution e < 1
$e=0$ for a perfectly inelastic collision since the colliding particles stick together after collision.
Hence, option (D) is correct.

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

A mass $m$ moves with velocity $v$ and collides inelastically with another identical mass. After collision, the 1st mass moves with velocity $\frac { v }{ \sqrt { 3 }  } $ in a direction perpendicualr to the initial direction of motion. find the speed of the second mass after collision.

  1. $v$
  2. $\sqrt { 3v } $
  3. $\frac { 2 }{ \sqrt { 3 } } v$
  4. $\frac { v }{ \sqrt { 3 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Conservation of momentum: m*v*i = m*v1*j + m*v2*cos(theta)*i + m*v2*sin(theta)*j. Since v1 = v/sqrt(3) in the y-direction, the x-component of momentum is conserved: m*v = m*v2*cos(theta), and y-component: 0 = m*v1 + m*v2*sin(theta). Solving these gives v2 = 2*v/sqrt(3).

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

A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed v. What is the mass of other body and the speed of the center of mass of two bodies ?

  1. $ 1.0 kg and \frac {2}{3}v $
  2. $ 1.2 kg and \frac{5}{8} $
  3. $ 1.4 kg and \frac {10}{17} v $
  4. $ 1.5 kg and \frac {4}{7} v $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$m _{1}=2\ kg$
$u _{1}=4\ m/s$
$v _{1}=\dfrac {u _{1}}{4}=\dfrac {4}{4}=1\ m/s$
$u _{2}=0$
since nothing is given so we would take it as elastic collision. 
hence conservation of momentum is given by:
$m _{1}u _{1}+m _{2}u _{2}=m _{1}v _{1}+m _{2}v _{2}$
$2\times 4+0=2\times 1+m _{2}v _{2}$
$m _{2}v _{2}=6$
according to the conservation kinetic energy:
$\dfrac {1}{2}m _{1}u _{1}^{2}+\dfrac {1}{2}m _{2}u _{2}^{2}=\dfrac {1}{2}m _{1}v _{1}^{2}+\dfrac {1}{2}m _{2}v _{2}^{2}$
$2\times 16+0=2\times 1+m _{2}v _{2}^{2}$
$m _{2}v _{2}^{2}=30$
$\dfrac {m _{2}v _{2}^{2}}{m _{2}v _{2}}=\dfrac {30}{6}$
$v _{2}=5\ m/s$
$m _{2}=1.2\ kg$


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

Two objects of mass 3 kg and 2 kg move along x and y ais with $ 4 m/s^2 $ and $ 3 m/s ^2 $ respectively on a horizontal smooth table.after collision the bodies stick together.then :

  1. Heat generated in the process is 15 joules

  2. Heat generated in the process is 18 joules
  3. direction of motion x-axis after collision is $ 60^0 $
  4. direction of x-axis after collision is $ tan{-1} \left( \frac { 1 }{ 3 } \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice modelling collisions collisions momentum work, energy and power physics

$n$ balls each of mass $m$ impinge elastically each second on a surface with velocity $u$. The average force experienced by the surface will be

  1. $mnu$
  2. $2\ mnu$
  3. $4\ mnu$
  4. $mnu/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
  • Elastically means the balls get bounced with $same$ speed i.e $u$
  • so the change in momentum will be $P _2-P _1=-mu-mu=-2mu$
  • so the momentum tranferred to the surface will be $-(-2mu)=2mu$
  • thus the total momentum transferred by $n$ balls will be $2mnu$
  • force $F=\dfrac{Momentum}{time}=\dfrac{2mnu}{1sec}=2mnu$
Multiple choice modelling collisions collisions momentum work, energy and power physics

A body moving towards a finite body at rest collides with it. It is possible that :

  1. both the bodies come to rest

  2. both the bodies move after collision

  3. the moving body comes to rest and stationary body starts moving

  4. the stationary body remain stationary and moving body changes its direction
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

If a body moving towards a finite body at rest collides with it then momentum will be conserved and hence, the velocities after collision are may be in inverse proportion to their masses or may get interchanged. Hence, both the bodies move after collision or the moving body comes to rest and stationary body starts moving.

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

A body moving towards a finite body at rest collides with it. It is possible that:

  1. both the bodies come to rest

  2. both the bodies moves after collision

  3. the moving body comes to rest and the stationary body starts moving

  4. the stationary body remains stationary, the moving body changes its velocity
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

both bodies cannot come to rest as  it will violate the law of conservation of momentum.
Option (D) will also violate conservation of momentum.