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Elastic And Inelastic Collision

Elastic And Inelastic Collision

Edited By Vishal kumar | Updated on Sep 03, 2024 12:35 PM IST

A collision is a fundamental aspect of physical interaction, existing in two primary forms: elastic and inelastic. In elastic collisions, both momentum and kinetic energy are conserved, such as with billiard balls or a rubber ball bouncing on a hard surface. These collisions idealize interactions without permanent deformation or heat generation. In contrast, inelastic collisions involve changes in the momentum of the objects post-collision.

This Story also Contains
  1. Collision
  2. In Perfectly Elastic Collision
  3. Special Cases of Head-on Elastic Collision
  4. Perfectly Elastic Oblique Collision
  5. Head-on Inelastic Collision
  6. Perfectly Inelastic collision
  7. Collision Between Bullet and Vertically Suspended Block
  8. Solved Example Based on Elastic And Inelastic Collision
  9. Summary
Elastic And Inelastic Collision
Elastic And Inelastic Collision

This article covers Elastic and Inelastic Collisions, including Perfectly Elastic Head-on and Oblique Collisions, Head-on Inelastic Collisions, Perfectly Inelastic Collisions, and Collisions Between a Bullet and a Vertically Suspended Block. Which belongs to the chapter work, energy, and power, which is an important chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), more than ten questions have been asked on this concept. And for NEET three questions were asked from this concept.

Background wave

Collision

The interaction between two or more objects is called a collision. And during this interaction strong force acts between two or more bodies for a short time as a result of which the energy and momentum of the interacting particle change.

Stages of Collision

There are three distinct identifiable stages in collision

  1. Before the collision-

The interaction forces are zero

  1. During the collision-

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The interaction forces are very large and these forces act for a very short time. And because of these interaction forces the energy and momentum of the interacting particle change.

  1. After the collision-

The interaction forces are zero

In Perfectly Elastic Collision

The law of conservation of momentum and that of Kinetic Energy hold good.

12m1u12+12m2u22=12m1v12+12m2v22..(1)m1u1+m2u2=m1v1+m2v2.....(2)m1,m2: masses u1,v1 : initial and final velocity of the mass m1u2,v2 : initial and final velocity of the mass m2

From equation (1) and (2)

 We get, u1u2=v2v1 …..(3)

Or, we can say Relative velocity of approach = Relative velocity of separation

And

e=v2v1u1u2= Relative velocity of separation  Relative velocity of approach 

So in Perfectly Elastic Collision

e = 1

From equations (1),(2), (3)

We get

v1=(m1m2m1+m2)u1+2m2u2m1+m2.......(4)

Similarly,

v2=(m2m1m1+m2)u2+2m1u1m1+m2........(5)

Now, let's understand this concept better from the example given below:

Example: A body of mass m1 moving with an unknown velocity of v1i^, undergoes a collinear collision with a body of mass m2 moving with a velocity v2i^. After collision m1 and m2 move with velocities of v3i^ and v4i^, respectively. If m2=0.5m1 and v3=0.5v1 the v1 is : 

1) v4v22
2) v4v2
3) v4v24
4) v4+v2

Solution:

Perfectly Elastic Collision -

The law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

12m1u12+12m2u22=12m1v12+12m2v22m1u1+m2u2=m1v1+m2v2m1,m2: masses 
u1,v1 : initial and final velocity of the mass m1
u2,v2 : initial and final velocity of the mass m2

before collision:

after collision:

(Pinitial )total =(Pfinal )total m1v1i^+m2v2i^=m1v3i^+m2v4i^m2=0.5m1&v3=0.5v1i^(m1v1+0.5m1v2)=i^(m1v3+0.5m1v4)v1+0.5v2=v3+0.5v4v1+0.5v2=0.5v1+0.5v4(v3=0.5v1)0.5v1+0.5v2=0.5v4v1+v2=v4v1=v4v2
Hence, the answer is option (2).

Special Cases of Head-on Elastic Collision

  1. Equal mass in case of perfectly elastic collision

Then, v1=u2 and v2=u1

Or, Velocity mutually interchange

  1. If a massive projectile collides with a light target  (i.e m1>>m2 ) 

Since m1>>m2 so we use m2=0
Putting m2=0 in equation (4) and (5)
We get v1=u1 and v2=2u1u2

  1. If the target particle is massive in case of elastic collision  (i.e; m2>>m1 ) 

 Since m2>>m1

So, the lighter particle recoils with the same speed and the massive target particle remains practically at rest.

i.e; v¯2=u¯2
v¯1=u¯1

Perfectly Elastic Oblique Collision

Let two bodies move as shown in the figure.

By the law of conservation of momentum

Along x-axis-

m1u1+m2u2=m1v1cosθ+m2v2cosϕ......(1)

Along y-axis-
0=m1v1sinθm2v2sinϕ............(2)

By the law of conservation of kinetic energy
12m1u12+12m2u22=12m1v12+12m2v22........(3)

In Perfectly Elastic Oblique Collision

Value of e=1

So along the line of impact (here along in the direction of v2)

We apply e=1

And we get e=1=v2v1cos(θ+ϕ)u1cosϕu2cosϕ ….. (4)

So we solve these equations (1),(2),(3),(4) to get unknown.

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Special Condition

if m1=m2 and u2=0
Then, from equation (1), (2) and (3)
We get, θ+ϕ=π2

i.e.; after perfectly elastic oblique collision of two bodies of equal masses (if the second body is at rest), the scattering angle θ+ϕ would be 90.

Head-on Inelastic Collision

In Inelastic Collision the Law of conservation of momentum holds good but kinetic energy is not conserved.

12m1u12+12m2u2212m1v12+12m2v22m1u1+m2u2=m1v1+m2v2........(1)m1,m2: masses 
u1,v1 : initial and final velocities of mass m1
u2,v2 : initial and final velocities of mass m2

In Inelastic Collision (0 < e < 1)

e=v2v1u1u2 ….. (2)

From equations (1),(2)

We get,

v1=(m1em2m1+m2)u1+(1+e)m2m1+m2u2......(3)

Similarly,
v2=(m2em1m1+m2)u2+(1+e)m1m1+m2u1.........(4)

Special Case

A sphere of mass m moving with velocity u hits inelastically with another stationary sphere of same mass.

As, e=v2v1u1u2
So, e=v2v1u
or,
ue=v2v1 ....(5)

By conservation of momentum

As, m1u1+m2u2=m1v1+m2

So v2+v1=u ….(6)

From equation (5) and (6)

 We get, v1v2=1e1+e


Loss in kinetic Energy

Loss in K.E = Total initial kinetic energy – Total final kinetic energy

ΔK.E.=(12m1u12+12m2u22)(12m1v12+12m2v22) …. (7)

From equation (3), (4) and (7)

We can write, Loss in kinetic energy in terms of e as

K.E.=12(m1m2m1+m2)(1e2)(u1u2)2

Perfectly Inelastic collision

In a perfectly inelastic collision, two bodies stick together after the collision, so there will be a final common velocity (v).

  1. When the colliding bodies are moving in the same direction

By the law of conservation of momentum

m1u1+m2u2=(m1+m2)vv=m1u1+m2u2(m1+m2)

Loss in kinetic energy

ΔKE=(12m1u12+12m2u22)(12(m1+m2)V2)ΔKE=12(m1m2m1+m2)(u1u2)2

  1. When the colliding bodies are moving in the opposite direction

By the law of conservation of momentum

m1u1+m2(u2)=(m1+m2)vv=m1u1m2u2m1+m2

If v is positive then the combined body will move along the direction of motion of mass m1

If v is negative then the combined body will move in a direction opposite to the motion of mass m1

Loss in kinetic energy

ΔKE=(12m1u12+12m2u22)(12(m1+m2)V2)ΔKE=12(m1m2m1+m2)(u1+u2)2


Understand this concept better from the solved example given below:

Example: Two particles of equal mass m have respective initial velocities ui^ and u(i^+j^2) They collide completely inelastically. The energy lost in the process is:

1) 38mu2
2) 23mu2
3) 13mu2
4) 18mu2

Solution:

Conserving Momentum-

mui^+m(u2i^+u2j^)=2m(u1i^+u2j^)

On solving

u1=3u4 and u1=u4

Change in K.E
[12mu2+12m(u22)2][12(2m)(9u216+u216)]=3mu245mu28=mu28

Hence, the answer is the option (4).

Collision Between Bullet and Vertically Suspended Block

A block of mass M is suspended by a vertical thread.

A bullet of mass m is fired horizontally with velocity u in the block.

After the collision bullet gets embedded in the block. And, the combined system is raised up to height h where the string makes an angle θ with the vertical.

The Common Velocity of the System Just After the Collision (V)

Here, the system is (block + bullet)

P= momentum
Pbullet +Pblock =Psystem mu+0=(m+M)VV=mum+M .....(1)

The Initial Velocity of the Bullet in Terms of h

By the conservation of mechanical energy

(T.E of system ) Just after collision =(T.E of system ) At height h
12(m+M)V2=(m+M)ghV=2gh

Equating (1) and (2)
We get V=2gh=mum+M
u=(m+Mm)2gh

Loss in Kinetic Energy

Loss of kinetic energy in a perfectly inelastic collision when given by

ΔKE=12(mMm+M)u2

Value of angle θ

From u=(m+Mm)2gh
We can write
h=(u22g)(mm+M)2

And from figure
Or,cosθ=LhL=1hL=1((u22gL)(mm+M)2)


Solved Example Based on Elastic And Inelastic Collision

Example 1: A bullet of mass m moving with a velocity v strikes a suspended wooden block of mass M as shown in the figure and sticks to it. If the block rises to a height h then the initial velocity of the bullet is:

1) mmM2gh
2) mm+M2gh
3) m+Mm2gh
4) mMm2gh

Solution:

The common velocity of the system just after the collision (V)

Here, the system is (block + bullet)

P= momentum Pbullet +Pblock =Psystem mu+0=(m+M)VV=mum+M .....(1)

And, the initial velocity of the bullet in terms of h. By the conservation of mechanical energy

(T.E of system ) Just after collision =(T.E of system) At height h
12(m+M)V2=(m+M)ghV=2gh

Equating (1) and (2)
We get V=2gh=mum+M
u=(m+Mm)2gh

So, the answer is -

m+Mm2gh

Example 2: A ball of mass ' m ' moving with velocity ' v ', collides inelastically with another identical ball. After the collision, the 1st ball moves with velocity v/2 in a direction perpendicular to the initial direction of motion. Find the speed of the second ball after the collision:

1) 32v
2) 12v
3) 32v
4) 2v

Solution:

Momentum conservation along X-axis

mv=mv1cosθv=v1cosθ(1)
along y-axis
0=mv/2mv1sinθv2=v1sinθ(2)v2+v22=v12(sin2θ+cos2θ)v12=3v22v1=32v

Hence, the answer is option (3).

Example 3: Two particles A and B of equal mass M are moving with the same speed v as shown in the figure. They collide completely inelastically and move as a single particle C. The angle θ that the path of C makes with the X-axis is given by :

1) tanΘ=3+212
2) tanΘ=3212
tanΘ=122(1+3)
4) tanΘ=131+2

Solution:

As we know,

Inelastic Collision -

The law of conservation of momentum holds good but kinetic energy is not conserved

i.e

12m1u12+12m2u2212m1v12+12m2v22 but m1u1+m2u2=m1v1+m2
where
m1,m2: masses 
u1,v1 : initional and final velocities of mass m1
u2,v2 : initional and final velocities of mass m2

So apply the Law of conservation of momentum

(i) Along x-direction

mvsin30mvsin45=2mvcosθvcosθ=12(1212)

(ii) Along y-direction

mvcos30+mvcos45=2mvsinθvsinθ=12(32+12)

Divide (i) with (ii):

tanθ=3+212

Hence, the answer is option (1).

Summary

Grasping the basics of elastic and inelastic collisions is key to understanding how objects interact and transfer energy. Real-world examples, like billiard games, often illustrate these principles. In inelastic collisions, objects can lose a lot of energy or stick together. Knowing these concepts is vital for explaining how things behave when they collide.

Frequently Asked Questions (FAQs)

1. What meaning does a collision hold in physics?

A collision in physics is the situation of two or even more than two bodies, for a short span of time, applying forces on each other, thus leading to a change in one's motion.

2. What is the main difference between elastic and inelastic collisions?

Both the momentum of striking bodies is preserved in an elastic collision and momentum in an inelastic collision, but in an elastic collision, kinetic energy is also preserved as a standard concept, while kinetic energy in an inelastic collision is not preserved, because some part is spent somewhere else.

3. What is an elastic collision, for example?

A classical example of an elastic collision is billiard balls in action, in which the balls, after hitting one another, rebound from each other without losing their kinetic energy.

4. What occurs in a perfectly inelastic collision?

The homes have their maximum kinetic energy loss because after a collision, the striking bodies bond together.

5. Why are most real-world collisions inelastic?

Most of the real-world collisions are inelastic because of the reason is that they dissipate energy into some other forms, e.g., heat, sound energy, and deformation. For instance, the case of car crashes, the balls of clay hitting the floor, and more.

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