Collisions - Physics Class 11 - NEET PDF Download

Introduction

Collisions are events in which two or more bodies exert large forces on each other for a very short time, causing abrupt changes in their velocities, kinetic energies and momenta. Collisions illustrate the application of conservation laws - especially the conservation of linear momentum - and provide clear examples of how energy may be redistributed between kinetic and other forms (for example, internal energy or potential energy during deformation).

Stages of a Collision

A collision may be conceptually divided into three stages:

• Before the collision: Bodies move under ordinary forces (if any) and their separations are larger than the interaction range.
• During the collision: Very large, short-duration interaction forces act between the bodies; these forces usually dominate other external forces during this short interval.
• After the collision: Bodies separate (or move together) and again other forces (if any) govern the subsequent motion.

Conservation Laws in Collisions

Two conservation principles are central to collision problems:

• Conservation of linear momentum: For an isolated system (no net external impulse), the vector sum of momenta of all particles remains constant during the collision. This law applies to both elastic and inelastic collisions.
• Conservation of mechanical energy (kinetic energy): Kinetic energy is conserved only in elastic collisions. In inelastic collisions some kinetic energy is transformed into other forms (internal energy, heat, deformation), though total energy is always conserved.

Types of Collisions

Collisions are classified in two common ways:

• On the basis of conservation of kinetic energy: elastic and inelastic.
• On the basis of geometry of motion: one-dimensional (head-on) and two-dimensional (oblique).

(i) Elastic collision

The collision in which both the total linear momentum and the total kinetic energy of the system remain conserved is called an elastic collision.

• During an elastic collision kinetic energy before = kinetic energy after (mechanical energy is conserved).
• The forces between colliding bodies can be considered effectively conservative during the short contact interval.

(ii) Inelastic collision

The collision in which only the total linear momentum is conserved but kinetic energy is not conserved (some is dissipated or converted into other forms) is called an inelastic collision.

• During an inelastic collision, some mechanical (kinetic) energy is converted into internal energy, heat, sound, permanent deformation, etc.
• If the colliding bodies stick together and move as one mass after impact, the collision is called perfectly (completely) inelastic.

One-Dimensional or Head-on Collisions

If the initial and final velocities of colliding bodies lie along the same line, then the collision is called one dimensional or head-on collision.

Inelastic One Dimensional Collision

Applying Newton’s experimental law, we have

• Velocities after collision:
  
• When masses of two colliding bodies are equal, then after the collision, the bodies exchange their velocities.
  v₁ = u₂ and v₂ = u₁
• If the second body of same mass (m₁ = m₂) is at rest, then after collision first body comes to rest and second body starts moving with the initial velocity of the first body.
  v₁ = 0 and v₂ = u₁
• If a light body of mass m₁ collides with a very heavy body of mass m₂ at rest, then after a collision.
  v₁ = – u₁ and v₂ = 0
  It means the light body will be rebound with its own velocity and heavy body will continue to be at rest.
• If a very heavy body of mass m₁ collides with a light body of mass m₂(m₁ > > m₂) at rest, then after collision
  v₁ = u₁ and v₂ = 2u₁
• Loss of kinetic energy:
  Equivalent form using reduced mass 

Example 1:

One sphere collides with another sphere of the same mass at rest inelastically. If the coefficient of restitution is $e=\frac{1}{2}$, the ratio of their speeds after the collision is:

(a) $1:2$
(b) $2:1$
(c) $1:3$
(d) 3:3 : 13:1

Sol: (c) 1 : 3

Explanation:

Let masses be equal: m₁ = m₂ = m, initial velocities: u₁ = u, u₂ = 0. After the collision, the speeds are v₁ and v₂.

Momentum: 

Restitution: 

Solving, we get v₂ = 1 + e2 u,    v₁ = 1 − e2 u.

Therefore,

Perfectly Inelastic One-Dimensional Collision

In a perfectly inelastic collision the bodies move independently before impact but after collision they stick together and move as a single combined mass.

(i) When the colliding bodies move in the same direction:

By conservation of momentum:

Loss in kinetic energy:

(ii) When the colliding bodies move in opposite directions:

By conservation of momentum:

When m₁u₁ > m₂u₂, then v_comb > 0 (positive), i.e., the combined body moves along the direction of motion of mass m₁.

When m₁u₁ < m₂u₂, then v_comb < 0, i.e., the combined body moves opposite to the motion of mass m₁.

Loss in kinetic energy:

Example 2: A body of mass 2 kg is placed on a horizontal frictionless surface. It is connected to one end of a spring whose force constant is 250 N/m. The other end of the spring is joined with the wall. A particle of mass 0.15 kg moving horizontally with speed v sticks to the body after collision. If it compresses the spring by 10 cm, the velocity of the particle is

(a) 3 m/s

(b) 5 m/s

(c) 10 m/s

(d) 15 m/s

Sol:

By the conservation of linear momentum: initial momentum of particle = final momentum of the combined mass.

By conservation of energy: compression of the spring stores the kinetic energy of the combined mass after collision.

Substituting and solving:

Putting numerical values:

Perfectly Elastic Head-on Collision

Let two bodies of masses m₁ and m₂ moving with initial velocities u₁ and u₂  in the same direction and they collide such that after collision their final velocities are v₁ and v₂ respectively.

Conservation of momentum:

This implies:

Conservation of kinetic energy:

This implies:

Dividing the energy equation by the momentum equation leads to the relation between relative velocities before and after collision (equivalent to the restitution relation for elastic collisions):

This implies:

Further, from the derived relations we obtain expressions for v1 and v2 in terms of u1, u2, m1 and m2:

Substituting and rearranging, we get:

Similarly:

Special Cases of Head-on Elastic Collision

(i) Equal masses (m1 = m2):

(ii) Very heavy projectile on a light target (m1 ≫ m2):

Substituting m₂ = 0, we get:

v₁ = u₁ and v₂ = 2u₁ − u₂

Sub case: u₂ = 0, i.e. the target is at rest.

Example: Collision of a truck with a cyclist.

(iii) Light projectile colliding with a very heavy target (m1 ≪ m2):

• For the velocities after the collision, we have:v₁ = −u₁ + 2u₂v₂ = u₂

Example: Collision of a ball with a massive wall.

• Before collision: The ball of mass m₁ = 50 gm is moving with velocity u₁ = 30 m/s, and the wall is stationary.

• After collision: The ball rebounds with velocity v₁ = −26 m/s, and the wall remains stationary with v₂ = 2 m/s.

Sub case:

• If u₂ = 0 (i.e., the target is at rest), then v₁ = −u₁ and v₂ = 0.

• This means the ball rebounds with the same speed in the opposite direction when the target is stationary.

Example 3:
Assertion (A): In an elastic collision of two billiard balls, the total kinetic energy is conserved during the short time of collision of the balls (i.e., when they are in contact).

Reason (R): Energy spent against friction does not follow the law of conservation of energy.

(a) Both A and R are true, and R is a correct explanation of A

(b) Both A and R are true, but R is not a correct explanation of A

(c) A is true, but R is false

(d) Both A and R are false

Sol: (d) Both A and R are false

Explanation:

(i) During the short contact time some kinetic energy may be temporarily stored as elastic potential energy of deformation; if rebound is perfectly elastic that stored energy returns to kinetic energy, but during contact the instantaneous kinetic energy of individual bodies may change. Thus the statement that kinetic energy is conserved at every instant during contact is not strictly correct.

(ii) The law of conservation of energy is always valid; energy "spent" against friction is converted into other forms (heat, sound), but total energy remains conserved.

Two-Dimensional or Oblique Collision

If the initial and final velocities of colliding bodies do not lie along the same line, then the collision is called two dimensional or oblique Collision.

In horizontal direction,

m₁u₁ cos α₁ + m₂u₂ cos α₂= m₁v₁ cos β₁ + m₂v₂ cos β₂

In vertical direction.

m₁u₁ sin α₁ – m₂u₂ sin α₂ = m₁u₁ sin β₁ – m₂u₂ sin β₂

If m₁ = m₂ and α₁ + α₂ = 90°

then β₁ + β₂ = 90°

If a particle A of mass m₁ moving along z-axis with a speed u makes an elastic collision with another stationary body B of mass m₂

From conservation law of momentum,

m₁u = m₁v₁ cos α + m₂v₂ cos β

sin β₂v₂ sin α – m₁v₁O = m

Example 4 : Three particles A, B, and C of equal mass are moving with the same velocity v along the medians of an equilateral triangle. These particles collide at the center G of the triangle. After the collision, A becomes stationary, B retraces its path with velocity v, then the magnitude and direction of the velocity of C will be:

(a) v and opposite to B
(b) v and in the direction of A
(c) v and in the direction of C
(d)  v and in the direction of B

Sol: 

From the figure (I), it is clear that before the collision, the initial momentum of the system is zero.

After the collision, A becomes stationary, B retraces its path with velocity v. Let C move with velocity V making an angle θ from the horizontal. As the initial momentum of the system is zero, therefore horizontal and vertical momentum after the collision should also be equal to zero.

From figure (II):

• Horizontal momentum: v cos⁡θ + v cos⁡30^∘ = 0 ……(i)

• Vertical momentum: v sin⁡θ − v sin⁡30^∘ = 0  ……(ii)

By solving (i) and (ii), we get θ = −30^∘ and V = v, i.e., C will move with velocity v in the direction of B.