Collisions | Physics Class 11 - NEET PDF Download

 Collisions

Collisions are fascinating events where a strong force acts between two or more bodies for a very short time, leading to changes in their energy and momentum.

 Whether it's a collision between billiard balls or an interaction on a subatomic level, collisions showcase the key principles of physics.

1. Stages of Collision: 

Collisions can be broken into three stages—before, during, and after the interaction.

• Before and after the collision, forces are absent. 
• During the collision, large forces dominate and govern the motion.

2. Conservation of Momentum and Energy: 

• In an ideal system, the total momentum and energy remain conserved throughout the collision. 
• This principle helps us relate initial and final velocities of the colliding objects, providing a deep insight into how motion evolves.

Types of Collisions

(i) On the basis of conservation of kinetic energy.

(ii) On the basis of the direction of colliding bodies:

1. Elastic collision

The collision in which both the momentum and the kinetic energy of the system remains conserved are called elastic collisions.

• In an elastic collision all the involved forces are conservative forces.
• Total energy remains conserved.

2. Inelastic collision

The collision in which only the momentum remains conserved but kinetic energy does not remain conserved are called inelastic collisions.

• In an inelastic collision some or all the involved forces are non-conservative forces.
• Total energy of the system remains conserved.
• If after the collision two bodies stick to each other, then the collision is said to be perfectly 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:
  v₁ = (m₁ – m₂) u₁ + 2m₂u₂ / (m₁ + m₂)
   v₂ = (m₂ – m₁) u₂ + 2m₁u₁ / (m₁ + m₂)
• 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:
  ΔE = m₁m₂ / 2(m₁ + m₂) (u₁ – u₂)² (1 – e²)

Example 1:

Sol: (c) 1:3

Explanation:

 Perfectly Inelastic One Dimensional Collision

In such types of collisions, the bodies move independently before the collision but after the collision, they move as one single body.

(i) When the colliding bodies are moving in the same direction:

Loss in kinetic energy:

(ii) When the colliding bodies are moving in the opposite direction:

Before Collision

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:

^{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.

According to the law of conservation of momentum:

This implies:

According to the law of conservation of kinetic energy:

This implies:

Dividing equation (iv) by equation (ii):

This implies:

Further, from equation (v) we get:

Substituting this value of $v2v\_2$v2 in equation (i) and rearranging, we get:

Similarly, we get:

Special cases of Head on Elastic Collision

(i) If projectile and target are of the same mass i.e. $m\_1\; =\; m\_2$m₁ = m₂

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

(iii) If a light projectile collides with a very heavy target, i.e. m₁ << m₂:

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) When they are in contact, some part of the kinetic energy may convert into potential energy, so it is not conserved during the short time of collision.

(ii) The law of conservation of energy is always true.

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 $A$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, $B$B retraces its path with velocity v, then the magnitude and direction of the velocity of $C$C will be:

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

Sol: 

Fig 1 Fig 2