Elastic Collision Formula: Definition & Solved Examples

Ans. Given that, 

• Mass of First Ball (m₁) = 5 Kg
• Mass of Second Ball (m₂) = 7 Kg
• Initial Velocity of First Ball (u₁) = 12 m/s
• Initial Velocity of Second Ball (u₂) = 0 m/s
• Final Velocity of First Ball (v₁) = 0 m/s
• Final Velocity of Second Ball (v₂) =?

According to the Elastic Collision Formula, 

1/2 m₁u₁² + 1/2 m₂u₂²  = 1/2 m₁v₁²  +1/2 m₂ v₂²

{1 × 5 × (12)2 }/2+ (1 x 7 × 0) /2 = (1 × 5× 0)/2 + (1 × 7)/2 × v₂²

360 = 3.5 v²

v² = 102.85

v = √102.85 = 10.141 m/s

Thus, the final velocity of the second ball is 10.141 m/s.

Ans. Given that, 

• Mass of First Block (m₁) = 10 Kg
• Initial Velocity of First Block (u₁) = 12 m/s
• Final Velocity of First Block (v₁) =?
• Mass of Second Block (m₂) = 8 Kg
• Initial Velocity of Second Block (u₂) = 4 m/s
• Final Velocity of Second Block (v₂) = 0 m/s

Elastic Collision Formula is given as: 

1/2 m₁u₁² + 1/2 m₂u₂²  = 1/2 m₁v₁²  +1/2 m₂ v₂²

(10 × 12) + (8 × 4 )= (10 × v₁) + (8 × 0)

120 + 32 = 10 v₁ + 0

152 = 10 v₁

v₁ = 15.2 m/s

Thus, the final velocity of the first block is 15.2 m/s.

Ans. Given parameters are, 

• Mass of Blue Ball (m₁) = 0.2 kg 
• Mass of Red Ball (m₂) = 0.25 kg
• Initial Velocity of Blue Ball (u₁) = 5 m/s
• Initial Velocity of Red Ball (u₂) = 0 m/s
• Final Velocity of Blue Ball (v₁) = 0 m/s
• Final Velocity of Red Ball (v₂) =?

According to the Elastic Collision Formula, 

1/2 m₁u₁² + 1/2 m₂u₂²  = 1/2 m₁v₁²  +1/2 m₂ v₂²

(0.2 kg) x (5 m/s) + (0.25 kg) x (0 m/s) = (0.2 kg) x (0) + (0.25 kg) x (v2)

1.0 kg.m/s + 0 = 0 + (0.25 kg) x (v₂)

1.0 kg.m/s = (0.25 kg) x (v₂)

(1.0 kg.m/s) / 0.25 kg = (v₂)

v₂ = 4 m/s 

Thus, the final velocity of the red ball is 4 m/s.

Ans. The characteristics of an elastic collision are: 

• The linear momentum of an object is conserved in an elastic collision.
• The total kinetic energy of the object is also conserved.
• Conservative forces are involved in an elastic collision.
• The mechanical energy is not converted into heat in an elastic collision.

Ans. Given that, 

• Mass of First Ball (m₁) = 5 Kg
• Mass of Second Ball (m₂) = 2 Kg
• Initial Velocity of First Ball (u1) = 6 m/s
• Initial Velocity of Second Ball (u₂) = 0 m/s
• Final Velocity of First Ball (v₁) =?
• Final Velocity of Second Ball (v₂) =?

Using the Elastic Collision Formula, 

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Substituting the values, 

(5 × 6) + (2 × 0) = 5v₁ + 2v₂

30 + 0 = 5v₁ + 2v₂

30 = 5v₁ + 2v₂ ⇢ (Equ. 1)

u₁ + v₁ = u₂ + v₂

6 + v₁ = 0 + v₂

6 = -v₁ + v₂ ⇢ (Equ. 2)

Using equations 1 and 2, 

∴ v₂ = 8.57 m/s

Put v₂ = 8.57 in Equation 2,

-v₁ + 8.57 = 6

v₁ = 2.57 m/s

Thus, the final velocities of the first and the second ball are 2.57 m/s and 8.57 m/s respectively. 

Ans. Given parameters are: 

• m₁ = 4 kg
• v₁ = u1/3
• u₂ = 0
• m₂ =?

Rearranging the elastic collision formula, 

v₁ = ((m₁ – m_​2) / (m₁ + m_​2)) × u₁ + (2m_​2u_​2) / (m₁ + m_​2)

u₁/3 = ((4 – m₂) / (4 + m₂)) × u₁

4 + m₂ = 12 – 3m₂

4m₂ = 12 – 4

m₂ = 2 kg

Thus, the mass of the second body is 2 kg. 

Ans. In order to make sure whether a collision is elastic or inelastic, we need to equate their total kinetic energy. This means, that if the kinetic energy remains the same before and after the said collision, it can be called an elastic collision. On the other hand, if there is a shift in the total kinetic energy, it means that the category of collision is inelastic. The loss of energy in an inelastic collision usually dissipates into sound and heat energy. 

Ans. Here are a few examples of elastic collision: 

• On dropping a ball on the floor, it bounces back to us immediately. In this case, the ball in motion preserves its overall momentum and kinetic energy, which is why it bounces back. 
• When two atomic particles collide with each other, they experience an elastic collision. In case of no loss of energy after contact, it can be classified as a perfectly elastic collision. 

Ans. Given information is that, 

• m₁ = 3 kg
• m₂ = 5 kg
• u₂ = 0
• v₁ = 2.2 m/s
• v₂ = 2 m/s
• u₁ =?

Using the elastic collision formula, 

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

3 × u1 + 5 × 0 = 3 × 2.2 + 5 × 2

3u₁ = 6.6 + 10

u₁ = 5.53 m/s

Thus, the initial velocity of the first body is 5.53 m/s.

Ans. Elastic Collision refers to a type of collision between two bodies in which the total kinetic energy of both bodies before and after the collision remains the same. No kinetic energy is lost in this particular type of collision neither it is converted into any other form of energy.