An elastic collision is a collision in which the total kinetic energy and momentum of the system remain conserved. There is no net loss of kinetic energy during the collision.
Various examples of the elastic collision are
- The collision between two steel balls on a smooth surface can be approximately elastic when very little kinetic energy is lost.
- Collisions between atoms can be elastic when their total kinetic energy is conserved. However, some atomic collisions are inelastic if energy is transferred to internal excitation or other processes.
- The collision between two pool or snooker balls is an example of an approximately elastic collision because most of their kinetic energy is retained, although a small amount may be lost as sound, heat, or deformation.
Formula
The momentum formula for elastic collision is
m1u1 + m2u2 = m1v1 + m2v2
where,
- m1 is the Mass of 1st body
- m2 is the Mass of 2nd body
- u1 is the Initial Velocity of 1st body
- u2 is the Initial Velocity of 2nd body
- v1 is the Final Velocity of 1st body
- v2 is the Final Velocity of 2nd body
The kinetic energy formula for elastic collisions is
1/2(m1u12) + 1/2(m2u22) = 1/2(m1v12) + 1/2(m2v22)
Derivation
Momentum formula for Elastic Collision is:
m1u1 + m2u2 = m1v1 + m2v2
∴ m(u1-v1) = m(v2-u2)  ⇢ (Equation A)
Kinetic energy formula for elastic collisions is:
1/2(m1u12) + 1/2(m2u22) = 1/2(m1v12) + 1/2(m2v22)
∴ m1(u12-v12) = m2(v22-u22)
∴ m1(u1+v1)(u1-v1) = m2(v2+u2)(v2-u2)  ⇢ (Equation B)
Divide Equation B to Equation A,
u1 + v1 = v2 + u2
∴ u1 - u2 = -(v1 - v2)
Relative velocity of approach = relative velocity of receding
Momentum formula for Elastic Collision is:
m1u1 + m2u2 = m1v1 + m2v2
We have,
u1 + v1 = v2 + u2
∴ v2 = u1 + v1 - u2
Put v2 in the momentum formula for Elastic Collision,
∴ m1u1 + m2u2 = m1v1 + m2(u1 + v1 - u2)
∴ m1u1 + m2u2 = m1v1 + m2u1 + m2v1 - m2u2
∴ m1u1 - m2u1 +2m2u2 = v1(m1 + m2)
v1 = ((m1 - m2)u1 + 2m2u2) / (m1 + m2)
v2 = ((m2 - m1)u2 + 2m1u1) / (m1 + m2)
Applications
The following are some of the applications for elastic collisions:
- The amount of force experienced by the body during the collision is affected by the amount of time involved in the collision. In other words, the smaller the force acting on the body, the longer the impact takes. As a result, in order to optimize the force, the collision duration must be reduced.
- However, in order to reduce the force, the contact duration must be increased. There are numerous examples of this. One of them is that airbags in cars reduce the amount of time spent in a collision and reduce the force applied to the item. The phenomenon behind this procedure is that the airbags reduce the force while increasing the object's crash time.
- Although an elastic collision can occur in a variety of locations or settings, inelastic collisions are more common.
Examples
Example 1: A 5 kg ball moving east at a speed of 6 m/s strikes a 2 kg ball at rest. Calculate the velocities of the two balls assuming a perfectly elastic collision.
Solution:
Given,
- m1 = 5 kg,
- u1 = 6 m/s,
- m2 = 2 kg,
- u2 = 0
To Find, v1, v2
We have,
m1u1 + m2u2 = m1v1 + m2v2
∴ (5 × 6) + (2 × 0) = 5v1 + 2v2
∴ 30 + 0 = 5v1 + 2v2
∴ 30 = 5v1 + 2v2  ⇢ (Equation 1)
u1 + v1 = u2 + v2
∴ 6 + v1 = 0 + v2
∴ 6 = -v1 + v2  ⇢ (Equation 2)
From Equation 1 and 2,
∴ v2 = 8.57 m/s
Put v2 = 8.57 in Equation 2,
-v1 + 8.57 = 6
∴ v1 = 2.57 m/s
Example 2: When a body of mass 4 kg collides with another body at rest, it continues to move in the same direction at a speed of one-third of its original speed. Determine the second body's mass.
Solution:
Given,
- m1 = 4 kg
- v1 = u1/3
- u2 = 0
To Find, m2
We have,
v1 = ((m1 - m2) / (m1 + m2)) × u1 + (2m2u2) / (m1 + m2)
∴ u1/3 = ((4 - m2) / (4 + m2)) × u1
∴ 4 + m2 = 12 - 3m2
∴ 4m2 = 12 - 4
∴ m2 = 2 kg
Example 3: A 1 kg ball travelling at 15 m/s collides head-on with another ball of unknown mass at rest, rebounding with a velocity of 10 m/s. What is the other ball's mass?
Solution:
Given,
- m1 = 1 kg
- u1 = 15 m/s
- v1 = -10 m/s (rebounds)
- u2 = 0
to Find, m2
We have,
v1 = ((m1 - m2) / (m1 + m2)) × u1 + (2m2u2) / (m1 + m2)
∴ -10 = ((1 - m2) / (1 + m2)) × 15
∴ -2 + (-2m2) = 3 - 3m2
∴ m2 = 5 kg
Example 4: Assume that m1 is 3 kg, m2 is 5 kg, u2 is at rest, v1 is 2.2 m/s, and v2 is 2 m/s in a specific example. What is the value of u1?
Solution:
Given,
- m1 = 3 kg
- m2 = 5 kg
- u2 = 0
- v1 = 2.2 m/s
- v2 = 2 m/s
To Find, u1
We have,
m1u1 + m2u2 = m1v1 + m2v2
∴ 3 × u1 + 5 × 0 = 3 × 2.2 + 5 × 2
∴ 3u1 = 6.6 + 10
∴ u1 = 5.53 m/s
Example 5: If u1 is 4 m/s, u2 is 2 m/s, and v2 is 6 m/s, the following equation can be used. What is the significance of v1?
Solution:
Given,
- u1 = 4 m/s
- u2 = 2 m/s
- v2 = 6 m/s
To Find, v1
u1 + v1 = v2 + u2
∴ 4 + v1 = 6 + 2
∴ v1 = 8 - 4
∴ v1 = 4 m/s
Practice Questions
Q1. If u1 is 12 m/s, u2 is 5 m/s, and v2 is 8 m/s, then find the v1 of the collision if the mass of the object is m.
Q2. If u1 is 8 m/s, u2 is 10 m/s, and v2 is 6 m/s, the following equation can be used. What is the significance of v1?
Q3. A box of mass 12 kg is sliding at 15 m/s and collides head-on with another box of unknown mass at rest, rebounding with a velocity of 10 m/s. What is the mass of the other box?
Q4. When a body of mass 12 kg collides with another body at rest, it continues to move in the same direction at a speed of one-fourth of its original speed. Determine the mass of the second body.
