Elastic Collision Formula

In physics, each collision between two objects is either elastic or inelastic. The elastic collision happens when there is no energy loss in the system of the two colliding items, whereas the inelastic collision happens when there is energy loss in the system of the two colliding items. 

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Elastic Collision

When two bodies hit, the overall kinetic energy is not lost, which is known as an elastic collision. In other terms, it is a collision between two bodies in which their total kinetic energy is equal. Furthermore, it might be one-dimensional or two-dimensional. As a result, in the actual world, totally elastic collisions are unlikely to occur since some energy exchange, however minor, is necessary. An elastic collision is a collision between two bodies in which their total kinetic energy stays constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

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When tiny objects collide, kinetic energy is first changed to potential energy associated with a repulsive or attractive force between the particles (when the particles move against the force, hence the angle between the force and the velocity is obtuse), and then returned to kinetic energy.

Elastic Collision Formula

The perfectly elastic collision formula of momentum is followed as,

m\[_{1}\]u\[_{1}\] + m\[_{2}\]u\[_{2}\] = m\[_{1}\]v\[_{1}\] + m\[_{2}\]v\[_{2}\]

Likewise, the conservation of the total kinetic energy is represented as, 

\[\frac{1}{2}\]m\[_{1}\]u\[_{1}^{2}\] + \[\frac{1}{2}\]m\[_{2}\]u\[_{2}^{2}\] = \[\frac{1}{2}\]m\[_{1}\]v\[_{1}^{2}\] + \[\frac{1}{2}\]m\[_{2}\]v\[_{2}^{2}\]

These equations can be solved directly to find v\[_{1}\], v\[_{2}\] when u\[_{1}\], u\[_{2}\] are known,

v\[_{1}\] = \[\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\]u\[_{1}\] + \[\frac{2m_{2}}{m_{1} + m_{2}}\]u\[_{2}\] 

v\[_{2}\] = \[\frac{2m_{1}}{m_{1} - m_{2}}\]u\[_{1}\] + \[\frac{m_{2} - m_{1}}{m_{1} + m_{2}}\]u\[_{2}\]

If both masses are the same, we get a trivial solution:

v\[_{1}\] = u\[_{2}\]  

v\[_{2}\] = u\[_{1}\]

Where,

• m\[_{1}\] is the mass of 1st body

• m\[_{2}\] is the mass of 2nd body

• u\[_{1}\] is the initial velocity of 1st body

• u\[_{2}\] is the initial velocity of the 2nd body

• v\[_{1}\] is the final velocity of the 1st body

• v\[_{2}\] is the final velocity of the 2nd body

Solved Examples

Ex.1. If the ball has a mass is 8 Kg and moves with a velocity of 16 m/s collides with a stationary ball of mass is 10 kg and comes to rest. Calculate the final velocity of the ball of mass 7 Kg ball after the elastic collision.

Solution:

Given parameters are

The mass (m\[_{1}\]) of 1st ball is 8 kg

The initial velocity (u\[_{1}\]) of the first ball  is 16 m/s

The mass (m\[_{2}\]) of the second ball is 10 kg

The initial velocity (u\[_{2}\]) of the second ball is 0

The final velocity (v\[_{1}\]) of the first ball is 0

Find the final velocity (v\[_{2}\]) of the second ball?

The elastic collision formula is given as,

\[\frac{1}{2}\]m\[_{1}\]u\[_{1}^{2}\] + \[\frac{1}{2}\]m\[_{2}\]u\[_{2}^{2}\] = \[\frac{1}{2}\]m\[_{1}\]v\[_{1}^{2}\] + \[\frac{1}{2}\]m\[_{2}\]v\[_{2}^{2}\]

\[\frac{1}{2}\] x 8 x 16\[^{2}\] + \[\frac{1}{2}\] x 10 x 0 = \[\frac{1}{2}\] x 8 x 0 + \[\frac{1}{2}\] x 10 x v\[_{2}^{2}\]  

1024 = 5 x v\[_{2}^{2}\] 

v\[_{2}^{2}\] = 204.8 

v\[_{2}\] = 14.31 m/s

Hence the velocity after elastic collision for second ball is 14.31 m/s

Ex.2. A 15 Kg block is moving with an initial velocity of 16 m/s with 10 Kg wooden block moving towards the first block with a velocity of 6 m/s. The 2nd body comes to rest after the collision. Determine the final velocity of the first body

Solution:

Given parameters are

The mass (m\[_{1}\]) of 1st ball is 15 kg

The initial velocity (u\[_{1}\]) of the first ball is 16 m/s

Mass (m\[_{2}\]) of the second ball is 10 kg

The initial velocity (u\[_{1}\]) of the second ball is 6 m/s

The final velocity (v\[_{1}\]) of the first ball is 0

Find the final velocity (v\[_{2}\]) of the first ball is?

The elastic collision formula is given as

m\[_{1}\]u\[_{1}\] + m\[_{2}\]u\[_{2}\] = m\[_{1}\]v\[_{1}\] + m\[_{2}\]v\[_{2}\]

15 x 16 + 10 x 6 = 15 x 0 + 10 x v\[_{2}\]

240 + 60 = 10v\[_{2}\]

v\[_{2}\] = \[\frac{300}{10}\]

v\[_{2}\] = 30 m/s

Hence the velocity after elastic collision for second ball is 30 m/s