## An Elastic Collision Shortcut

- Category: Shortcuts
- Written by fisikastudycenter

physics.fisikastudycenter.com - A physics shortcut to collision problem wih conditions the collision should be elastic and the masses of the two bodies involved are the same.

**Problem**

Given two bodies in an elastic collision above.

What are the velocities of the two bodies after collision?

**Answer**

The ordinary way used for this question

**Step 1 **

Get the problem data include the signs velocities, in here to the right s (+), and (−) when to the left:

m1 = m2 = 2 kg

e = 1 → elastic collision

v1 = 4 m/s (right)

v2 = − 6 m/s (left)

**Step 2**

Get the first equation using the conservation of momentum

m_{1}v_{1} + m_{2}v_{2} = m_{1}v_{1}' + m_{2}v_{2}'

2v_{1} + 2v_{2} = 2v_{1}' + 2v_{2}'

Canceling 2

v_{1} + v_{2} = v_{1}' + v_{2}'

Inserting the data

4 + (−6) = v_{1}' + v_{2}'

v_{2}' + v_{1}'= −2 ........ (Equation 1)

**Step 3**

Get the second equation from restitution coefficient e

The coefficient of restitution e for a head-on collision is defined as the ratio of the relative speed after the collision to the relative speed before.

Elastic collision e = 1

where

e = − ^{v1' − v2'} / _{v1 − v2}

1 = − ^{v1' − v2'} / _{v1 − v2}

v_{1} − v_{2} = v_{2}' − v_{1}'

4 −(−6) = v_{2}' − v_{1}'

v_{2}' − v_{1}' = `10.......(Equation 2)

**Step 4**

Combine equations 1 and 2

v_{2}' + v_{1}'= −2

v_{2}' − v_{1}' = `10

---------------------------------- +

2v_{2}' = 8

v_{2}' = 4 m/s

v_{2}' + v_{1}'= −2 .

4 + v_{1}'= −2

v_{1}'= −2 −4 = −6 m/s

**Caution:**

Again take a note the conditions for this shortcut:

- elastic collision, do not use in unelastic collision

- m_{1} and m_{2} are the same in mass, do not use for different masses.