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### Course: Class 11 Physics (India)>Unit 10

Lesson 9: Elastic and inelastic collisions

# Deriving the shortcut to solve elastic collision problems

In this video, David derives the expression that we can use as a shortcut to solve for finding the velocities in an elastic collision problem. Created by David SantoPietro.

## Want to join the conversation?

• Does this equation (Vti + Vtf) = (Vgi + Vgf) also work for elastic collisions in two dimensions? Or do you need to break up the velocities into the x and y components and then use the equation on each component?
• Can this method be used on the AP Physics 1 test for full credit, or is it necessary to show all the steps with conservation of Kinetic Energy and Conservation of Momentum?
• Why not take this derivation a step further and solve for one of the final velocities? Doing this, you would get v1f=(v1i(m1-m2)+2m2v2i)/(m1+m2). Using this formula, you could solve for the final velocities of the two objects a lot quicker than using a system of equations.
• Just out of curiosity, could you say that the velocity is conserved, or that won't be the case in this scenario.
• Can't he have just divided both sides by mass on ? It would've been quicker to derive it.
(1 vote)
• I still dont understand how u could solve it because dont u still have 2 unknowns?
(1 vote)
• You still need to use simultaneous eauation to solve it, but it would be WAY easier now since you have eliminated all the unnecessary calculations, and it wouldbe faster to solve it
• can I actually apply this symbolic method to all other equations?
(1 vote)
• if this is the case of finding the velocity in an elastic collision problem? then what is the use of using Vaf=(ma-mb)/(ma+mb)*Vai+(2mb)/(ma+mb)*Vbi ? is the equation meant for perfect elastic problems ?
(1 vote)
• i used those formula for this problem too
it's a lot faster and somehow more intuitive for me

but i guess David's intention for that might be in deriving the equation of v_i1+v_f1 = v_i2+v_f2 and giving us the nature of elastic collisions that the sum velocity (or net velocity if you will) of an object must be same as the other object in a collision

and this intuition couldn't be derived (at least directly or simply) from the equation we two used only to get the final velocities for each object

in sum,
1. your way might be better to get a faster answer
2. David's way better to grasp another intuition about an elastic collision
3. perfectness must be assumed in both cases, i believe. otherwise, kinetic energy must be lost somewhere. then all of the equations here and in video might not work
(1 vote)
• Would this equation (Vti + Vtf = Vgi + Vgf) be applicable for real world examples? I learned that kinetic energy is not conserved in a system in elastic collisions. Some of the kinetic energy is lost to heat, light, sound, etc.
• I tried solving this problem by eliminating Vgf:

0,07 = 0,058*Vtf + 0,045*Vgf => ( multiply by 0,023)
102,65 = 0,029*Vbf + 0,023*Vgf => ( multiply by 0,045)

Subtracting the equations from each other and solving for Vbf gave me a velocity of:
159310,34 m/s
something went wrong.....
Can anyone point out my mistake ?