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### Course: Mechanics (Essentials) - Class 11th>Unit 8

Lesson 4: How to measure the speed of a bullet using a block of wood and a string?

# Elastic and inelastic collisions

David explains what it means for a collision to be elastic or inelastic. Created by David SantoPietro.

## Want to join the conversation?

• If momentum is conserved ?

how can we describe this :
you throw something towards the wall, it bounces back with less speed or stuck at the wall ( arrows game for example ) .. HOW COME the initial total momentum becomes greater than the final total momentum ( sometimes it's value is ZERO ) !!

please explain ,, thanks a lot !
• In the situation you describe, the momentum of the individual object thrown at the wall is not conserved, but the total momentum of object, wall/Earth is conserved. Essentially the momentum gets transferred into the Earth, but since the Earth is so massive, it doesn't change its motion much.
• If in inelastic collision, the lost kinetic energy after collision is converted into other forms of energy such as sound and thermal energy, does it mean in elastic collision no sound and thermal energy is generated by the collision? I wonder if it is possible to have a soundless and heat-less (which reminds me of absolute zero where no atoms move microscopically) collision. It seems no collision is truly elastic in this sense.
• True there are no really elastic collisions in nature. It is an ideal case.
• I still don't get how even in inelastic collision, the law of conservation of momentum still holds. Wouldn't the conservation of momentum be invalid since the initial velocities of objects would be greater than the final velocities of objects, thus making the whole equation invalid?
• Actually, the total momentum of the objects that collide is conserved, not of an individual one. If two objects of mass 1kg are moving towards each other at 10m/s, the total momentum they have is 0, since their momentums cancel out. Let's say that after the collision, they bounce off and travel at 5m/s in opposite directions. The individual momentum of the objects is not conserved. But, the total momentum is still 0.
Hope that helps... after 8 years.
• Isn't every action inelastic, because some, even negligible, amounts of energy, become potential energy?
• If you write thermal energy instead of potential energy, then yes. But again, if you can neglect this amount, it means that calling it elastic is good enough
• Can a collision involving a robber ball be elastic? If yes, does it mean a material can be either hard like iron is or soft like robber is to cause elastic collision?
• Steel is much more elastic than rubber, but neither of them is perfectly elastic.
• I couldn't understand how the kinetic energy gets converted into thermal energy. Can anyone pls explain it ? And what is the thermal energy ?
• Thermal energy is heat. If the collision is inelastic, the objects are going to deform a little bit when they collide. Deforming an object involves energy absorption by that object. KE gets absorbed by the object and the object becomes a bit warmer. Also some KE will get converted to sound, and the sound will dissipate in the air, making the air a little warmer.
• How can we define the elasticity of an object in physics?
• Kartik Nagpure,

Good question! One method of determining an object's elasticity is to determine its "coefficient of restitution". In a physics class, this is commonly done by dropping random objects like a plastic ball and comparing the height it was dropped to the height of each consecutive bounce. By determining how much the max height was lowered after each bounce, one can calculate the amount of energy that was lost in the collisions.
• Doesn't an elastic collision imply that the two objects never actually touch?
• Elastic means that no energy is converted into heat during the collision so kinetic energy before and after the collision remains constant. In both elastic and inelastic collisions, momentum is always conserved.
• Hi,

In the example in the video the momentum is not conserved :

Vib*mb + vis*ms != vfb*mb +vfs*ms
10*0.65 + 0*0.45 != 1*0.65 + 8*0.45

Is it a mistake(e.g I shouldn't focus on this because it's random values) or am I missing something here?
(1 vote)
• Momentum is conserved. Look more carefully at the collision data.
10*0.65 - 8*.45 = 1*.65 + 5*.45
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