I learnt that, there are two types of weight plates in strength training, cast iron plates and bumper plates.

The bumper plates are coated with rubber material so that if dropped comparatively lesser damage is caused. I know this has something to do with the rubber absorbing energy but can't understand it.

If we take one in each type with same mass and drop it from the same height (neglecting air resistance and assume same contact area during impact), what is the difference between the two impacts?

  • 1
    $\begingroup$ The rate of deceleration or the rebound energy... $\endgroup$
    – Solar Mike
    Jan 30 '19 at 13:43

There is no difference in the rebound energy unless the coefficient of restitution is different. In brief - a perfectly elastic collision, whether with cold steel or sponge rubber, will return all the energy to the object which collided.

Rubber bumpers or other devices designed to reduce damage or rebound do so by being inelastic - either the energy is used to deform (permanently) the material, or it's dispersed in the form of random energy (heat, sonic waves, etc), or in some designs redirected to safe locations. An example of the last is an airbag, which responds to a collision (with a person...) by forcing the compressed air out numerous exit holes in all directions.

  • $\begingroup$ Could you please suggest any resource to study further about this(book,website,etc..)? $\endgroup$
    – user17332
    Jan 30 '19 at 19:15
  • 1
    $\begingroup$ I think you mean coefficient of restitution $\endgroup$
    – Phil Sweet
    Jul 8 '19 at 2:58
  • $\begingroup$ @PhilSweet oops. I'll correct that. $\endgroup$ Jul 8 '19 at 13:04

Its question of reducing impact acceleration, or in some texts G force. If you have two elastic collision with same mass but different elasticity, one though one flexible.

Both of them rebound reducing their velocity to zero and reverse, but the softer one over a longer time and smaller acceleration.

$ \alpha = m\frac{ V_1 - V_2}{\delta t} $

Smaller acceleration means smaller force, $ F=m*\alpha$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy