Loading....
Coupon Accepted Successfully!

 

Kinetic Energy Transfer During Head- on Elastic Collision

  • Kinetic energy of projectile before collision,
     
    40788.png
  • Kinetic energy of projectile after collision
     
    40782.png
  • Kinetic energy transferred from projectile to target, ΔK = decrease in kinetic energy in projectile
     
    ∴ 40775.png
     
    Fractional decrease in kinetic energy
     
    40769.png
     
    We can substitute the value of v1 from the equation
     
    40763.png
     
    If the target is at rest, i.e., u2 = 0, then
     
    40757.png
     
    From (9),
     
    40751.png
Note:
  • Greater the difference in masses,lesser will be the transfer of kinetic energy and vice versa.
  • The transfer of kinetic energy will be maximum when the difference in masses is minimum.
     
    i.e., m1 – m2 = 0 or m1 = m2,
     
    then 41237.png
     
    So the transfer of kinetic energy in head on elastic collision (when target is at rest) is maximum when the masses of particles are equal i.e., mass ratio is 1 and the transfer of kinetic energy is 100%.
  • If m2 = n m1 then from equation (iii) we get 41227.png
  • Kinetic energy retained by the projectile 41221.png kinetic energy transferred by projectile
     
    ⇒ 41215.png
     
    41209.png

Velocity, momentum, and kinetic energy of stationary target after head-on elastic collision

Velocity of target: We know
40676.png
⇒ 40670.png
42847.png40664.png 
∴ 40658.png
 
Momentum of target: P2 = m2v2 = 40652.png
40646.png
∴ 40640.png
 
Kinetic energy of target:
40634.png
 
40627.png
40621.png 




Test Your Skills Now!
Take a Quiz now
Reviewer Name