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A 0.25 kg ball rolling at a 1m/s and overtakes a 0.3kg ball rolling in the same direction at 0.5m/s. The ball sticks together on impact. What is the velocity of the two balls after collision?​

Sagot :

The velocity of the two balls after the collision is 0.73 m/s.

The velocity of the two balls after the collision can be calculated using the formula below.

Formula:

  • mu+m'u' = V(m+m')............... Equation 1

Where:

  • m = mass of the first ball
  • m' = mass of the second ball
  • u = initial velocity of the first ball
  • u' = initial velocity of the second ball
  • V = velocity of the two balls after the collision.

make V the subject of the equation

  • V = (mu+m'u')/(m+m')................ Equation 2

From the question,

Given:

  • m = 0.25 kg
  • m' = 0.3 kg
  • u = 1 m/s
  • u' = 0.5 m/s

Substitute these values into equation 2

  • V = [(0.25×1)+(0.3×0.5)](0.25+0.3)
  • V = 0.4/0.55
  • V = 0.73 m/s.

Hence, the velocity of the two balls after the collision is 0.73 m/s

Learn more about collision here: https://brainly.com/question/7694106