Question

a ball of mass 4 kg moving on smooth horizontal surface makes an elastic collision with another ball of mass m at rest in the line of motion of first ball. if after collision first ball moves in the same direction with one fourth of its velocity before collision, then mass of second ball is​

Answer 1

• Mass of the second ball is 2.4 kg.

Given,

Mass of ball A= 4 kg

Mass of second ball B = m kg

Elastic collision takes place between them so by applying momentum conservation is this we get-

4V = 4 × V/4 + m V₀

⇒  3V = mV₀

⇒ V₀= 3V/m

Now we have to apply energy conservation which states that initial energy should be equal to the final energy so that there is no net gain or loss of energy.

1/2 × 4× V² = 1/2× 4× (V/4)² + 1/2 m V₀²

⇒ 2 V² = V²/8 + 1/2 m 9V²/m²

⇒ m = 12/5 = 2.4 kg .

Hence mass of second ball is 2.4 kg.

Answer 2

Answer:

The mass of the second ball is 2.4 kg.

Explanation:

Given;-

     Mass of first ball , m₁ = 4 kg

     Mass of second ball = m₂

    Initial velocity of first ball = u₁  (let )

    Initial velocity of second ball = u₂ = 0

    Final velocity of first ball = v₁= u/4

    Final velocity of second ball = v₂ ( let )

Now, we know by formula for elastic collision that the velocity is given by;

      v₁ = (m₁ - m₂) u₁/ ( m₁+m₂) + 2m₂ u₂/ (m₁+m₂)

Putting the given values in the question, we get;-

      u/4 = (4 - m₂)u / (4 + m₂) + 0

          1/ 4 = 4 - m₂/ 4 + m₂

     4 + m₂ = 16 - 4m₂

 4m₂ + m₂ = 16 - 4

         5m₂ = 12

           m₂ = 12/5

       m₂ = 2.4 kg.

Hope it helps! ;-))