Question

26.
A ball of mass m falls vertically to the ground from a heighth and rebound to
a height h2. The change in momentum of the ball on striking the ground is
​

Answer 1

answer is

∆P = m ( √2gH2 - √2gH1 )

Explanation ,

Given ,

• A ball is falling vertically down from a height say H1 and rebound to a Height H2.
• So the heights given are , H1 and H2.

To Find change in momentum of the ball on striking ground ,

We will use the formula ,

• Momentum = Mass x velocity

As the mass is constant and the change will take place only with velocities.

• CHANGE IN MOMENTUM = MASS x CHANGE IN VELOCITY

So ,

• ∆P = m ∆V

For velocities ,

First velocity ,

• if 'V1' be the first velocity height = H1 before striking the ground then ,
• V1 = √2gH1

Second velocity ,

• Velocity after rebounding with respect to height "H2" will be ,
• V2 = √2gH2

So , the change in momentum is ,

• ∆P = m ∆V
• ∆P = m ( √2gH2 - √2gH1 )