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
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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 )
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