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Grade: 9
        A rubber ball of mass 50g falls from a height of 5m and rebounces to a height of 1.25m.the average force between ball and ground for which they are in contact was 0.1s
one year ago

Answers : (1)

Vikas TU
6874 Points
							
Change in momentum = > Mvf – Mvi
that is :
deltaP = M(vf – vi)
vi = root(2gh) = > 9.899 m/s
vf = vi -gt => 9.899 – 9.8*0.1 = > 8.919 m/s
 
Hence, deltaP = (50/1000)*(8.919 – 9.899) => -0.048975
That is equal to F.dt
Hence,
J =F.dt = deltaP
F = 0.048975/0.1 => 0.48975 N.
one year ago
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