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`        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`
2 years ago

Vikas TU
8454 Points
```							Change in momentum = > Mvf – Mvithat is :deltaP = M(vf – vi)vi = root(2gh) = > 9.899 m/svf = vi -gt => 9.899 – 9.8*0.1 = > 8.919 m/s Hence, deltaP = (50/1000)*(8.919 – 9.899) => -0.048975That is equal to F.dtHence,J =F.dt = deltaPF = 0.048975/0.1 => 0.48975 N.
```
2 years ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

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