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Impulse of a falling ball under gravity onto a solid surface

  • Thread starter GatorJ
  • Start date
  • #1
3
0

Homework Statement


A 0.5-kg rubber ball is dropped from rest a height of 19.6 m above the surface of the Earth. It strikes the sidewalk below
and rebounds up to a maximum height of 4.9 m. What is the magnitude of the impulse due to the collision with the sidewalk?

Homework Equations


J=[tex]\Delta[/tex]P


The Attempt at a Solution


mgh=.5mv2
v=[tex]\sqrt{}2gh[/tex]

Just before collision:
v=[tex]\sqrt{}2*9.8*19.6[/tex]
v=19.6 m/s

Just after collision:
v=[tex]\sqrt{}2*9.8*4.9[/tex]
v=9.8 m/s

J=m(vafter-vbefore)=.5(9.8-19.6)=4.9 kg m/s

The answer is 14.7 N.s, so I'm not sure what I'm missing.
 

Answers and Replies

  • #2
430
2


Well velocity is a vector - is this how you would subtract them?
 
  • #3
3
0


J=.5[9.8-(-19.6)]=14.7 kg m/s

Wow that was a stupid mistake, thanks!
 

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